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© Boardworks Ltd 2005 2 of 58
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S4.1 Sin, cos and tan of any angle
S4 Further trigonometry
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
© Boardworks Ltd 2005 3 of 58
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.
© Boardworks Ltd 2005 4 of 58
The opposite and adjacent sides
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
© Boardworks Ltd 2005 5 of 58
The opposite and adjacent sides
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
© Boardworks Ltd 2005 7 of 58
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)
© Boardworks Ltd 2005 8 of 58
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°
© Boardworks Ltd 2005 9 of 58
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)
© Boardworks Ltd 2005 11 of 58
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.
In the second quadrant, 90° < θ < 180°.In the second quadrant, 90° < θ < 180°.
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)
© Boardworks Ltd 2005 12 of 58
Cosine of angles in the third quadrant
In the third quadrant, 180° < θ < 270°.In the third quadrant, 180° < θ < 270°.
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°
© Boardworks Ltd 2005 13 of 58
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°
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.)
© Boardworks Ltd 2005 16 of 58
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.
In the second quadrant, 90° < θ < 180°.In the second quadrant, 90° < θ < 180°.
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)
© Boardworks Ltd 2005 17 of 58
Tangent of angles in the third quadrant
In the third quadrant, 180° < θ < 270°.In the third quadrant, 180° < θ < 270°.
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°
© Boardworks Ltd 2005 18 of 58
Tangent 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°
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)
© Boardworks Ltd 2005 19 of 58
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)
© Boardworks Ltd 2005 20 of 58
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
© Boardworks Ltd 2005 24 of 58
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S4.2 Sin, cos and tan of 30°, 45° and 60°
Contents
S4 Further trigonometry
S4.3 Graphs of trigonometric functions
S3.4 Area of a triangle using ½ab sin C
S4.6 The cosine rule
S3.5 The sine rule
S4.1 Sin, cos and tan of any angle
© Boardworks Ltd 2005 25 of 58
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
© Boardworks Ltd 2005 26 of 58
Sin, cos and tan of 30°
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°.
Using Pythagoras’ theorem,
= 3
3
The height of the triangle = 2² – 1²
13
© Boardworks Ltd 2005 27 of 58
Sin, cos and tan of 60°
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°.
Using Pythagoras’ theorem,
= 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
© Boardworks Ltd 2005 28 of 58
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
© Boardworks Ltd 2005 29 of 58
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 cos 135°:
–1
2cos 135° =
31
3
© Boardworks Ltd 2005 30 of 58
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
311
3
32
12
Use this table to write the exact value of tan 120°
tan 120° = –3
© Boardworks Ltd 2005 31 of 58
Sin, cos and tan of 30°, 45° and 60°
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
© Boardworks Ltd 2005 32 of 58
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S4.3 Graphs of trigonometric functions
Contents
S3.4 Area of a triangle using ½ab sin C
S4.6 The cosine rule
S3.5 The sine rule
S4 Further trigonometry
S4.1 Sin, cos and tan of any angle
S4.2 Sin, cos and tan of 30°, 45° and 60°
© Boardworks Ltd 2005 37 of 58
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S3.4 Area of a triangle using ½ab sin C
Contents
S4.6 The cosine rule
S3.5 The sine rule
S4 Further trigonometry
S4.1 Sin, cos and tan of any angle
S4.3 Graphs of trigonometric functions
S4.2 Sin, cos and tan of 30°, 45° and 60°
© Boardworks Ltd 2005 38 of 58
The area of a triangle
Remember,Remember,
b
h
Area of a triangle = bh12
© Boardworks Ltd 2005 39 of 58
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.)
© Boardworks Ltd 2005 40 of 58
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
© Boardworks Ltd 2005 42 of 58
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S3.5 The sine rule
Contents
S4.6 The cosine rule
S3.4 Area of a triangle using ½ab sin C
S4 Further trigonometry
S4.1 Sin, cos and tan of any angle
S4.3 Graphs of trigonometric functions
S4.2 Sin, cos and tan of 30°, 45° and 60°
© Boardworks Ltd 2005 43 of 58
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,
© Boardworks Ltd 2005 44 of 58
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
© Boardworks Ltd 2005 45 of 58
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
© Boardworks Ltd 2005 46 of 58
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
© Boardworks Ltd 2005 48 of 58
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
© Boardworks Ltd 2005 49 of 58
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
© Boardworks Ltd 2005 51 of 58
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AS4.6 The cosine rule
Contents
S3.5 The sine rule
S3.4 Area of a triangle using ½ab sin C
S4 Further trigonometry
S4.1 Sin, cos and tan of any angle
S4.3 Graphs of trigonometric functions
S4.2 Sin, cos and tan of 30°, 45° and 60°
© Boardworks Ltd 2005 52 of 58
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
© Boardworks Ltd 2005 53 of 58
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
1
2
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.
© Boardworks Ltd 2005 54 of 58
The cosine rule
For any triangle ABC,For any triangle ABC,
A
B C
c
a
b
a2 = b2 + c2 – 2bc cos Aa2 = b2 + c2 – 2bc cos A
or
cos A = b2 + c2 – a2
2bc
© Boardworks Ltd 2005 55 of 58
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 = b2 + c2 – 2bc cos Aa2 = b2 + c2 – 2bc cos A
a2 = 72 + 42 – 2 × 7 × 4 × cos 48°
a2 = 27.53 (to 2 d.p.)
a = 5.25 cm (to 2 d.p.)
© Boardworks Ltd 2005 57 of 58
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.