MEI C2 Study Resources Core2 Trigonometry 1-5

MEI Core 2 Trigonometry Section 1: Trigonometric Functions and Identities Study Plan Background At G.C.S.E. you encountered the basic trigonometric ratios of sine, cosine, and tangent as well as key words such as angle of depression, angle of elevation. You learned the definitions of sin , cos  and tan  in terms of the lengths of the sides of a right-angled triangle. By the nature of this definition, it only applies to angles between 0° and 90°, since you cannot have a rightangled triangle which also contains an angle greater than 90°. Also, this definition does not make sense if you try to apply it to negative angles. In this section you will learn to extend the definitions of sin , cos  and tan  so that they can be applied to any angle. This allows you to solve many more problems involving trigonometry, some of which will be introduced in later sections.

Detailed work plan 1. Read the introductory section on pages 271 and 272, and the section “Trigonometrical functions” on pages 271-275. This revises GCSE work, and also looks at the values of sin , cos  and tan  for certain common angles. You should aim to memorise these values. Write them out in a table and include 0 and 90 then leave some space for the table to be extended to include angles greater than 90. A summary is given in the Notes and Examples, together with a worked example. 2. You can test yourself using the Flash resource Standard values of trig ratios and the interactive questions Exact trig function values of standard angles. 3. Read the section “Trigonometrical functions for angles of any size” on pages 275-276. Activity 10.1 is very important, and an explanation is given in the Notes and Examples, together with a useful diagram which is worth memorising. 4. The Mathcentre video Trigonometric ratios in all quadrants may also be useful. 5. Read the section “Identities involving sin , cos  and tan  on page 276. There is a worked example in the Notes and Examples. 6. These identities are demonstrated in the Flash resources Identities: sin x / cos x = tan x and Identities: sin²x + cos²x = 1.

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MEI C2 Trigonometry Section 1 Study plan 7. Read pages 277-279, on the sine, cosine and tangent graphs. You should be able to remember what these graphs look like, as you will often need to sketch them, particularly in Section 2 when you will be looking at solving trigonometrical equations. Look at the graphs of the trig functions and make a note of the ranges where each of them is positive and where they are negative. This will help later when you are solving problems. Figure 10.16 will also help. 8. The Flash resources The graph of y = sin x, The graph of y = cos x and The graph of y = tan x show dynamically how the graphs relate to the unit circle. The graphs and some notes about them are given in the Notes and Examples. You may also find the Mathcentre video Trigonometric functions useful. 9. Now you can extend your table of trig values for 0 to 90 to include the equivalent angles in each of the four quadrants. Check this against the sample version on the website shown at the beginning of the worked solutions for exercise 10A. There is an example using these values in the Notes and Examples. Leave pages 280-283 for now. This work is covered in Section 2. 10. You should now be ready for parts of Exercise 10A. Try at least questions 2, 3, 5, 6*, 8, 11*. (Questions 4 and 10 are covered in Section 2).

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MEI Core 2 Trigonometry Section 1: Trigonometric Functions and Identities Notes and Examples In this section you learn about the trigonometric functions and some trigonometric identities. These notes contain sub-sections on:  The trigonometric functions for angles between 0° and 90°  Common values of sin θ, cos θ and tan θ  Trigonometric functions for angles of any size  Trigonometric identities  Graphs of trigonometric functions

The trigonometric functions for angles between 0° and 90° From GCSE, you know that for a right-angled triangle the trigonometric functions are defined as:

opposite hypotenuse adjacent cos θ  hypotenuse opposite tan θ  adjacent

sin θ  hypotenuse opposite

θ adjacent You can also write: sin(90  θ )  So:

adjacent opposite and cos(90  θ)  hypotenuse hypotenuse

cos θ  sin(90  θ ) and sin θ  cos(90  θ)

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For example: cos 70° = sin 20° sin 15° = cos 75°

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MEI C2 Trigonometry Section 1 Notes and Examples Common values of sin θ, cos θ and tan θ You should learn the values of sin θ, cos θ and tan θ for θ = 0°, 30°, 45°, 60° and 90°. θ

0°

cos θ

1

sin θ

0

tan θ

0

30°

45°

3 2 1 2

1 2  2 2

60° 1 2

1 2  2 2

3 2

1

1

3

undefined

1 3  3 3

90° 0

You can see how these values are found on page 272 of the textbook. If you find it hard to remember the values for 30°, 45° and 60° at first, you can sketch the relevant triangles …. B

B

30° 2

2

A

45°

60° 1

D

1

C

A

2 45° 45° 1

C

… and then you can simply read off the value you need.

You can test yourself on these values using the Flash resource Standard values of trig ratios. Try to remember the exact value of each trig ratio, and click on the box to check your answer. In the example below you need to substitute exact values into an expression.

Example 1 Show that sin 2 30  cos2 30  1 Solution

sin 30 

1 3 and cos 30  2 2

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MEI C2 Trigonometry Section 1 Notes and Examples So substituting these values into sin 2 30  cos2 30 : 2

2 1  3 1 3    1     4 4  2   2 

as required.

Trigonometric functions for angles of any size The textbook shows how the definitions for sine, cosine and tangent can be extended to angles of any size using a diagram like the one below. y

This gives the definitions:

P (x, y)

y sin θ   y 1 x cos θ   x 1 y tan θ  x

1

y

θ O

x

x

In Activity 10.1, you are asked to investigate the sign of sin , cos  and tan  from the definitions above. Try this before looking at the explanation below.    

In the first quadrant,  lies between 0° and 90°. The values of x and y are both positive, so the values of sin , cos  and tan  are all positive. In the second quadrant,  lies between 90° and 180°. The value of x is negative and the value of y is positive, so sin  is positive but cos  and tan  are both negative. In the third quadrant,  lies between 180° and 270° (or alternatively, between -90° and -180°). The values of x and y are both negative, so tan  is positive but sin  and cos  are both negative. In the fourth quadrant,  lies between 270° and 360° (or alternatively, between 0° and -90°). The values of x is positive and the value of y is negtive, so cos  is positive but sin  and tan  are both negative.

This can be summarised in the diagram below.

Sine positive

All positive or, more simply

Tangent positive

Cosine

S

A

T

C

positive

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MEI C2 Trigonometry Section 1 Notes and Examples This is often known as the “CAST” diagram: some people remember it using the word CAST (make sure you start the word in the right place!); others use a phrase such as “All School Teachers Criticise” (which, while rather unfair to the teaching profession, does have the advantage of starting in the first quadrant). The symmetry of the diagram allows you to find relate trig rations of angles in the second, third and fourth quadrant to trig ratios angles in the first quadrant. 180° – 



 

 

360° – 

180° + 

To find a trig ratio of an angle in the second, third or fourth quadrant in terms of the trig ratio of an acute angle, you need to find the equivalent acute angle using symmetry, and then use the CAST diagram to find whether the trig ratio is positive or negative. You may find the Mathcentre video Trigonometric ratios in all quadrants useful to help you understand this work.

Trigonometric identities You need to learn the following identities:

sin  cos  2 sin   cos2   1 tan  

e.g. tan 30 

sin 30 cos 30

An identity is true for all values of θ. Proofs of these two identities are given on page 276 of the textbook.

These identities are demonstrated in the Flash resources Identities: sin x / cos x = tan x and Identities: sin²x + cos²x = 1.

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MEI C2 Trigonometry Section 1 Notes and Examples In the next example you need to use the trigonometric identities to rewrite an expression.

Example 2 Show that (sin   cos  )(sin   cos  )  2sin 2  1 Solution Working with the LHS and expanding the brackets gives: (sin   cos  )(sin   cos  )  sin 2   cos 2  Since sin 2   cos2   1 then cos2   1  sin 2   Substituting  into  gives: (sin   cos  )(sin   cos  )  sin 2   (1  sin 2  ) Simplifying: (sin   cos  )(sin   cos  )  2sin 2  1



as required.

Graphs of trigonometric functions Now that you can define sin θ, cos θ and tan θ for any value of θ you can draw the graphs of these functions: Graph of y = sin x y

1

y = sin x x

–360

–270

–180

–90

90

180

270

360

–1

Note: 1. The graph of y = sin x has a period of 360°. So it repeats every 360°.

e.g. sin 15° = sin 375° and sin 15° = sin -345°

2. It has rotational symmetry about the origin. 3. 1  sin x  1 So y = sin x lies between 1 and -1. 4. There is a line of symmetry at x = 90° and x = -90°.

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e.g. sin 30°= 0.5 so sin -30° = -0.5

e.g. sin x = sin (180° - x)

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MEI C2 Trigonometry Section 1 Notes and Examples You can see how this graph relates to the circle diagram on page 3 of these notes using the Flash resource The graph of y = sin x.

Graph of y = cos x y

1

y = cos x x –360

–270

–180

–90

90

180

270

360

–1

e.g. cos 25° = cos 385° and cos 25° = cos -335°

Note: 1. The graph of y = cos x has a period of 360°. So it repeats every 360°.

e.g. cos 30°= cos -30°

2. It has a line of symmetry in the y-axis. 3. 1  cos x  1 So y = cos x lies between 1 and -1. 4. The graph of y = cos x is a translation of the graph y = sin x by 90° to the left. In general, cos  sin(  90) e.g. cos 0° = sin 90° and cos 10° = sin 100°

You can see how this graph relates to the circle diagram on page 3 of these notes using the Flash resource The graph of y = cos x.

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MEI C2 Trigonometry Section 1 Notes and Examples Graph of y = tanx 3

y

y = tan x 2

1 x –360

–270

–180

–90

90

180

270

360

–1

–2

–3

e.g.tan 15° = tan 195° and tan 15° = tan -165°

Note: 1. The graph of y = tan x has a period of 180°. So it repeats every 180°.

3 so tan -60° = - 3 e.g. tan 60°=

2. It has rotational symmetry about the origin.

3.   tan x   So tan x can be any value not just those between -1 and 1. 4. There are asymptotes at x = 90 , x = 270 , x = 450 … where the value of tan x is undefined (since cos x = 0 for these values of x)

You can see how this graph relates to the circle diagram on page 3 of these notes using the Flash resource The graph of y = tan x. You may also find the Mathcentre video Trigonometric functions useful.

In the next example you need to use the symmetries of the trigonometrical graphs.

Example 3 Write down the exact values of (i) tan 120° (ii) sin (-135°) (iii) cos 300°

„Exact‟ means leave as a fraction or a surd – don‟t just give all the decimal places from your calculator!

Solution (i) y = tan θ has period of 180° so tan 120° tan (120° - 180°) = tan (-60°).

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MEI C2 Trigonometry Section 1 Notes and Examples y = tan θ has rotational symmetry about the origin so tan (-60°) = -tan 60°   3 So tan120   3 (ii) y = sin  has rotational symmetry so sin (-135°) = -sin 135°. y = sin  has a line of symmetry at  = 90° so sin 135° = sin (180° - 135°) = sin 45° So sin (-135°) = -sin 45°  

1 2

(iii) y = cos  has period of 360° so cos 300° = cos (300° - 360°) = cos (-60°). y = cos  is symmetrical about the y-axis so cos (-60°) = cos 60°. So cos 300° = cos 60° = 12 .

For practice in questions like the one above, try the interactive resource Exact trig function values of standard angles.

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Core 2 Trigonometry Section 1: Trigonometric Functions and Identities Crucial points 1. You should know the exact values for certain angles Make sure that you use the exact values of sin θ, cos θ and tan θ when θ = 0°, 30°, 45°, 60° or 90°, not the rounded values from your calculator. 2. You must be able to accurately draw trigonometrical graphs and know their properties Make sure you can draw accurate sketches of the graphs of y = cos θ, y = sin θ and y = tan θ, and that you know their symmetries and periodic properties.

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Core 2 Trigonometry Section 1: Trigonometric functions and identities Exercise 1. Triangle ABC is right angled at B. AB = 10cm and AC = 26 cm. (i) Calculate the length of BC. (ii) Write down the values of sin A, cos A, and tan A leaving your answers as fractions. (iii) Write down the values of sin C, cos C, and tan C leaving your answers as fractions. (iv) Write down three separate equations connecting the trig ratios for angle A to those for angle C. (v) In general, what conclusions can you draw from your answers to iv)? 2. (i) Sketch the curve of y = tan x for angles between 0° and 360°. (ii) Solve the equation for tan x = 1 and illustrate the roots on your sketch. (iii) Write down two angles that have tan x = –1 in the interval 0° to 360° without using your calculator. 3. Using a sketch of y = sin x, write down all of the angles between 90° and 540° (i) that have the same sin as 40°; (ii) that have the same sine as 160°. 4. Find all of the values of x between 0° to 360° such that (i) cos x = cos 25° (ii) sin x = sin 50° (iii) tan x = tan 120° (iv) sinx = –sin 60° (v) cos x = –cos 20° 5. Write the following as fractions or using square roots. You should not need your calculator. (i) sin 120° (ii) cos (–120°) (iii) tan 135° (iv) sin 300° (v) cos 270° 6. In the following give your answers as fractions (i) θ is acute and sinθ = 12 13 . Write down the value of cos θ. (ii) θ is obtuse and sin θ =

7 25

. write down the values of cos θ and tan θ.

(iii) θ is obtuse and tanθ = − . Write down the values of sin θ and cos θ. 8 15

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Core 2 7. Using the identities sin2 x + cos2 x = 1 and/or tan x = a)

1 − cos 2 x tan x

b)

sin x 1 − sin 2 x

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sin x , simplify cos x cos 2 x c) 1 + sin x

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Core 2 Trigonometry Section 1: Trigonometric functions and identities Solutions to Exercise 1.

C 26

B (i)

(ii)

10

A

BC 2 = AC 2 − AB 2 = 26 2 − 10 2 = 576 BC = 24 cm

24 12 = 26 13 10 5 = cos A = 26 13 24 12 = tan A = 10 5 sin A =

10 5 = 26 13 24 12 = cos C = 26 13 10 5 = tan C = 24 12

(iii) sin C =

(iv) sin A = cos C cos A = sin C 1 tan A = tan C (v)

Since C = 90° – A, this can be generalised to sin x = cos (90° – x) cos x = sin (90° – x) 1 tan x = tan( 90° − x )

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Core 2 2. (i) 4

y

3 2 1

x 90

180

270

360

−1 −2 −3 −4

tan x = 1

(ii)

x = 45° or 180° + 45° x = 45° or 225° (iii) By symmetry, angles are 180° - 45° = 135° and 360° - 45° = 315° 3.

2

y

1 x 90

180

270

360

450

540

−1

−2

(i)

180° - 40° = 140° 360° + 40° = 400° 540° - 40° = 500°

(ii)

360° + 20° = 380° 540° - 20° = 520°

4. (i) (ii) (iii) (iv) (v)

x = 360° - 25° = 335° x = 180° - 50° = 130° x = 180° + 120° = 300° x = 180° + 60° = 240° and x = 360° - 60° = 300° x = 180° - 20° = 160° and x = 180° + 20° = 200°

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Core 2

5. (i) (ii)

3 2

sin 120° = sin 60° =

cos( −120 °) = cos 120° = − cos 60° = −

(iii) tan 135° = − tan 45° = −1 (iv)

sin 300° = − sin 60° = −

(v)

cos 270° = − cos 90 ° = 0

6. (i)

cos θ =

5 13

1 2

3 2

13

12

θ 5 (ii)

Since θ is in the second quadrant, cos θ and tan θ are both negative. 25

7

θ 24 24 cos θ = − 25 7 tan θ = − 24

(iii) Since θ is in the second quadrant, sin θ is positive and cos θ is negative. 17

8

θ 15 sin θ = 178 cos θ = − 15 17

7. (i)

1 − cos 2 x sin 2 x = tan x tan x cos x = sin x × sin x = cos x

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Core 2 (ii)

(iii)

sin x

sin x cos x 1 − sin x sin x = cos x = tan x 2

=

cos 2 x 1 − sin 2 x = 1 + sin x 1 + sin x (1 + sin x )(1 − sin x ) = 1 + sin x = 1 − sin x

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MEI Core 2 Trigonometry Section 1: Trigonometric functions and identities Multiple Choice Test Do not use a calculator for this test. 1) What is the exact value of cos 120° ? (a) − 12

(b)

(c) − 23 (e) I don’t know

(d)

1 2 3 2

2) What is the exact value of sin 120° ? (a)

(b) − 12

1 2

(d) −

(c) 23 (e) I don’t know

3 2

3) What is the exact value of tan120° ? (a)

(b) − 3

1 3

(c) −

1 3

(d)

3

(e) I don’t know 4) What is the exact value of tan 2 30° + tan 2 60° ?

(a) 1

(b)

(c) 103 (e) I don’t know

(d)

4 3 4 3

5) What is the exact value of sin 2 30° − cos 2 30° ? (a) 12 (c) -1 (e) I don’t know

(b) 1 (d) − 12

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MEI C2 Trigonometry Section 1 MC test 6) What is the exact value of sin 2 60° + cos 2 150° ? (a) 0 (c) 1 (e) I don’t know

(b) (d)

7) For which of these values of x is sin x =

3 2

3 2 1 2

?

(a) 510° (c) 870° (e) I don’t know

(b) 840° (d) 660°

8) For which of these values of x is cos x =

3 2

(a) 1050° (c) 780° (e) I don’t know

(b) 1020° (d) 1080°

?

9) Which of the following statements are true? (i) If cos x = a then cos( − x ) = a (ii) If sin x = a then sin(180° − x) = a (iii) If tan x = a then tan(− x) = a (a) (i) and (iii) only (c) (i), (ii) and (iii) (e) I don’t know

(b) (ii) and (iii) only (d) (i) and (ii) only

10) Which of the following statements are true? (i) If cos x = a then cos(− x) = − a (ii) If sin x = a then sin(− x) = − a (iii) If tan x = a then tan(− x ) = − a (a) (i), (ii) and (iii) (c) (i) and (ii) only (e) I don’t know

(b) (ii) and (iii) only (d) (i) and (iii) only

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MEI Core 2 Trigonometry Section 2: Trigonometric Equations Study Plan Background In this section you will learn how to solve simple equations involving trigonometric functions.

Detailed work plan 1. Read page 280, including Example 10.2. This shows you how to solve the most simple type of trigonometric equation. There are some additional notes on principal values in the Notes and Examples, together with a further example (Example 1). 2. You can see further examples using the Flash resource Solving basic trig equations. 3. Example 10.3, 10.4 and 10.5, and Examples 2, 3, 4 and 5 in the Notes and Examples, look at different techniques which you may need to use to solve trigonometric equations. 4. You can see some more examples using the Flash resources Solving trig equations using identities (1) and Solving trig equations using identities (2). You may also find the Mathcentre video Trigonometric equations helpful. (Some of the examples in this video involve the use of radians, which are covered in a later section). 5. Now try the following questions from Exercise 10A: 4*, 10*

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MEI Core 2 Trigonometry Section 2: Trigonometric equations Notes and Examples In this section you learn how to solve trigonometric equations. These notes contain subsections on  Principal values  Solving simple trigonometrical equations  More complicated examples of trigonometrical equations.

Principal values 1 . 2 Your calculator will only give one solution – the principal value. You find this by pressing the calculator keys for arcsin 0.5 (or sin-1 0.5 or invsin 0.5). Check that you can get the answer of 30°. There are infinitely many roots to an equation like sin  

You can find other roots by looking at the symmetry of the appropriate graph. y 1

x –360

–270

–180

–90

90

180

270

360

–1

A second solution in a 360° cycle can be found by 180°- θ

y = cos θ

When you use the inverse cosine function, your calculator will always give you an answer from 0° to 180°.

y 1

x –360

A second solution in a 360° cycle can be found by 360°- θ

–270

–180

–90

90

180

270

360

–1

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MEI C2 Trigonometry Section 2 Notes and Examples y = tan θ 4

When you use the inverse tan function, your calculator will always give you an answer between -90° and 90°.

y

3 2 1 x –360

–270

–180

–90

90

180

270

360

–1 –2

A second solution in a 360° cycle can be found by θ + 180° or θ - 180°

–3 –4

Alternatively, you can use the quadrant diagram to find other solutions, by thinking about which quadrants the solutions will be in.

Solving simple trigonometrical equations Because there are infinitely many solutions to a trigonometric equation you are only ever asked to find a few of them! Any question at this level asking you to solve a trigonometric equation will also give you the interval or range of values in which the solutions must lie, e.g. you might be asked to solve tan   2 for 0    360 . You can only directly solve trigonometric equations like sin   12 or cos   14 or tan   2 . Here is an example.

Example 1 Solve sin  

3 for 360    360 . 2

Solution

3    60 2 There will be a second solution in the second quadrant. 180° - 60° = 120° is also a solution. Since y  sin  has a period of 360° any other solutions can be found by adding/subtracting 360° to these two solutions. So the other solutions are: 60° - 360° = -300° and 120° - 360° = -240° 3 So the values of θ for which sin   are -300°, -240°, 60°, 120°. 2 sin  

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MEI C2 Trigonometry Section 2 Notes and Examples You can see some more examples like the one above using the Flash resource Solving basic trigonometric equations. Look at the graph each time.

More complicated trigonometrical equations Any more complicated equations need to be manipulated algebraically before they can be solved. There are a number of techniques you can use: 1. 2.

3.

4.

Rearrange the equation to make cos  , sin  or tan  the subject. Check to see if the equation factorises to give two (or more) equations which involve just one trigonometric function (see Example 2). If it is a quadratic in either sin θ, cos θ, or tan θ it can either be factorised or solved using the formula for solving quadratic equations (see Example 3). If the equation involves just sin θ and cos θ (and no powers), check to sin  see if you can use the identity tan   (see Example 4). cos  If the equation contains a mixture of trigonometric functions (e.g. cos2 θ and sin θ) then you may need to use the identity sin 2   cos2   1 to make it a quadratic in either sin θ, cos θ, or tan θ (see Example 5).

Example 2 Solve 2cos sin   cos  0 for 0    360 . Solution 2cos sin   cos  0 can be factorised as there is cos  in both terms on the LHS. cos (2sin   1)  0 Factorise: It is wrong to divide So either cos   0 or 2sin   1  0

cos  0    90 360° - 90° = 270° is also a solution.

through by cos because you lose the solutions to cos = 0.

2sin   1  0  sin    12 This has solutions in the third and fourth quadrants. The solutions are 180° + 30° = 210° and 360° - 30° = 330°. So the values of θ for which 2cos sin   cos  0 are 90°, 210°, 270° and 330°.

In Example 3 you need to solve a quadratic equation.

Example 3 Solve 2cos2   3cos   2 for 0    360 .

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You can replace cos with x to make things simpler! Or factorise straightaway to get: (2cos – 1)(cos + 2) = 0 and then solve.

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MEI C2 Trigonometry Section 2 Notes and Examples Solution 2cos2   3cos   2 is a quadratic equation in cos  Rearrange the quadratic: 2cos2   3cos   2  0 Let cos  = x: 2 x2  3x  2  0 Factorise: (2 x 1)( x  2)  0 x  12 or x  2  cos   12 or cos  2 cos  2 has no solutions.

cos   12  cos  60 There is also a solution in the 4th quadrant, so 360° - 60° = 300° is also a solution. So we need to solve

So the values of θ for which 2cos2   3cos   2 are 60° and 300°.

In the next example you need to use the identity tan  

sin  . cos 

Example 4 Solve sin   2cos  0 for 0    360 . Solution You need to rearrange the equation. sin   2cos  0 sin  2 0 Dividing by cos : cos  sin  Since tan   : tan   2  0 cos   tan   2    63.4 to 1 d.p. There is also a solution in the 3rd quadrant. So 63.4° + 180° = 243.4° is also a solution.

You can safely divide by cos θ because it can’t be equal to 0. If it were then sin θ would also have to be 0 and cos θ and sin θ are never both 0 for the same value of θ.

So the values of θ for which sin   2cos  0 are 63.4° and 243.4° to 1 d.p.

You can see more examples like this one using the Flash resource Solving trig equations using identities (1).

In the next example you need to use the trigonometric identity sin 2   cos2   1 .

Example 5 Solve sin 2 x  sin x  cos2 x for 0  x  360

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MEI C2 Trigonometry Section 2 Notes and Examples Solution Rearranging the identity gives:

sin 2   cos2   1 cos2 x  1  sin 2 x



Substituting  into the equation sin 2 x  sin x  cos2 x gives: sin 2 x  sin x  1  sin 2 x This is a quadratic in sin x . Rearranging: 2sin 2 x  sin x 1  0 Rearranging: 2sin 2 x  sin x 1  0 This factorises to give: (2sin x 1)(sin x  1)  0 So either: or 2sin x 1  0 sin x  1  0 1  sin x  2  sin x  1  x  30 or 150  x  270 2 2 So the solutions to sin x  sin x  cos x are x  30,150 or 270

You can see more examples like this one using the Flash resource Solving trig equations using identities (2). You may also find the Mathcentre video Trigonometric equations helpful. (Some of the examples in this video involve the use of radians, which are covered in a later section).

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Core 2 Trigonometry Section 2: Trigonometric Equations Crucial points 1. Check that the solutions are in the range asked for Make sure that you check what range the solutions should lie in. 2. Remember to factorise instead of cancelling Never cancel terms like sin θ or cos θ . Always factorise instead. For example, in an equation like sin θ − sin θ cos θ = 0 , do not cancel out the term sin θ because you will lose the roots to the equation sin θ = 0 . Instead, take out the factor sin θ to give sin θ (1 − cos θ ) = 0 . 3. You must be able to draw trigonometrical graphs accurately and know their properties Make sure you know how to sketch the graphs of y = cos x, y = sin x and y = tan x and their properties.

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Core 2 Trigonometry Section 2: Trigonometric equations Exercise 1. Solve the following equations for 0° ≤ x ≤ 360°. (i) sin x = 0.3 (ii) tan x = 1.5 (iii) cos x = -0.7 (iv) sin x = -0.6 2. Find all of the values for x between -180° and 180° such that (i) sin x = 0.6 (ii) cos x = 0.8 (iii) tan x = - 0.6 (iv) cos x = -0.3 3. Without using a calculator find values for x between 0° and 360° such that 1 (i) sin x = 2 1 (ii) sin x = − 2 3 (iii) cos x = 2 1 (iv) cos x = − 2

4. Solve the following equations for 0° ≤ x ≤ 360°. (i) sin 2x = 1 (ii) cos 12 x = 0.5 (iii) tan 3x =

3

5. Solve the following equations for x between 0° and 360°. (i) 3 sin x = 4 cos x (ii) 2 cos x = –3 sin x 6. Solve the following equations for θ in the range from 0° to 360°. (i) 4 cos2 θ = 3 (ii) 2 cos2 θ = cos θ (iii) 4 sin θ cos θ = sin θ (iv) cos2 θ – cos θ – 2 = 0 (v) 3 sin2 θ + 5 cos θ – 1 = 0 (vi) 3 tan θ – 2 cos θ = 0

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Core 2 Trigonometry Section 2: Trigonometric equations Solutions to Exercise 1. (i)

sin x = 0.3 Solutions are in the first and second quadrants x = 17.5° or 180° − 17.5°

x = 17.5° or 162.5° (ii)

tan x = 1.5 Solutions are in the first and third quadrants x = 56.3° or 180° + 56.3°

x = 56.3° or 236.3° (iii) cos x = −0.7 Solutions are in the second and third quadrant x = 180° − 45.6° or 180° + 45.6°

x = 134.4° or 225.6° (iv)

sin x = −0.6 Solutions are in the third and fourth quadrant x = 180° + 36.9° or 360° − 36.9°

x = 216.9° or 323.1°

2. (i)

sin x = 0.6 Solutions are in the first and second quadrants x = 36.9° or 180° − 36.9°

x = 36.9° or 143.1° (ii)

cos x = 0.8 Solutions are in the first and fourth quadrants x = 36.9 ° or − 36.9 °

(iii) tan x = −0.6 Solutions are in the second and fourth quadrants x = 180° − 31.0° or − 31.0°

x = 149.0° or − 31.0° (iv)

cos x = −0.3 Solutions are in the second and third quadrants

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Core 2 x = 180° − 72.5° or − (180° − 72.5 )° x = 107.5° or − 107.5°

3. (i)

1 2

sin x =

1 , and solutions are in the first and second quadrants 2 so x = 45° or 180° − 45° sin 45° =

x = 45° or 135° (ii)

1 2 1 sin 30° = , and solutions are in the third and fourth quadrants 2 so x = 180° + 30° or 360° − 30° sin x = −

x = 210° or 330° (iii) cos x =

3 2

3 , and solutions are in the first and fourth quadrants 2 so x = 30° or 360° − 30° cos 30° =

x = 30° or 330° (iv)

1 2 1 cos 45° = , and solutions are in the second and third quadrants 2 so x = 180° − 45° or 180° + 45° cos x = −

x = 135° or 225° 4. (i)

sin 2 x = 1 2 x = 90° or 450°

x = 45° or 225 ° (ii)

cos 21 x = 0.5 1 2

x = 60°

x = 120°

As we are going to divide by 2, we need to use 360° + 90°

As we are going to multiply by 2, we do not need to consider the value of 1 2 x in the fourth quadrant.

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Core 2 (iii) tan 3 x = 3

3 x = 60°, 240°, 420°, 600°, 780°, 960°

x = 20°, 80°, 140°, 200°, 260°, 320°

5. (i)

3 sin x = 4cos x tan x =

4 3

x = 53.1° or 180° + 53.1° x = 53.1° or 223.1° (ii)

Using tan θ =

sin θ cos θ

2 cos x = −3 sin x tan x = − 23

x = 180° − 33.7° or 360° − 33.7° x = 146.3° or 326.3° 6. (i)

As we are going to divide by 3, we need to add 360° and 720° to each of the two basic solutions

Using tan θ =

sin θ cos θ

4cos 2 θ = 3 cos 2 θ =

3 4

cos θ = ± cos θ =

3 2

⇒ θ = 30° or 330°

3 2

cos θ = −

3 2

⇒ θ = 150° or 210°

θ = 30°, 150°, 210, 330° (ii)

2 cos 2 θ = cos θ 2 cos 2 θ − cos θ = 0 cos θ(2 cos θ − 1) = 0 cos θ = 0 cos θ = 0 cos θ = 21 θ = 60°,

or cos θ = 21 ⇒ θ = 90° or 270° ⇒ θ = 60° or 300° 90°, 270, 300°

(iii) 4 sin θ cos θ = sin θ

4 sin θ cos θ − sin θ = 0 sin θ(4cos θ − 1) = 0 sin θ = 0 or cos θ = 41 sin θ = 0 ⇒ θ = 0° or 180° or 360° cos θ = 41 ⇒ θ = 75.5° or 284.5° θ = 0°, 75.5°, 180°, 284.5 , 360°

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Core 2 (iv)

cos 2 θ − cos θ − 2 = 0 (cos θ − 2)(cos θ + 1) = 0 cos θ = 2 or cos θ = −1 There are no real values of θ for which cos θ = 2 cos θ = −1 ⇒ θ = 180°

(v)

3 sin 2 θ + 5 cos θ − 1 = 0 3(1 − cos 2 θ ) + 5 cos θ − 1 = 0

Using sin

2

3 − 3 cos 2 θ + 5 cos θ − 1 = 0

θ + cos 2 θ = 1

3 cos 2 θ − 5 cos θ − 2 = 0 (3 cos θ + 1)(cos θ − 2) = 0 cos θ = − 31 or cos θ = 2 There are no real values of θ for which cos θ = 2 cos θ = − 31 ⇒ θ = 109.5° or 250.5°

(vi)

3 tan θ − 2 cos θ = 0 3 sin θ − 2 cos θ = 0 cos θ 3 sin θ − 2 cos 2 θ = 0

Using tan θ =

sin θ cos θ

3 sin θ − 2(1 − sin 2 θ ) = 0 2 sin 2 θ + 3 sin θ − 2 = 0 (2 sin θ − 1)(sin θ + 2) = 0

Using sin

2

θ + cos 2 θ = 1

sin θ = 21 or sin θ = −2 There are no real values of θ for which sin θ = −2 sin θ = 21 ⇒ θ = 30° or 150°

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MEI Core 2 Trigonometry Section 2: Trigonometric equations Multiple Choice Test 1) Solve cos  0.3 for 0    360 . (a) 72.5° and 107.5° (c) 17.5° and 342.5° (e) I don’t know

(b) 72.5° and 252.5° (d) 72.5° and 287.5°

2) Solve tan x  0.5 for 0    360 . (a) 60° and 240° (c) 30° and 210° (e) I don’t know

(b) 26.6° and 206.6° (d) 26.6°

3) Solve 2sin   1  0 for 0    360 . (a) 210° and 330° (c) -30° and 330° (e) I don’t know

(b) 150° and 210° (d) 30° and 150°

4) Solve 3tan   2  0 for 0    360 . (a) 63.4° and 243.4° (c) 33.7° and 326.3° (e) I don’t know

(b) 146.3° and 326.3° (d) 33.7° and 213.7°

5) Solve 2cos  sin   0 for 0    360 . (a) 26.6° and 206.6° (c) 63.4° and 243.4° (e) I don’t know

(b) 116.6° and 296.6° (d) 153.4° and 333.4°

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MEI C2 Trigonometry Section 2 MC test 6) Solve (2cos   1)(cos   2)  0 for 0    360 . (a) 60° and 300° (c) 0°, 120°, 240° and 360° (e) I don’t know

(b) 120° and 240° (d) 0°, 60°, 300° and 360°

7) Solve 2sin 2   sin  1  0 for 180    180 . (a) 90°, 210° and 330° (c) 30°, 90°, and 150° (e) I don’t know

(b) -150°, -30° and 90° (d) -90° and 90°

8) Solve 2sin  cos  sin   0 for 0    360 . (a) 0°, 120°, 240° and 360° (c) 0° and 360° (e) I don’t know

(b) 60° and 300° (d) 0°, 60°, 180°, 300° and 360°

9) Solve tan 2   1  0 for 180    180 . (a) -135°, -45°, 45° and 135° (c) 30° and 150° (e) I don’t know

(b) -135° and 45° (d) -30°, -150°, 30° and 150°

10) Solve cos2   cos  sin 2  for 0    360 . (a) 30°, 180° and 330° (c) 30° and 330° (e) I don’t know

(b) 60° and 300° (d) 60°, 180° and 300°

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MEI Core 2 Trigonometry Section 3: The sine and cosine rules Study Plan Background Now that you can find trig ratios for angles that are bigger than 90, you can deal with triangles that are not right angled. The sine and cosine rules allow you to calculate sides and angles for any triangle. You should learn these rules so that they are your fingertips. You are familiar with the formula “half the base multiplied by the height” for the area of a triangle. This section introduces you to another formula that can be used when the height is not obvious. The formula relies on two sides and the angle included between them.

Detailed work plan 1. Read pages 286 to 287 paying particular note to example 10.7. This is sometimes called the ambiguous case because there are two possible answers. When tackling questions it is always wise to draw and label a diagram. There are some additional examples in the Notes and Examples (Examples 1 and 2). 2. You can see further examples of finding an unknown side using the Flash resource The sine rule, and further examples of finding an unknown angle using the Geogebra resource The sine rule – finding an angle. 3. Exercise 10B Attempt all three questions. Question 3 has a worked solution for you to consult. 4. Read pages 289 to 290. Again, note that you should always draw a labelled diagram. There are some additional examples in the Notes and Examples (Examples 3 and 4). 5. You can see further examples of finding an unknown side using the Flash resource The cosine rule, and further examples of finding an unknown angle using the Geogebra resource The cosine rule – finding an angle. 6. Exercise 10C Attempt all three questions. Question 3 has a worked solution for you to consult.

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MEI C2 Trigonometry Section 3 Study plan 7. For further practice try the interactive questions Finding an angle using the cosine rule, in which you are given the coordinates of all three vertices of a triangle. 8. Read Example 10.10 on page 292. This shows how the two rules can be applied together to solve problems. Also look at example 5 in the Notes and Examples. 9. You may also find the Mathcentre video The sine and cosine formulae useful. 10. Exercise 10D Try questions 2*, 4, 5*, 7, 8, 9*, 11*, 12. 11. The next part is about finding the area of a triangle. Read the proof on page 296 and example 10.11. There is another example in the Notes and Examples (Example 6). 12. You can see further examples using the Flash resource Area of a triangle. 13. Exercise 10E Try questions 1, 2*, 4, 6, 7*. 14. For further practice, try the interactive questions The area of a triangle given 3 sides and The area of a triangle given two sides and an angle. Note that in some cases you will need to use the sine rule or cosine rule first. 15. For an extra challenge, try the extension worksheet Cosy Cubes.

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MEI Core 2 Trigonometry Section 3: Sine and Cosine Rules Notes and Examples In this unit you learn about finding an unknown side or angle in any triangle. You will also learn a new formula for finding the area of a triangle. These notes contain subsections on:  The sine rule  The cosine rule  Choosing which rule to use  The area of a triangle

The sine rule The sine rule: A

a b c   sin A sin B sin C

b

c

This form is easier to use when finding an unknown side. The sine rule can also be written as:

C

B

a

sin A sin B sin C   a b c This form is easier to use when finding an unknown angle. Note: When you use the sine rule to find a missing angle, θ , always check whether 180° - θ is a possible solution as well. Example 1 shows a straightforward application of the sine rule to find an unknown side. B Example 1 Find the side BC in the triangle ABC.

30° x

c

A

10 cm

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MEI C2 Trigonometry Section 3 Notes and Examples Solution

180  30  75 2 x b  sin A sin B x 10  sin 75 sin 30 10sin 75  x sin 30 x = 19.3 cm (to 3 sig.fig.)

The triangle is isosceles so  BAC is By the sine rule: So:

so

You can see more examples like this using the Flash resource The sine rule.

Example 2 shows a straightforward application of the sine rule to find an unknown angle. Example 2 A, B and C are three points on a level plane. B is 6 km due west of A. C is 5 km from B and is on a bearing of 285° from A. Find  ACB. Due west is a bearing of 270°, so this angle must be 15°.

Solution First draw a diagram:

Your diagram doesn’t need to be accurate – just large enough to show all the information

N

C b a

5 km

15° B

By the sine rule: So:



A

6 km c sin A sin C  a c sin15 sin C  5 6 6sin15 sin C  5 sin C  0.310.... C = 18.1° to 1 d.p.

Don’t round here! Store the number in your calculator.

Check whether 180°- C is also a solution: 180°- 18.1° = 161.9° to 1 d.p. This also works so  ACB is 18.1° to 1 d.p. or 161.9° to 1 d.p

© MEI, 27/05/10

Angles A and C still add up to less than 180°

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MEI C2 Trigonometry Section 3 Notes and Examples You can see examples similar to this using the Geogebra resource The sine rule – finding an angle. This resource also shows geometrically what is happening when there is more than one possible solution.

The cosine rule A

The cosine rule:

a 2  b2  c 2  2bc cos A

b

c

This form is easier to use when finding an unknown side. The cosine rule can also be written as:

C

b2  c 2  a 2 cos A  2bc

B

a

This form is easier to use when finding an unknown angle.

Example 3 shows an application of the cosine rule to find an unknown side.

Example 3 Find the side YZ in the triangle XYZ. X y 7 cm

95°

z 6 cm Y

Z

x

Solution The cosine rule for this triangle is: So:

x2  y 2  z 2  2 yz cos X x 2  7 2  62  2  7  6cos 95 x 2  92.32... x = 9.61 cm to 3 sig. fig.

You can see more examples like this using the Flash resource The cosine rule.

Example 4 shows a straightforward application of the cosine rule to find an unknown angle.

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MEI C2 Trigonometry Section 3 Notes and Examples Example 4 Find the angle θ in the triangle ABC. A c 7m b

5m

θ C

B 4m a

Solution The cosine rule for this triangle is:

a 2  b2  c 2 cos C  2ab 42  52  7 2 cos C  2 45 cos C  0.2 C = 101.5° to 1 d.p.

You can see examples similar to this using the Geogebra resource The cosine rule – finding an angle.

You can test yourself using the interactive questions Finding an angle using the cosine rule. You are given the coordinates of all three vertices of a triangle.

Choosing which rule to use Use the sine rule when:  

you know 2 sides and 1 angle (not between the two sides) and want a 2nd angle (3rd angle is now obvious!) you know 2 angles and 1 side and want a 2nd side

Use the cosine rule when:  

you know 3 sides and want any angle you know 2 sides and the angle between them and want the 3rd side

You may find the Mathcentre video The sine and cosine formulae useful.

Example 5 shows how to decide whether to use the sine or the cosine rule.

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MEI C2 Trigonometry Section 3 Notes and Examples Example 5 A ship sails from a port, P, 6 km due East to a lighthouse, L, 6 km away. The ship then sails 10 km on a bearing of 030° to an island, A. Find: (i) The distance AP (ii) The bearing of P from A Solution First draw a diagram:

N A

N 30°

l

p 10 km 120° P

6 km a

L

(i) You know 2 sides and the angle between them so you need the cosine rule. l 2  a 2  p 2  2ap cos l The cosine rule for this triangle is: l 2  62  102  2  6 10cos120 l 2  196 l  14 km So the distance AP is 14 km. (ii) You can now use either the cosine rule or the sine rule to find the angle PAL. sin A sin L  The sine rule for this triangle is: a l Check whether 180° - 21.8°= 158.2° is sin A sin120  So: also a solution. 6 14 It isn’t because the angles 6sin120 in the triangle would total  sin A  more than 180°. 14 sin A  0.371…. A = 21.8° So the bearing is 180° + 30° + 21.8° = 231.8° to 1 d.p.

The area of a triangle To find the area of any triangle you can use the rule: Area of triangle ABC 

So you need two sides and the angle between them.

1 ab sin C 2

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MEI C2 Trigonometry Section 3 Notes and Examples Example 6 shows how to use this formula.

Example 6 Find the area of triangle ABC from Example 4. Solution A c 7m 5m

θ

b

B 4m a In Example 4, angle C was found to be 101.5° to 1 d.p. 1 Using the formula Area = ab sin C gives: 2 1 Area of triangle ABC =  4  5  sin101.5...  9.80 m2 2 C

You can see more examples like this using the Flash resource Area of a triangle. For practice in these techniques, try the interactive questions The area of a triangle given 2 sides and an angle and The area of a triangle given three sides. In the first one, the angle given may not always be the angle between the two given sides, so always draw a diagram and decide whether you need to use the sine rule or cosine rule first. For the second one, you will need to use the cosine rule to find an angle before finding the area.

For an additional challenge, try the extension worksheet Cosy Cubes.

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Core 2 Trigonometry Section 3: Sine and Cosine Rules Crucial points 1. Try geometry first when finding angles Make sure that you check whether any missing angles can be found using geometry first. 2. When finding angles, check for equivalent values Make sure when you use the sine rule to find a missing angle, θ, that you check to see whether 180° – θ is also a solution. 3. Check that all units are the same Make sure you check that all the units are the same so you are not mixing, say, kilometres and metres. 4. Check your calculator mode Make sure your calculator is in degrees mode and not in radians mode. 5. Be careful not to lose accuracy through rounding Make sure you never round a number until you reach your final answer.

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Core 2 Trigonometry Section 3: The sine and cosine rules Exercise 1. Solve the triangle ABC in which A = 66°, B = 42° and c = 12 cm. 2. Find two possible values of c in triangle ABC given that a = 16 cm, b = 10 cm, and B = 30°. 3. Solve the triangle ABC in which a = 6 cm, b = 9 cm and C = 97°. 4. Solve the triangle PQR in which p = 8 cm, q = 9 cm and r = 10 cm. 5. In triangle XYZ, X = 100°, Y = 30° and XY = 10 cm. Calculate the area of the triangle. 6. The area of a triangle is 12 cm2. Two of the sides are of lengths 6 cm and 7 cm. Calculate possible lengths for the third side 7. A ship S is 6.8 km from a lighthouse on a bearing of 310°. A second ship T is 8.4 km from the lighthouse on a bearing 075°. Calculate ST and the bearing of T from S correct to the nearest degree. 8. A golfer hits a ball B a distance of 170 m on a hole that measures 195 m from tee to hole. If his shot is directed 10° away from the direct line to the hole, find how far his ball is from the hole. 9. Calculate AB in the diagram below given that CD is 15 m, angle BCA = 40° and angle BDA = 20°. A

B C

D

10. A tower stands on a slope inclined at 18° to the horizontal. From a point lower down the slope and 150 m from the base of the tower, the angle of elevation of the top of the tower is 9.5°. Find the height of the tower. 11. A barge is moving at a constant speed along a straight canal. The angle of elevation of a bridge is 10°. After 10 minutes the angle of elevation is 15°. After how much longer does the barge reach the bridge? Give your answer to the nearest second.

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Core 2 Trigonometry Section 3: The sine and cosine rules Solutions to Exercise 1.

C

A

66°

42°

B

12

Angle C = 180° − 66° − 42° = 72° Using the sine rule:

a sin A

=

c

=

c

=

sin B

sin C 12 = sin 66° sin 72° 12 sin 66° a= = 11.53 cm sin 72°

a

Using the sine rule:

b sin B

sin C 12 = sin 42° sin 72° 12 sin 42° = 8.44 cm b= sin 72°

b

2. Using the sine rule:

sin A

a

b

sin A sin 30° = 16 10 16 sin 30° sin A = = 0.8 10 A = 53.1° or 126.9° C = 180 ° − 30° − A = 96.1° or 23.1°

Using the sine rule:

c sin C

=

b

sin B 10 = sin C sin 30° 10 sin C c= = 19.9 cm or 7.9 cm sin 30°

c

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Core 2 3.

C 9

97°

6

A

B

Using the cosine rule:

c 2 = a 2 + b 2 − 2ab cos C = 92 + 6 2 − 2 × 9 × 6 cos 97° c = 11.4 cm

Using the sine rule:

sin A

a

=

sin C

c

sin A sin 97 = 6 11.4 6 sin 97 sin A = 11.4 A = 31.5°

B = 180 ° − 97° − 31.5° = 51.5° . 4.

R 8

9 P

Q

10

Using the cosine rule:

q2 + r 2 − p2 92 + 10 2 − 8 2 cos P = = 2qr 2 × 9 × 10 P = 49.5°

Using the cosine rule:

cos Q =

p 2 + r 2 − q 2 8 2 + 10 2 − 92 = 2 pr 2 × 8 × 10

Q = 58.8° R = 180 ° − 49.46° − 58.75 ° = 71.8°

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Core 2 5.

X 10 Y

100°

30°

Z

Angle Z = 180° - 100° - 30° = 50°

x

Using the sine rule:

sin X

x

Area of triangle =

1 2

=

z sin z

10 sin 100° sin 50° 10 sin 100° x= = 12.86 sin 50° xz sinY =

= 21 × 12.86 × 10 sin 30° = 32.1 cm 2

6. Let a = 6 and b = 7 Area of triangle = 21 ab sin C 12 = 21 × 6 × 7 sin C

C = 34.85° or 145.15° c 2 = a 2 + b 2 − 2ab cos C

Using the cosine rule:

= 6 2 + 7 2 − 2 × 6 × 7 cos C = 85 − 84cos C

If C = 34.85°, c = 4.01 cm If C = 145.14°, c = 12.41 cm 7.

S 50° 6.8

T 50°

75°

8.4

L Using cosine rule:

ST2 = 6.8 2 + 8.42 − 2 × 6.8 × 8.4cos 125° ST = 13.5 km

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Core 2 sin S sin 125° = 8.4 13.5 8.4 sin 125° sin S = 13.5 S = 30.6° Bearing of T from S = 180° - 50° - 30.6° = 099.4°

Using sine rule:

B

8.

170 10°

T

H

195

t 2 = 170 2 + 195 2 − 2 × 170 × 195 cos 10° t = 40.4 It is 40.4 m from the hole. Using the cosine rule:

9.

A 50° 20°

B

20° 40° 140° 15 C

Using the sine rule on triangle ACD:

c sin C

c

For triangle ABD:

D =

a sin A

15 sin 140° sin 20° 15 sin 140° c= = 28.2 sin 20° AB = AD cos 70 ° = 28.2 cos 70° = 9.64 m

10.

=

62.5°

h 18° 9.5° 18°

150

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Core 2 h

150 sin 9.5° sin 62.5° 150 sin 9.5° h= = 27.9 m sin 62.5°

Using the sine rule:

=

11.

h 10°

tan 10° =

tan 15° =

h

15° 600v

(600 + t )v

h vt

vt

⇒ h = v (600 + t )tan 10°

⇒ h = vt tan 15°

v (600 + t )tan 10° = vt tan 15° 600 tan 10° + t tan 10° = t tan 15° 600 tan 10° = 1155 seconds tan 15° − tan 10° Time taken = 19 mins 15 seconds

t =

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Cosy Cubes Extension exercise involving the Sine and Cosine rules ‘Cosy Cubes’ ice cube manufacturer is marketing its new triangular ice cube, made especially for cocktails. This new brand has two triangular faces and a fixed thickness d, as shown:

d The triangular faces have sides a, b, c and corresponding angles θ , B, C, where θ is the largest angle. There are three possibilities for θ : (i) θ > 90D (ii) θ < 90D (iii) θ = 90D

The ice cubes each have a fixed volume of ice, so that they cool the drinks equally, and a fixed thickness d. The triangular faces therefore have a fixed area A. Cosy Cubes wants to find the optimal shape of their new cubes, by investigating the relationship between the area A, the angle θ and the lengths of the sides, a, b and c. For any of the three cases (i), (ii) or (iii), they drop a perpendicular from the vertex θ to its opposite side a. The length of this perpendicular is labelled h, and it meets side a at P. Lengths CP and BP are labelled x and y respectively. 1. (a) Draw a sketch of the triangle and write y in terms of a and x. (b) In the right–angled triangle containing angle C, write down a trigonometric equation relating h to b and c. (Label this equation (1)). (c) Write down the area A of the whole triangle θ BC, in terms of h and a, and hence in terms of a, b and C. (Label this equation (2)). (d) Referring to equation (2), for fixed area A, what is the effect of increasing angle C on b, for a fixed a, and vice versa. (e) Using symmetry, write down two more (similar) expressions for A.

Cosy Cube’s problem is ensuring that their ice cubes sit into as many glasses as possible, for a fixed volume of ice, without them getting stuck. Clearly, the longer the sides of the triangle, the more likely the ice is to get stuck, as the cubes turn around in the glasses. They, hence, decide to find an equation relating each angle to its opposite side, to see how, for example, varying θ will affect its opposite (and longer) side a.

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Cosy Cubes 2. (a) Using a similar method to that in 1 (b), write h in terms of c and B. (b) Use this result and equation (1) to prove the sine rule. (c) What does the sine rule show about the effect of increasing θ on a. [This is important to Cosy Cubes, as a is the longest side in the triangle]. Cosy Cubes also want to relate angle θ to sides a, b and c. Consider case (i) θ > 90D : Drop a perpendicular from vertex B to side b extended, as shown. Label this extra length x, and the perpendicular length y. Let D be the angle between sides x and c.

b C

3.

θ a

x D

y

c B

(a) Write down D in terms of θ . (b) Using Pythagoras, find an equation relating x, y and c. (Label this 3). (c) Also, (in another right-angled triangle), use Pythagoras to find an equation relating b, x, y and a. (Label this (4)). (d) Write down a trigonometric equation relating x to D and c. (Label this (5)). (e) Use (3), (4) and (5) to eliminate x and y, and form an equation relating a, b, c and D. (f) Write down cosD in terms of cos θ [Hint: think of the CAST diagram], and hence give the two forms of the cosine rule, with: (i) a 2 as the subject (ii) cos θ as the subject

The problem with the ice when θ > 90D is that the longest side of the triangle is longer, for a given volume of ice, and so does not fit into as many of Cosy’s Clients’ glasses. The new wine bar ‘The Triangle’ sells champagne cocktails in a particularly narrow glass, of diameter 5cm, and Cosy Cubes need to make sure it sells ice that fits into these glasses. So, they consider making θ less than 90D . 4. Follow the same procedure as in question 3, but for θ , and, thus, all angles, less than 90D (case (ii)). [Hint: drop a perpendicular from B to b to make two right-angled triangles].

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Cosy Cubes 5. (a) Verify that the cosine rule is satisfied when the triangle is: (i) Right-angled ( θ = 90D ) (ii) Equilateral (b) If the triangle is isosceles. What happens to the cosine rule when: (i) B=C θ =C (ii) (Remember, θ is the biggest angle) (c) Which type of triangle do you think will fit into most glasses and hence, be best to use at The Triangle? Why?

6. The volume of ice used in all of Cosy’s Cubes is equal to 20cm 3 and they all have a thickness, d, of 2cm. Thus, the area of the triangles, A, is 10cm 2 . (a) Find a triangle of the correct area with θ > 90D . (b) Find a triangle of the correct area with θ < 90D . (c) Is it possible to construct an equilateral triangle with the correct volume of ice, which will fit into The Triangle’s champagne glasses? (d) Try to construct a non-equilateral isosceles triangle of area 10cm 2 that fits into The Triangle’s champagne glasses. (e) Find a right-angled triangle with area 10cm 2 . Can you find one that fits into The Triangle’s champagne glasses?

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MEI Core 2 Trigonometry Section 3: Sine and Cosine Rules Multiple Choice Test Questions 1 and 2 refer to the triangle ABC. B θ

12 cm

x

C 30°

8 cm

A 1) The angle ABC is (a) 19.5° (c) 48.6° or 131.4° (e) I don’t know

(b) 48.6° (d) 19.5° or 160.5°

2) The side AB is (a) 18.2 cm (c) 9.12 cm (e) I don’t know

(b) 27.4 cm (d) 2.03 cm Z 25°

Questions 3 and 4 refer to the triangle XYZ. 15 cm

13 cm

X

θ x

3) The side XY is (a) 40.5 cm (c) 14.7 cm (e) I don’t know

Y

(b) 27.3 cm (d) 6.37 cm

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MEI C2 Trigonometry Section 3 MC test 4) The angle θ is (a) 31.2° (c) 59.6° (e) I don’t know

(b) 84.6° (d) 78.4°

Questions 4-7 refer to three points A, B and C, which lie on a level plane. B is 7 km from A on a bearing of 030°. C is 5 km from B on a bearing of 280°.

5) The distance AC is (a) 9.90 km (c) 6.24 km (e) I don’t know

(b) 9.28 km (d) 7.08 km

6) The bearing of A from C is (a) 68.4° (c) 148.4° (e) I don’t know

(b) 168.4° (d) 191.6°

7) The area of the triangle ABC is (a) 14.3 km2 (c) 16.4 km2 (e) I don’t know

(b) 12.2 km2 (d) 9.90 km2

Questions 8-10 are about an island on which some pirates have buried some treasure. The treasure, X, is buried on a bearing of 050° from the swamp, S, and on a bearing of 300° from the caves, C. The caves are 680 m from the swamp on a bearing of 085°.

8) The distance CX is (a) 415 m (c) 300 m (e) I don’t know

(b) 721 m (d) 604 m

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MEI C2 Trigonometry Section 3 MC test The harbour is 650 m south-west of the caves.

9) How many metres do the pirates need to walk to find the treasure from the harbour? (a) 713 m (c) 462 m (e) I don’t know

(b) 570 m (d) 675 m

10) What is the bearing of the treasure from the harbour to the nearest degree? (a) 015° (c) 009° (e) I don’t know

(b) 045° (d) 035°

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MEI Core 2 Trigonometry Section 4: Circular Measure Study plan Background Up until now you have always measured angles in degrees. However, as you can read on page 299, degrees have a basis in legend rather than physical fact. For many applications in the sciences and mathematics this is not good enough. A measure is required that has some basis in physical reality and hence radians are used.

Detailed work plan 1. Read pages 229-302, including examples 10.12, 10.13, 10.14. Don’t be put off by the fact that some angles are specified in terms of . This is to aid calculation and accuracy. The Notes and Examples give additional examples of converting between degrees and radians, and solving a variety of trig equations using radians. 2. It is worth taking the time to become very familiar with the radian equivalents of the common angles such as 30, 45, 60, 90 etc. Note these down beside your table of trig values in surd form so that you become equally familiar with angles in radians as with angles in degrees. You can also use the Flash resource Standard values of trig ratios, pressing the radian button. 3. For further examples you can look at the Flash resources Converting degrees to radians and Converting radians to degrees. 4. Exercise 10F Try at least questions 1, 2, 3, 4*, 6*. 5. For extra practice try the interactive questions Convert degrees to radians and Convert radians to degrees. 6. Read pages 303-304. This introduces you to the calculation of lengths of arc and areas of sectors of circles. You need to learn the appropriate formulae so that you can use them almost without having to think. There are two additional examples in the Notes and Examples. 7. To see further examples use the Flash resources Arc length and Area of sector. You may also find the Mathcentre video Radian measure helpful. 8. Exercise 10G Try at least questions 1, 2*, 3*, 5, 6*, 7.

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MEI Core 2 Trigonometry Section 4: Circular Measure Notes and Examples In this section you learn about radians and circular measure. These notes contain subsections on:  Radians  Common values of trig functions in radians  Trigonometric graphs in radians  Solving trigonometric equations using radians  Sectors of circles

Radians Often mathematicians use radians rather than degrees to measure angles as they sometimes make calculations easier. To convert degrees to radians you multiply by To convert radians to degrees you multiply by

 180 180



In the example below you need to convert degrees into radians.

Example 1 Convert these degrees to radians: (a) 60° (b) 270° Solution (a) 60° = 60 ×

 180

(b) 270° = 270 × (c) 173° = 173 ×

3



180 180

173°



=



(c)

=

3 2

= 3.02 rads Page: 1

You can see further examples like this one using the Flash resource Converting degrees to radians.

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MEI C2 Trigonometry Section 4 Notes and Examples For extra practice in questions like the one above, try the interactive questions Convert degrees to radians.

In the next example you need to convert radians into degrees.

Example 2 Convert these radians to degrees:  7 (a) (b) 10 4 Solution (a)



=



×

(c)

0.65 rads

180

= 45° 4 4  7 7 180 (b) = × = 126° 10 10  180 (c) 0.65 rads = 0.65 × = 37.2°



You can see further examples like this one using the Flash resource Converting radians to degrees.

For extra practice in questions like the one above, try the interactive resource Convert radians to degrees.

Common values of trig functions in radians You should soon be able to remember the radian equivalent of many common angles:

 radians 6  90° = radians 2 30° =

45° =

 radians 4

180° =  radians

60° =

 radians 3

360° = 2 radians.

You should normally express these as fractions of  rather than as decimals. Just as when working in degrees, you should be able to recognise the values of sin , cos  and tan  for common values of . This table reminds you of the common values you should know, with the angles given in radians.

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MEI C2 Trigonometry Section 4 Notes and Examples θ

0

cos θ

1

sin θ

0

tan θ

0

 6 3 2 1 2

 4 1 2  2 2 1 2  2 2 1

1 3  3 3

 3 1 2

 2 0

3 2

1

3

undefined

You can test yourself on these values using the Flash resource Standard values of trig ratios. Switch to radians, then try to remember the exact value of each trig ratio, and click on the box to check your answer.

Trigonometric graphs in radians You need to be familiar with the graphs of the trig functions when working in radians as well as in degrees. Here are reminders of these graphs. The graph of y = sin θ where θ is in radians: y

1

x –2¤

–¤

¤

2¤

–1

It repeats every 2 radians



The graph of y = sin x has a period of 2 . So for example, sin 0.2  sin(0.2  2 )



It has rotational symmetry about the origin.    So for example, since sin  0.5 then sin     0.5 6  6



There is a line of symmetry at  



2

and   

© MEI, 23/12/10

 6

rad 

180  30 6



2

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MEI C2 Trigonometry Section 4 Notes and Examples The graph of y = cos θ where θ is in radians y

1

x –2¤

–¤

¤

2¤

–1

The graph of y = cos θ has a period of 2 . So for example, cos 0.2  cos(0.2  2 )  It has line symmetry about the y-axis.    So for example, cos  cos    3  3 

The graph of y = tan θ where θ is in radians 4

y

3 2 1 x –2¤

–¤

¤

2¤

–1 –2 –3 –4

 

The graph of y = tan x has a period of  . So for example, tan 0.2  tan(0.2  2 ) It has rotational symmetry about the origin.    So for example, since tan  3 then tan      3 3  3

 3

rad 

180  60 3

Solving trigonometric equations using radians In section 2 you looked at solving trig equations, giving your answers in degrees. The same methods are used when working in radians. Make sure that your calculator is set to Radians. However, remember that you should be able to recognise common values of the trig functions and give the exact angle in radians as a fraction of  where appropriate.

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MEI C2 Trigonometry Section 4 Notes and Examples In this example you need to find a solution to a trigonometric equation in radians. When the range is given in radians then you must find a solution in radians.

Example 3 Solve cos  0.5 for 0    2 Solution Page: 5 You know that cos 60°= 0.5

60  60 So  





180





3

radians

is one solution. 3 There is another solution in the 4th quadrant.  5 is also a solution. 2   3 3 So the values of θ for which cos  0.5 are

 5 and 3 3

In this example you need to use a trigonometric identity.

Example 4 Solve 2sin   cos for      Solution You can divide by cos  since we know that cos   0 as sin  and cos  are not both equal to 0 for the same value of  . 2sin  cos  2sin    1 So:  cos  cos  cos  sin  You know tan    cos  2 tan   1 Substituting  into : 1  tan   2    0.464 rads There is also a solution in the 3rd quadrant. 0.464 - π = -2.68 rads is also a solution. So the values of θ for which 2sin   cos are -2.68 rads and 0.464 rads. In this example you need to use the identity sin 2 θ  cos2 θ  1.

Example 5 Solve 3  3sin θ  2cos2 θ for 0  θ  2π

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MEI C2 Trigonometry Section 4 Notes and Examples Solution Using the identity sin 2 θ  cos2 θ  1 gives: Substituting  into 3  3sin θ  2cos2 θ :

 cos2 θ  1  sin 2 θ 2 3  3sin θ  2(1  sin θ )

3  3sin θ  2  2sin 2 θ 2sin 2 θ  3sin θ  1  0 (2sin θ  1)(sin θ  1)  0

This is a quadratic in sin θ which rearranges to: Factorising: 1 So either: (2sin θ  1)  0  sin θ   2 π  π  11π  θ   or 2π      6 6  6    7 or          6 6 or (sin θ  1)  0  sin θ  1 π π 3π  θ   or   2π  2 2 2 7π 11π 3π So the solutions to 3  3sin θ  2cos2 θ are θ  , or 6 6 2

Sectors of circles 



A sector of a circle is the shape enclosed by an arc of the circle and two radii. A minor sector is a sector which is smaller than a semi-circle. A major sector is a sector which is larger than a semi-circle.

minor sector

The formula for arc length is: Arc length = r



major sector

where  is in radians.

The formula for the area of a sector is:

1 2 r  where  is in radians. 2 In this example you need to use the formula for arc length. Area of a sector =

Example 6 A sector of a circle with radius 6 cm has an arc length of 2 . Find the angle subtended at the centre of the circle. Solution

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MEI C2 Trigonometry Section 4 Notes and Examples Page: 6 Use arc length = r . So 2π = 6θ 2  So    6 3 You can see more examples like this using the Flash resource Radians: arc length. In the next example you need to use the formula for sector area.

Example 7 Find the shaded area. This is called a segment.

A 5 120° 5 O

B

Solution Convert 120° to radians:

120° = 120

 180

=

2 3

Page: 7 Area of segment = Area of sector OAB – area of triangle OAB 1 Area of sector = r 2 where θ is in radians. 2 1 Area of triangle = r 2 sin  2 So area of segment  12 r 2  12 r 2 sin 

 12 r 2 (  sin  ) Area of segment =

1 2  2 2  2 5   sin  = 15.4 cm 2 3   3

Area of triangle is ½absin C but a and b are equal to the radius of the circle.

Make sure your calculator is in radians mode.

You can see more examples like this using the Flash resource Radians: sector area. You may also find the Mathcentre video Radian measure helpful.

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Core 2 Trigonometry Section 4: Circular Measure Crucial points 1. Where possible, leave your answer in terms of π When an angle is a simple multiple or fraction of 180°, then leave your answers in terms of π . 2. You must know the notation for angles in radians When an angle is given as a multiple or fraction of π you can assume it is in radians. Otherwise you write, say, 0.7 rads. In some older textbooks 0.7 rads might be written as 0.7c. 3. Make sure that your calculator is in the right mode π Remember when you want to work out, say, sin , make sure your 3 calculator is in ‘rad’ mode. 4. Make sure that angles are in radians before using sector formulae 1 The formulae for arc length ( rθ ) and sector area ( r 2θ ) can only be used 2 when the angle, θ , is in radians. π To change degrees to radians multiply by . 180° 5. Make sure that your solutions lie in the required range When you are solving a trigonometric equation make sure you check what range the solutions are to lie in and give all the solutions within that range. If the range is given in radians, e.g. 0 ≤ θ ≤ 2π , then give your answers in radians. If the range is given in degrees, e.g. 0 ≤ θ ≤ 360° , then give your answers in degrees. 6. You must know the relationship between the radius and tangent of a circle Remember that the radius of a circle and the tangent to a circle meet at right angles.

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Core 2 Trigonometry Section 4: Circular measure Exercise 1. Solve the following equations for 0 ≤ θ ≤ 2π giving your answer in terms of π. You should not need your calculator. 3 3 (i) sin θ = (ii) (iii) tan θ = 3 cos θ = − 2 2

2. Solve the following equations for 0 ≤ θ ≤ 2π giving your answers to 1 d.p. where necessary. (i) cos θ = 0.4 (ii) tan θ = -1.2 (iii) sin2 θ = 1 3. Find values for x in the range 0 ≤ x ≤ π. (i) 2 sin x cos x = sin x (ii) 2 sin2 x – cos x – 1 = 0 4. An arc AB subtends an angle of 1.5 radians at the centre O of a circle of diameter 20 cm. Find the length of arc AB and the area of sector AOB. 5. A chord AB subtends an angle of 0.75 radians at the centre of a circle of radius 20 cm. Find the area of the minor segment cut off by AB. 6. An equilateral triangle is inscribed in a circle of radius 10 cm. (i) Find the area of the circle (ii) Find the area of the triangle (iii) Find the area of the three segments surrounding the triangle. 7. How far is a chord AB from the centre of a circle of radius 5 cm given that the arc length AB is 9.3 cm? 8. Once upon a time a hermit found an island shaped like a triangle with straight shores of lengths 6 km, 8 km, and 10 km. Needing seclusion, he declared that no one should approach within 1 km of his shore. What was the area of his ‘exclusion’ zone? 9. Two circles with centres A and B intersect at points P and Q such that ∠APB is a right angle and ∠PAQ =

π

. If AB = 10 cm, find the length of the perimeter of 3 the region common to the two circles. 10. Two parallel chords of lengths 8 cm and 11 cm lie on the same side of a circle of radius 6 cm. Calculate the perimeter and area of the region enclosed by the two chords.

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MEI Core 2 Trigonometry Section 4: Circular measure Solutions to Exercise 1. (i)

3 2  3 and solutions are in the first and second quadrants sin  3 2

sin  

so  



3

or  



2   or 3 3 (ii)



3

3 2  3 and solutions are in the second and third quadrants cos   6 2

cos   

so    



or  

6 5 7  or 6 6



6

(iii) tan   3

tan



3

so  

 3 and solutions are in the first and third quadrants

 3



or  

4   or 3 3 2. (i)

 3

cos  0.4 Solutions are in the first and fourth quadrants   1.16 or 2  1.16

  1.2 or 5.1 radians (1 d.p.) (ii)

tan   1.2 Solutions and in the second and fourth quadrants     0.88 or 2  0.88

  2.3 or 5.4 radians (1 d.p.)

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MEI C2 Trigonometry Section 4 Exercise solutions (iii) sin 2   1

sin   1 sin   1

 

sin   1



3. (i)

 2

or

3 2



2 3   2

2 sin x cos x  sin x 2 sin x cos x  sin x  0 sin x(2 cos x  1)  0 sin x  0 or cos x  21 sin x  0  x  0 or 

cos x 

x  0, (ii)

x

1 2

 3

 3

,

2 sin 2 x  cos x  1  0 2(1  cos 2 x )  cos x  1  0 2  2 cos 2 x  cos x  1  0 2 cos 2 x  cos x  1  0 (2 cos x  1)(cos x  1)  0 cos x 

1 2

cos x 

1 2

or cos x  1

x

3  x 

cos x  1

x

 3



or 

4.

A 10 O 1.5

B Arc length  r  10  1.5  15 cm Area of sector  21 r 2  21  102  1.5  75 cm²

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MEI C2 Trigonometry Section 4 Exercise solutions 5.

A 20 O 0.75 B Area of triangle AOB  21  20  20 sin 0.75  136.33 Area of sector AOB  21 r 2  21  202  0.75  150 Area of segment  150  136.33  13.7 cm²

6.

A

O 10 120° 10 30°

B

(i)

Area of circle    102  314.16 cm²

(ii)

Area of triangle AOB  21  10  10 sin 120  43.30 Total area of triangle = 3  43.30  129.90 cm²

(iii) Area of segments  314.16  129.9  184.26 cm²

7.

A 5 O

d  B

Arc length  r 9.3  5  2

  0.93

d  5 cos 0.93  2.99 cm

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MEI C2 Trigonometry Section 4 Exercise solutions Since 6  8  10 , the triangle is right-angled. 2

8.

2

2

A 6 C

10 8

B

Angle at A is given by sin A  108  A  0.927 Angle in sector of circle at A =   0.927  2.214 Area of sector of circle at A  21  1 2  2.214  1.107 km² Angle at B is given by sin B  106  B  0.6435 Angle in sector of circle at B =   0.6435  2.498 Area of sector of circle at B  21  1 2  2.498  1.249 km² Area of quarter circle at C  21    1 2  0.785 Total area of exclusion zone  6  8  10  1.107  1.249  0.785

 27.14 km 2

9. P

P

rB

rA A

B

A

 6

 3

10

B

Q

r A  10cos 6  10  21 3  5 3 Arc length PQ for circle with centre A  r A 

rB  10cos 3  10  21  5 Arc length PQ for circle with centre B  rB 

 3



5 3  9.07 cm 3

2 2 5  10.47 cm 3 3

Total perimeter = 9.07 + 10.47 cm = 19.54 cm

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MEI C2 Trigonometry Section 4 Exercise solutions

10.

A B 11

8 C

D Let angle subtended by chord AD be  2  6 sin 21   11    2.319 Let angle subtended by chord BC be  2  6 sin 21   8    1.459 Length of arc AD  6 Length of arc BC  6  Length of arcs AB and CD  6(   )  5.16 Perimeter of region = 11 + 8 + 5.16 = 24.16 cm Area of sector OAD  21  62   18 Area of triangle OAD  21  62 sin   18 sin  Area of minor segment bounded by AD  18(  sin  ) Similarly, area of minor segment bounded by BC  18(   sin  ) Area of region ABCD  18(  sin  )  18(   sin  )  20.18 cm²

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MEI Core 2 Trigonometry Section 4: Circular Measure Multiple Choice Test 1) Convert 50° to radians.

5 rads 18 18π rads (c) 5 (e) I don’t know (a)

2) Convert

18 rads 5 5π (d) rads 18 (b)

3π to degrees. 2

(a) 270° (c) 0.08° (e) I don’t know

(b) 85.9° (d) 120°

3) Solve the equation sin θ = 0.5 for 0 ≤ θ ≤ 2π . π 11π , 6 6 π 5π (c) , 6 6 (e) I don’t know (a)

(b) 30° (d)

π 6

4) Solve the equation 2 cos θ = 0.2 for 0 ≤ θ ≤ 2π . (a) 0.685 rads, 2.46 rads (c) 1.47 rads, 1.67 rads (e) I don’t know

(b) 1.37 rads, 4.91 rads (d) 1.47 rads, 4.81 rads

5) Solve the equation tan θ = 3 for −π ≤ θ ≤ π . π 7π , 6 6 2π π (c) − , 3 3 (e) I don’t know (a)

π 4π , 3 3 5π π (d) − , 6 6

(b)

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MEI C2 Trigonometry Section 4 MC test 6) Solve the equation 3cos θ = 2sin 2 θ for 0 ≤ θ ≤ 2π . π 5π , 3 3 π 2π (c) , 3 3 (e) I don’t know

π 5π , 6 6 π 11π (d) , 6 6

(a)

(b)

7) Find the perimeter of this sector.

30° 5 cm (a) 2.62 cm (c) 12.62 cm (e) I don’t know

(b) 160 cm (d) 150 cm

8) Find the area of the shaded segment. 6 cm π/3 (a) 18.5 cm2 (c) 15.6 cm2 (e) I don’t know

(b) 18.8 cm2 (d) 3.26 cm2

9) The arc length of a sector of radius 8cm is 3π cm. Find the area of the sector. (a) 12π cm2 9π cm2 2 (e) I don’t know (c)

3π cm2 2 3π (d) cm2 4

(b)

10) The area of a sector with an angle of 45° is 2π cm2. Find its perimeter. (a) 4π cm (c) (π + 8) cm (e) I don’t know

(b) π cm (d) (4π + 32) cm

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MEI Core 2 Trigonometry Section 5: Transformations Study Plan Background Back in chapter 3 you looked at the way in which graphs could be sketched quickly using translations. This section applies these ideas to trigonometric graphs and extends them further to include one-way stretches and reflections. The work is very useful when applying mathematics either in science or in mechanics.

Detailed work plan 1. You can investigate stretches and translations of trigonometric graphs using the Geogebra resource Transformations of trigonometric graphs. 2. Remind yourself of the effect translations have on curves (page 311). This is summarised in the Notes and Examples and a worked example is given. 3. Read the section on stretches (pages 311 to 315). Again, this is summarised in the Notes and Examples, and further worked examples are given (Examples 2 and 3). 4. Stretches like these can, of course, be applied to any graphs, not just trigonometric graphs. Look at the Flash resources Horizontal stretches and Vertical stretches to see stretches applied to a polynomial graph. 5. You can use graphs like these to solve more trigonometric equations. There are some worked examples in the Notes and Examples (this is not covered in the textbook but you could be asked to solve an equation like this in your examination). 6. The section in the Notes and Examples on reflections is extension work which will be covered in C3. This looks at the special cases of stretches where the scale factor is -1, resulting in a reflection. 7. Exercise 10H Try questions 1, 2, 3, 4*, 5, 7, 8*, 9, 10*.

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MEI Core 2 Trigonometry Section 5: Transformations Notes and Examples These notes contain sub-sections on:  Translations  Stretches  Solving equations  Reflections (extension work)

In this section you learn how to sketch transformations of graphs of trigonometrical functions and other functions. This will allow you to sketch graphs like f( x)  2sin x , f( x)  tan 2x and f( x)  cos( x  30) . Work on translations has already been covered in chapter 3: this work is revised and stretches and reflections are introduced. You can investigate the effect of translations and stretches of the graphs of trigonometric functions using the Geogebra resource Transformations of trigonometric graphs.

Translations In a translation, the graphs of sin x , cos x or tan x are translated (or moved) a certain number of units in the x or y direction. A function of the form: f ( x)  cos x  f( x)  cos( x  t )    f( x)  sin( x  t )  translates of the graphs of f ( x)  sin x  f ( x)  tan x  f( x)  tan( x  t ) 

t by the vector   (i.e. t units in the positive x direction). 0 In general, any function of the form y = f(x – t) represents a translation of the graph of t y = f(x) by the vector   . 0 If t is negative, giving a function of the form y = f(x + t), then of course the translation is in the negative x direction, i.e. to the left. A function of the form: f ( x)  cos x  f ( x)  s  cos x    f ( x)  s  sin x  translates the graphs of f ( x)  sin x  f ( x)  tan x  f ( x)  s  tan x  0 by the vector   (i.e. s units in the positive y direction). s

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MEI C2 Trigonometry Section 5 Notes and Examples In general, any function of the form y = s + f(x) represents a translation of the graph of 0 y = f(x) by the vector   . s Combining these two transformations: Any function of the form y = s + f(x – t) represents a translation of the graph of t  y = f(x) by the vector   . s This example shows how to translate a graph. Example 1 Sketch the graphs of (i) y  1  sin x (ii) y  cos( x  60) Solution (i) This is a translation of the graph y  sin x by 1 unit vertically upwards. 2

y

1

y = 1 + sin x x

–360

–270

–180

–90

90

180

270

360

y = sin x –1

The graph of y = sin x has been moved one unit in the positive y direction.

–2

(ii) This is a translation of the graph y  cos x by 60° to the right. 1.5

y

y = cos x

1

y = cos (x -60°)

0.5 x –360

–270

–180

–90

90

180

270

360

–0.5

–1

The graph of y = cos x has been moved 60 units in the positive x direction.

–1.5

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MEI C2 Trigonometry Section 5 Notes and Examples Stretches In a stretch, the graphs of sin x , cos x or tan x are stretched by a certain scale factor parallel to the x or y axes. A function of the form: f ( x)  cos x  f ( x)  cos(ax)    f ( x)  sin(ax)  stretches the graphs of f ( x)  sin x  f ( x)  tan x  f ( x)  tan(ax)  parallel to the x axis with scale factor

1 . a

In general, a function of the form y = f(ax) represents a one-way stretch of the graph 1 of y = f(x) parallel to the x axis with scale factor . a A function of the form: f ( x)  cos x  f ( x)  a cos x    f ( x)  a sin x  stretches the graphs of f ( x)  sin x  f ( x)  tan x  f ( x)  a tan x  parallel to the y axis with scale factor a . In general, a function of the form y = af(x) represents a one-way stretch of the graph of y = f(x) parallel to the y axis with scale factor a.

This example shows how to stretch a graph.

Example 2 Sketch the graph of y  2cos x Solution This is a one-way stretch parallel to the y axis with scale factor 2. y y = 2 cos x 2 y = cos x

1 x –360

–270

–180

–90

90

180

270

360

–1

The y co-ordinate of every point on the original curve needs to be doubled.

–2

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MEI C2 Trigonometry Section 5 Notes and Examples The next example shows how to find the equation of a curve.

Example 3 Write down the equation of the following graph: 1

y

x –360

–270

–180

–90

90

180

270

360

–1

Solution Compare the graph with that of y = sin x: 2

y

y = sin x 1 x –360

–270

–180

–90

90

180

270

360

–1

–2

The x co-ordinate of every point on the graph of y = sin x has been halved. Therefore the equation is y = sin 2x.

Stretches like these can, of course, be applied to any graphs, not just trigonometric graphs. Look at the Flash resources Horizontal stretches and Vertical stretches to see stretches applied to a polynomial graph.

Solving equations In section 2 you solved simple equations like sin x  0.5 . As you know, almost all equations of this type have two solutions in the range 0  x  360 . Graphs like the ones above can be very useful in helping to solve equations like sin 2 x  0.5 . Notice that the graph of y  sin 2 x repeats itself every 180°, whereas the graph of y  sin x repeats itself every 360°. This means that the equation sin 2 x  0.5 has four solutions in the range 0  x  360 , as shown in the graph below.

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MEI C2 Trigonometry Section 5 Notes and Examples y

y  sin 2 x

1

x –90

90

180

270

360

450

–1

Since four solutions are expected in the range 0  x  360 , add 360° to each of the first two values for 2x to give two more.

sin 2 x  0.5 2 x  30, 150, 390, 510 x  15, 75, 195, 255

Halving each value for 2x gives four solutions for x in the range 0  x  360

Example 4 Solve each of the following equations in the given range: (i) tan 3x  1 for 0  x  2 (ii) cos(2 x  40)  0.5 for 180  x  180 Solution tan 3x  1 (i)  5 3x  , , 4 4  5 x , , 12 12 (ii)

9 13 17 21 , , , 4 4 4 4 3 13 17 7 , , , 4 12 12 4

cos(2 x  40)  0.5 2 x  40  60, 300, 420, 660 2 x  20, 260, 380, 620 x  10, 130, 190, 310

There will be six solutions in the range.

The range was given in radians, so the solutions must be in radians There will be four solutions in the range.

Subtract 40° from each value, then divide by 2.

Reflections This section is extension work. It will be covered in more depth in C3. Special cases of one-way stretches occur when the scale factor of the stretch is given by -1. A one-way stretch with scale factor -1 in the y direction results in the graph being reflected in the x axis.

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MEI C2 Trigonometry Section 5 Notes and Examples A function of the form: f ( x)  cos x  f ( x)   cos x    f ( x)   sin x  reflects the graphs of f ( x)  sin x  in the x axis. f ( x)  tan x  f ( x)   tan x  In general, any function of the form y = -f(x) represents a reflection of the graph of y = f(x) in the x axis. A one-way stretch with scale factor -1 in the x direction results in the graph being reflected in the y axis. A function of the form: f ( x)  cos( x)  f ( x)  cos x    f ( x)  sin( x)  reflects the graphs of f ( x)  sin x  in the y axis. f ( x)  tan( x)  f ( x)  tan x  In general, any function of the form y = f(-x) represents a reflection of the graph of y = f(x) in the y axis. This example shows how to find the equation of a curve. Example 5 Write down the equation of the following graph: 2

y

1 x –2¤

–¤

¤

2¤

–1

–2

Solution This graph is clearly something to do with tan x. Comparing the graph with y = tan x:

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MEI C2 Trigonometry Section 5 Notes and Examples 2

y

1 x –2¤

–¤

¤

2¤

–1

y = tan x –2

You can see that the graph is a reflection of y = tan x in the y axis. So the equation is y = -tan x

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Core 2 Trigonometry Section 5: Transformations Crucial points 1. Know how the graph of y = f(ax) is related to the graph of y = f(x) Remember that the graph of, say, y = sin 3 x is a one-way stretch of the 1 graph y = sin x with scale factor parallel to the x-axis. 3 2. Know how the graph of y = f(x + a) is related to the graph of y = f(x) Remember that the graph of, say, y = cos( x + 60°) is a translation of the ⎛ −60 ⎞ graph y = cos x by the vector ⎜ ⎟ ⎝ 0 ⎠ 3. Be careful when solving equations of the form sin ax = k Sketching a graph is useful. Think about the number of solutions you expect to obtain.

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Core 2 Trigonometry Section 5: Transformations Exercise 1. Starting with the graph of y = sin x, state the transformation which can be used to sketch (i) y = 3 sin x (ii) y = sin (x + 90°) (iii) y = sin 14 x (iv) y = sin x – 1 (v) y = sin (x – 180°) 2. Starting with the graph of y = cos x, state the transformation which can be used to sketch (i) 2y = cos x (ii) y = cos(x – 90°) (iii) y = cos 3x (iv) y = cos x – 2 (v) y = cos ( x + 30°) 3. Sketch the graphs of y = cos x and y = cos ½ x on the same set of axes for the range 0° ≤ x ≤ 360°. 4. Sketch the graph of y = tan x for the range –90° ≤ x ≤ 360°. On separate sets of axes sketch the graphs of (i) y = tan 2x (ii) y = tan x + 1 (iii) y = tan (x – 90°) 5. Solve the following equations for 0° < x < 360° . (i) cos 3x = 0.2 (ii) sin 12 x = 0.7 (iii) tan(2 x + 25°) = 3 6. Solve the following equations for 0 ≤ x < 2π . (i) tan 12 x = −1 (ii) sin ( 2 x − π6 ) = 0.5

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Core 2 7. Starting with the graph of y = x2, write down the equations of the curves obtained if the following transformations are applied to y = x2. (i) translation by 2 units in the positive x direction (ii) translation by 2 units in the negative y direction (iii) stretch by a scale factor ½ parallel to the y axis (iv) stretch by a scale factor of 3 parallel to the x axis ⎛1⎞ (v) translation by ⎜⎜ ⎟⎟ ⎝ 2⎠ 8. Find the equations of the curves obtained when the following transformations are applied to y = x3 + x2. (i) translation by 2 units in the positive x direction (ii) translation by 2 units parallel to Oy (iii) stretch scale factor 2 parallel to the y axis (iv) stretch scale factor 2 parallel to the x axis 9. State the transformation that must be applied to the graph of y = xn to obtain the graph of (i) y = xn – 4 (ii) y = (x – 3)n (iii) y = 2xn (iv) y = (x + 2)n – 3 (v) y = 3xn + 4 10. The graph of y = h(x) is shown below. y 2 1 2

4

x

On separate axes draw sketch graphs of the following: (i) y = h(x) + 1 (ii) y = 2 h(x) (iii) y = h(2x) (iv) y = h(x + 2) (v) y = h(x – 1)

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Core 2 11. For the following sketches, the graph of y = f(x) is shown as a solid line. Write down, in terms of f, the equation of the graph shown with a dotted line. y 4

(i)

2 x

y

(ii)

-2

-4 (iii)

2

x

4

y 4 3 2 2

4

x

6

12. The diagram below shows the graph of y = f(x) for -3 ≤ x ≤ 4. The graph has a maximum at (2, 1), an asymptote at x = -3 and passes through (4, 0). y

(2, 1) x -3

4

Sketch y = 2f(x) and y = f(2x) on separate axes.

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Core 2 Trigonometry Section 5: Transformations Solutions to Exercise 1. (i)

Stretch, parallel to the y-axis, scale factor 3. ⎛ −90° ⎞ Translation by ⎜ ⎟ ⎝ 0 ⎠

(ii)

(iii) Stretch, parallel to the x-axis, scale factor 4. ⎛ 0 ⎞ (iv) Translation by ⎜ ⎟ ⎝ −1 ⎠ ⎛ 180° ⎞ Translation by ⎜ ⎟ ⎝ 0 ⎠

(v)

2. (i)

Stretch, parallel to the y-axis, scale factor ⎛ 90 ° ⎞ (ii) Translation by ⎜ ⎟ ⎝ 0 ⎠ (iii) Stretch, parallel to the x-axis, scale factor 17 ⎛ 0 ⎞ 8 (iv) Translation by ⎜ ⎟ ⎝ −2 ⎠

1 2

1 3

θ

30° ⎞ ⎛ −15 Translation by ⎜ ⎟ ⎝ 0 ⎠

(v)

3.

2

y

1 x 90

180

270

360

−1

−2

© MEI, 04/04/07

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Core 2 4. y = tan x 4

y

3 2 1 −90

x 90

−1

180

270

360

−2 −3 −4

(i)

y = tan 2x 4

y

3 2 1 −90

x 90

−1

180

270

360

−2 −3 −4

(ii)

y = tan x + 1 4

y

3 2 1 −90

x 90

−1

180

270

360

−2 −3 −4

(iii) y = tan (x – 90°) 4

y

3 2 1 −90

−1

x 90

180

270

360

−2 −3 −4

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Core 2 5. (i) 2

y

y = cos 3 x 1

x −90

90

180

270

360

450

−1

−2

cos 3 x = 0.2 3 x = 78.5°, 281.5°, 438.5° , 641.5°, 798.5° , 1001.5°

x = 26.2°, 93.8°, 146.2°, 213.8°, 266.2°, 333.8° (ii) 2

y

y = sin 21 x 1

x −90

90

180

270

360

450

−1

−2

sin 21 x = 0.7 1 2

x = 44.4°, 135.6°

x = 88.8°, 171.2° (iii)

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Core 2 4

y = tan(2 x + 25°)

y

3

2

1 x −90

90

180

270

360

450

−1

−2

tan(2 x + 25°) = 3 2 x + 25° = 60°, 240°, 420°, 600 2 x = 35°, 215°, 395°, 575°

x = 17.5°, 107.5°, 197.5°, 287.5°

6. (i) 2

y

1

x −¤/2

¤/2

¤

3¤/2

2¤

5¤/2

−1

−2

tan 21 x = −1 3π 1 2 x = 4 3π x= 2

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Core 2 (ii) 2

y

1

x −¤/2

¤/2

¤

3¤/2

2¤

5¤/2

−1

−2

sin ( 2 x − π6 ) = 0.5 2 x − π6 = π6 ,

5π 6

,

2 x = π3 , π ,

7π 3

, 3π

x = π6 , 7. (i)

π

2

,

7π 6

,

13π 6

,

17 π 6

3π 2

y = ( x − 2)2 y = x 2 − 4x + 4

(ii)

y = x2 −2

(iii) y = 21 x 2 (iv)

y = ( 31 x )2 y=

(v)

1 9

x2

y = ( x − 1)2 + 2 y = x2 − 2x + 3

8. (i)

y = ( x − 2)3 + ( x − 2)2 y = ( x − 2)2( x − 2 + 1) y = ( x 2 − 4 x + 4)( x − 1) y = x3 − 5 x2 +8x − 4

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Core 2 (ii)

y = x3 + x2 + 2

(iii) y = 2( x 3 + x 2 )

y = 2x3 + 2x2 (iv)

y = ( 21 x )3 + ( 21 x )2 y = 81 x 3 + 41 x 2

9. (i) Translation by 4 units in the negative y direction (ii) Translation by 3 units in the positive x direction (iii) Stretch scale factor 2 parallel to the y axis ⎛ −2 ⎞ (iv) Translation of ⎜ ⎟ ⎝ −3 ⎠ (v) Stretch scale factor 3 parallel to the y axis followed by translation by 4 units in the positive y direction. 10. (i)

y

(ii)

y

4

3 2

2

1 2

4

x

(iii)

2 (iv)

y

y 2

2 1

1

x

1 2 (v)

x

4

-2

2

x

y 2 1 1

3

5

x

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Core 2 11. (i) y = 21 f( x ) (ii) y = f( 21 x ) (iii) y = 43 f( x )

y = 2f( x )

12.

y = f(2 x )

y

y (2, 2) (1, 1)

-3

4

x −

3 2

© MEI, 04/04/07

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x

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MEI Core 2 Trigonometry Section 5: Transformations Multiple Choice Test Do not use a graphics calculator or computer graph drawing package for this test. 1) What is the equation of this curve? y 3

2

1 x −360

−270

−180

−90

90

(a) y = 3 + sin x (c) y = 2 + sin x (e) I don’t know

180

270

360

(b) y = 1 + sin x (d) y = 2 sin x

2) What is the equation of this curve? 2

y

1 x −180

−90

90

180

−1

−2

(a) y = 3 + tan x (c) y = 3 tan x

(b) y = tan 3 x (d) y = tan 13 x

(e) I don’t know

© MEI, 12/01/09

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MEI C2 Trigonometry Section 5 MC test 3) What is the equation of this curve? 2

y

1 x −2¤

−¤

¤

2¤

−1

−2

(a) y = 2 + cos x (c) y = 2 cos x

(b) y = cos 2 x (d) y = cos 12 x

(e) I don’t know

4) Which of the following equations y = sin( x + π2 ) (i)

y = sin( x − π2 ) (ii) (iii) y = cos( x + π ) y = cos( x + π2 ) (iv) correctly describe the curve below? 2

y

1 x −2¤

−¤

¤

2¤

−1

−2

(a) (i) and (iii) (c) (ii) and (iv) (e) I don’t know

(b) (ii) and (iii) (d) (iii) only

5) The graph of y = cos 12 x is obtained from the graph of y = cos x by: (a) a one-way stretch, parallel to the x axis, scale factor 2 (b) a one-way stretch, parallel to the x axis, scale factor 12 (c) a one-way stretch, parallel to the y axis, scale factor 2 (d) a one-way stretch, parallel to the y axis, scale factor 12 (e) I don’t know

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MEI C2 Trigonometry Section 5 MC test 6) Which of the following graphs is the same as the graph of y = cos( x − 270°) ? (i) y = cos( x + 90°) (ii) y = sin( x + 270°) (iii) y = sin( x − 90°) (iv) y = sin( x − 180°) (a) (ii) and (iii) (c) (i) and (iv) (e) I don’t know

(b) (i) only (d) (i) and (iii)

Questions 7 – 10 relate to the following curves: y

2

P

1 x

−360

−270

−180

−90

90

180

270

360

−1

−2

2

Q

y

1 x

−360

−270

−180

−90

90

180

270

360

−1

−2

2

R

y

1 x

−360

−270

−180

−90

90

180

270

360

−1

−2

© MEI, 12/01/09

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MEI C2 Trigonometry Section 5 MC test 2

S

y

1 x

−360

−270

−180

−90

90

180

270

360

−1

−2

7) Which curve has the equation y = sin x − 1 ? (a) P (c) R (e) I don’t know

(b) Q (d) S

8) Which curve has the equation y = sin 12 x ? (a) P (c) R (e) I don’t know

(b) Q (d) S

9) Which curve has the equation y = sin( x + 90°) ? (a) P (c) R (e) I don’t know

(b) Q (d) S

10) Which curve has the equation y = 12 sin x ? (a) P (c) R (e) I don’t know

(b) Q (d) S

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MEI Core 2 Trigonometry Chapter Assessment 1. Find the angle θ and the length x in the triangle shown below. B 52° 12 cm x cm A 15 cm

θ C [7]

2. Andrew walks 5 km on a bearing of 140°, and then walks 3 km on a bearing of 025°. (i) How far is Andrew from his starting point? [3] (ii) On what bearing should Andrew walk to get back to his starting point? [4]

3. A belt is wrapped around a cylinder of radius 2.5 m as shown.

5m

Find the length of the belt.

[5]

4. Find the perimeter and area of the shaded sections of these shapes. (i) 3 cm 60°

[7] (ii) 8 cm 45°

[7] 10 cm

© MEI, 12/08/08

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MEI C2 Trigonometry Assessment 5. Solve these equations for 0° ≤ θ ≤ 360° (i) cos θ = 0.5 (ii) sin θ = −0.5 (iii) tan θ = 2

[6]

6. Solve these equations for 0 ≤ θ ≤ 2π . Give your answers as a multiple of π . 3 (i) cos θ = 2 (ii) sin θ = 0.5 (iii) tan θ = 3

[6]

7. Solve these equations for 0 ≤ θ ≤ 2π . Give your answers as a multiple of π . 3 (i) cos 2 θ = 4 2 (ii) 3 tan θ = 1

[6]

8. Solve these equations for 0° ≤ θ ≤ 360° (i) sin 2 x = − 12 3 (ii) cos 12 x = 0.3 (iii) tan 3x = 0.5

[9]

9. Solve these equations for 0° ≤ θ ≤ 360° (i) cos 2 θ + sin θ = 1 (ii) 2sin θ cos θ + sin θ = 0 (iii) 3 sin θ = cos θ

[4] [4] [3]

Total 70 marks

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