© boardworks ltd 2005 1 of 58 s4 further trigonometry ks4 mathematics

© Boardworks Ltd 2005 1 of 58

S4 Further trigonometry

KS4 Mathematics


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S4.1 Sin, cos and tan of any angle


Contents

S3.4 Area of a triangle using ½ab sin C

S3.5 The sine rule

S4.3 Graphs of trigonometric functions

S4.2 Sin, cos and tan of 30°, 45° and 60°

S4.6 The cosine rule


The opposite and adjacent sides

Suppose we have a right-angled triangle with hypotenuse h and acute angle θ.

θ

h

a) Write an expression for the length of the opposite side in terms of h and θ.

b) Write an expression for the length of the adjacent side in terms of h and θ.



Suppose we have a right-angled triangle with hypotenuse h and acute angle θ.

θ

h

a) sin θ =opphyp

opp = hyp × sin θ

opp = h sin θ

b) cos θ =adjhyp

adj = hyp × cos θ

adj = h cos θ



So, for any right-angled triangle with hypotenuse h and acute angle θ. We can label the opposite and adjacent sides as follows:

θ

hh sin θ

h cos θ

We can write, tan θ =h sin θ h cos θ

tan θ =sin θ cos θ

opposite

adjacent


The sine of any angle


Sine of angles in the second quadrant

We have seen that the sine of angles in the first and second quadrants are positive.

The sine of angles in the third and fourth quadrants are negative.

In the second quadrant, 90° < θ < 180°.In the second quadrant, 90° < θ < 180°.

sin θ = sin (180° – θ)sin θ = sin (180° – θ)

For example,sin 130° = sin (180° – 130°)

= sin 50°

= 0.766 (to 3 sig. figs)


Sine of angles in the third quadrant

In the third quadrant, 180° < θ < 270°.In the third quadrant, 180° < θ < 270°.

sin θ = –sin (θ – 180°)sin θ = –sin (θ – 180°)

For example,

sin 220° = – sin (220° – 180°)

= – sin 40°

= – 0.643 (to 3 sig. figs)

Verify, using a scientific calculator, that sin 220° = –sin 40°


Sine of angles in the fourth quadrant

In the fourth quadrant, 270° < θ < 360° or 0° > θ > –90° In the fourth quadrant, 270° < θ < 360° or 0° > θ > –90°

sin θ = –sin(360° – θ) or sin –θ = –sin θ sin θ = –sin(360° – θ) or sin –θ = –sin θ

For example,sin 300° = –sin (360° – 300°)

= –sin 60°

= –0.866 (to 3 sig. figs)

sin –35° = –sin 35°

= –0.574 (to 3 sig. figs)


The cosine of any angle


Cosine of angles in the second quadrant

We have seen that the cosines of angles in the first and fourth quadrants are positive.

The cosines of angles in the second and third quadrants are negative.


cos θ = –cos (180° – θ)cos θ = –cos (180° – θ)

For example,cos 100° = –cos (180° – 100°)

= –cos 80°

= –0.174 (to 3 sig. figs)


Cosine of angles in the third quadrant


cos θ = –cos (θ – 180°)cos θ = –cos (θ – 180°)

For example,

cos 250° = –cos (250° – 180°)

= –cos 70°

= –0.342 (to 3 sig. figs.)

Verify, using a scientific calculator, that cos 250° = –cos 70°


Sine of angles in the fourth quadrant


cos θ = cos(360° – θ) or cos –θ = cos θ cos θ = cos(360° – θ) or cos –θ = cos θ

For example,cos 317° = cos (360° – 317°)

= cos 43°

= 0.731 (to 3 sig. figs.)

cos –28° = cos 28°

= 0.883 (to 3 sig. figs.)


The tangent of any angle


Tangent of angles in the second quadrant

We have seen that the tangent of angles in the first and third quadrants are positive.

The tangent of angles in the second and fourth quadrants are negative.


tan θ = –tan (180° – θ)tan θ = –tan (180° – θ)

For example,tan 116° = –tan (180° – 116°)

= –tan 64°

= –2.05 (to 3 sig. figs)


Tangent of angles in the third quadrant


tan θ = tan (θ – 180°)tan θ = tan (θ – 180°)

For example,

tan 236° = tan (236° – 180°)

= tan 56°

= 1.48 (to 3 sig. figs)

Verify, using a scientific calculator, that tan 236° = tan 56°


Tangent of angles in the fourth quadrant


tan θ = –tan(360° – θ) or tan –θ = –tan θ tan θ = –tan(360° – θ) or tan –θ = –tan θ

For example,tan 278° = –tan (360° – 278°)

= –tan 82°

= –7.12 (to 3 sig. figs)

tan –16° = –tan 16°

= –0.287 (to 3 sig. figs)


Sin, cos and tan of angles between 0° and 360°

The sin, cos and tan of angles in the first quadrant are positive.The sin, cos and tan of angles in the first quadrant are positive.

In the second quadrant:In the second quadrant: sin θ = sin (180° – θ)cos θ = –cos (180° – θ)tan θ = –tan (180° – θ)

sin θ = sin (180° – θ)cos θ = –cos (180° – θ)tan θ = –tan (180° – θ)

In the third quadrant:In the third quadrant: sin θ = –sin (θ – 180°)cos θ = –cos (θ – 180°)tan θ = tan (θ – 180°)

sin θ = –sin (θ – 180°)cos θ = –cos (θ – 180°)tan θ = tan (θ – 180°)

In the fourth quadrant:In the fourth quadrant: sin θ = –sin (360° – θ)cos θ = cos (360° – θ)tan θ = –tan(180° – θ)

sin θ = –sin (360° – θ)cos θ = cos (360° – θ)tan θ = –tan(180° – θ)


3rd quadrant

2nd quadrant 1st quadrant

4th quadrant

Tangent is positiveTT

Sine is positiveSS All are positiveAA

Remember CAST

We can use CAST to remember in which quadrant each of the three ratios are positive.

Cosine is positiveCC


Positive or negative?


Find the equivalent ratio


Solving equations in θ


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S3.5 The sine rule



Sin, cos and tan of 45°

A right-angled isosceles triangle has two acute angles of 45°.

45°

45°Suppose the equal sides are of 1 unit length.

1

1

Using Pythagoras’ theorem,

= 2

2

We can use this triangle to write exact values for sin, cos and tan 45°:

sin 45° =12

cos 45° =12

tan 45° = 1

The hypotenuse = 1² + 1²



Suppose we have an equilateral triangle of side length 2.

We can use this triangle to write exact values for sin, cos and tan 30°:

sin 30° =12

cos 30° =32

tan 30° =

2 2

2

60° 60°

60°

2

60°

30°

1

If we cut the triangle in half then we have a right-angled triangle with acute angles of 30° and 60°.


= 3

3

The height of the triangle = 2² – 1²

13



Suppose we have an equilateral triangle of side length 2.

2

60°

30°

1

If we cut the triangle in half then we have a right-angled triangle with acute angles of 30° and 60°.


= 3

3

The height of the triangle = 2² – 1²

We can also use this triangle to write exact values for sin, cos and tan 60°:

sin 60° = cos 60° =32

tan 60° =12

3


Sin, cos and tan of 30°, 45° and 60°

The exact values of the sine, cosine and tangent of 30°, 45° and 60° can be summarized as follows:

30°

sin

cos

tan

45° 60°

121

2

12

32

1

32

12

Use this table to write the exact value of sin 150°:

sin 150° =12

31

3




30°

sin

cos

tan

45° 60°

121

2

12

32

1

32

12

Use this table to write the exact value of cos 135°:

–1

2cos 135° =

31

3




30°

sin

cos

tan

45° 60°

121

2

12

32

311

3

32

12

Use this table to write the exact value of tan 120°

tan 120° = –3



Write the following ratios exactly:

1) cos 300° =12

3) tan 240° =

5) cos –30° =

7) sin 210° =

2) tan 315° =

4) sin –330° =

6) tan –135° =

8) cos 315° =

–1

312

32

1

–12

1

2


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S3.5 The sine rule





The graph of sin θ


The graph of cos θ


The graph of tan θ


Transforming trigonometric graphs


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S3.5 The sine rule






The area of a triangle

Remember,Remember,

b

h

Area of a triangle = bh12


The area of a triangle

Suppose that instead of the height of a triangle, we are given the base, one of the sides and the included angle. For example,

What is the area of triangle ABC?

A

B C7 cm

4 cm

47°

Let’s call the height of the triangle h.

hWe can find h using the sine ratio.

h4

= sin 47°

h = 4 sin 47°

Area of triangle ABC = ½ × base × height

= ½ × 7 × 4 sin 47°

= 10.2 cm2 (to 1 d.p.)


The area of a triangle using ½ ab sin C

The area of a triangle is equal to half the product of two of the sides and the sine of the included angle.The area of a triangle is equal to half the product of two of the sides and the sine of the included angle.

A

B C

c

a

b

Area of triangle ABC = ab sin C12


The area of a triangle using ½ ab sin C


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S3.5 The sine rule

Contents








The sine rule

Consider any triangle ABC,

C

A B

b a

If we drop a perpendicular line, h from C to AB, we can divide the triangle into two right-angled triangles, ACD and BDC.

D

sin A =

h

a is the side opposite A and b is the side opposite B.

hb

h = b sin A

sin B =ha

h = a sin B

b sin A = a sin Bb sin A = a sin BSo,


The sine rule

b sin A = a sin Bb sin A = a sin B

Dividing both sides of the equation by sin A and then by sin Bwe have:

bsin B

=a

sin A

If we had dropped a perpendicular from A to BC we would have found that:

b sin C = c sin Bb sin C = c sin B

Rearranging:

bsin B

=c

sin C


The sine rule

For any triangle ABC,For any triangle ABC,

C

A B

b

c

a

asin A

=b

sin B=

csin C

orsin A sin B sin C

a=

b=

c


Using the sine rule to find side lengths

If we are given two angles in a triangle and the length of a side opposite one of the angles, we can use the sine rule to find the length of the side opposite the other angle. For example,

Find the length of side a

Using the sine rule,

asin 118°

=7

sin 39°

a =7 sin 118°

sin 39°

a = 9.82 (to 2 d.p.)

a

7 cm

118°

39°

A

B

C


Using the sine rule to find side lengths


Using the sine rule to find angles

If we are given two side lengths in a triangle and the angle opposite one of the given sides, we can use the sine rule to find the angle opposite the other given side. For example,

Find the angle at B

Using the sine rule,

sin B8

=6

sin 46°

sin B =8 sin 46°

6

B = 73.56° (to 2 d.p.)

sin–1B =8 sin 46°

6

6 cm

46°B

8 cm

A

C


Finding the second possible value

Suppose that in the last example we had not been given a diagram but had only been told that AC = 8 cm, CB = 6 cm and that the angle at A = 46°.

6 cm

46°B

8 cm

A

C

There is a second possible value for the angle at B.Instead of this triangle …

… we could have this triangle.

Remember, sin θ = sin (180° – θ)

So for every acute solution, there is a corresponding obtuse solution.

B = 73.56° (to 2 d.p.)

or

B = 180° – 73.56°

= 106.44° (to 2 d.p.)

46°

6 cm

B


Using the sine rule to find angles


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AS4.6 The cosine rule

Contents

S3.5 The sine rule







The cosine rule

Consider any triangle ABC.

C

A B

b a

If we drop a perpendicular line, h from C to AB, we can divide the triangle into two right-angled triangles, ACD and BDC.

D

h

a is the side opposite A and b is the side opposite B.

c is the side opposite C. If we call the length AD x, then the length BD can be written as c – x.

c – xx


The cosine rule

C

A B

b a

Using Pythagoras’ theorem in triangle ACD,

D

hb2 = x2 + h2

c – xx

Also,

cos A = xb

In triangle BCD,x = b cos A

a2 = (c – x)2 + h2

a2 = c2 – 2cx + x2 + h2

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Substituting and , 1 2

a2 = c2 – 2cb cos A + b2

a2 = b2 + c2 – 2bc cos Aa2 = b2 + c2 – 2bc cos A

a2 = c2 – 2cx + x2 + h2

This is the cosine rule.


The cosine rule

For any triangle ABC,For any triangle ABC,

A

B C

c

a

b


or

cos A = b2 + c2 – a2

2bc


Using the cosine rule to find side lengths

If we are given the length of two sides in a triangle and the size of the angle between them, we can use the cosine rule to find the length of the other side. For example,

Find the length of side a.

B

C A7 cm

4 cm

48°

a


a2 = 72 + 42 – 2 × 7 × 4 × cos 48°

a2 = 27.53 (to 2 d.p.)

a = 5.25 cm (to 2 d.p.)


Using the cosine rule to find side lengths


Using the cosine rule to find angles

If we are given the lengths of all three sides in a triangle, we can use the cosine rule to find the size of any one of the angles in the triangle. For example,

Find the size of the angle at A.

4 cm

8 cm6 cm

A

B

C

cos A = b2 + c2 – a2

2bc

cos A = 42 + 62 – 82

2 × 4 × 6

cos A = –0.25

A = cos–1 –0.25

A = 104.48° (to 2 d.p.)

This is negative so A must be obtuse.


Using the cosine rule to find angles

© boardworks ltd 2005 1 of 58 s4 further trigonometry ks4 mathematics

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