Transcript S4 Further trigonometry

Maths Age 14-16
S4 Further trigonometry
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Contents
S4 Further trigonometry
A S4.1 Sin, cos and tan of any angle
A S4.2 Sin, cos and tan of 30°, 45° and 60°
A S4.3 Graphs of trigonometric functions
A S3.4 Area of a triangle using ½ab sin C
A S3.5 The sine rule
A S4.6 The cosine rule
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The opposite and adjacent sides
Suppose we have a right-angled triangle with hypotenuse
h and acute angle θ.
h
θ
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.
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The opposite and adjacent sides
Suppose we have a right-angled triangle with hypotenuse
h and acute angle θ.
h
θ
a)
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opp
sin θ =
hyp
b)
adj
cos θ =
hyp
opp = hyp × sin θ
adj = hyp × cos θ
opp = h sin θ
adj = h cos θ
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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:
h
h sin θ
θ
h cos θ
We can write,
h sin θ
tan θ =
h cos θ
opposite
adjacent
sin θ
tan θ =
cos θ
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The sine of any angle
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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°.
sin θ = sin (180° – θ)
For example,
sin 130° = sin (180° – 130°)
= sin 50°
= 0.766 (to 3 sig. figs)
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Sine of angles in the third quadrant
In the third quadrant, 180° < θ < 270°.
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°
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Sine of angles in the fourth quadrant
In the fourth quadrant, 270° < θ < 360° or 0° > θ > –90°
sin θ = –sin(360° – θ)
For example,
or
sin –θ = –sin θ
sin 300° = –sin (360° – 300°)
= –sin 60°
= –0.866 (to 3 sig. figs)
sin –35° = –sin 35°
= –0.574 (to 3 sig. figs)
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The cosine of any angle
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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°.
cos θ = –cos (180° – θ)
For example,
cos 100° = –cos (180° – 100°)
= –cos 80°
= –0.174 (to 3 sig. figs)
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Cosine of angles in the third quadrant
In the third quadrant, 180° < θ < 270°.
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°
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Sine of angles in the fourth quadrant
In the fourth quadrant, 270° < θ < 360° or 0° > θ > –90°
cos θ = cos(360° – θ)
For example,
or
cos –θ = cos θ
cos 317° = cos (360° – 317°)
= cos 43°
= 0.731 (to 3 sig. figs.)
cos –28° = cos 28°
= 0.883 (to 3 sig. figs.)
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The tangent of any angle
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The tangent of any angle
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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°.
tan θ = –tan (180° – θ)
For example,
tan 116° = –tan (180° – 116°)
= –tan 64°
= –2.05 (to 3 sig. figs)
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Tangent of angles in the third quadrant
In the third quadrant, 180° < θ < 270°.
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°
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Tangent of angles in the fourth quadrant
In the fourth quadrant, 270° < θ < 360° or 0° > θ > –90°
tan θ = –tan(360° – θ)
For example,
or
tan –θ = –tan θ
tan 278° = –tan (360° – 278°)
= –tan 82°
= –7.12 (to 3 sig. figs)
tan –16° = –tan 16°
= –0.287 (to 3 sig. figs)
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Sin, cos and tan of angles between 0° and 360°
The sin, cos and tan of angles in the first quadrant are positive.
In the second quadrant:
sin θ = sin (180° – θ)
cos θ = –cos (180° – θ)
tan θ = –tan (180° – θ)
In the third quadrant:
sin θ = –sin (θ – 180°)
cos θ = –cos (θ – 180°)
tan θ = tan (θ – 180°)
In the fourth quadrant:
sin θ = –sin (360° – θ)
cos θ = cos (360° – θ)
tan θ = –tan(180° – θ)
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Remember CAST
We can use CAST to remember in which quadrant each of
the three ratios are positive.
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2nd quadrant
1st quadrant
S
Sine is positive
A
All are positive
3rd quadrant
4th quadrant
T
Tangent is positive
C
Cosine is positive
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Positive or negative?
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Find the equivalent ratio
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Solving equations in θ
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Contents
S4 Further trigonometry
A S4.1 Sin, cos and tan of any angle
A S4.2 Sin, cos and tan of 30°, 45° and 60°
A S4.3 Graphs of trigonometric functions
A S3.4 Area of a triangle using ½ab sin C
A S3.5 The sine rule
A S4.6 The cosine rule
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Sin, cos and tan of 45°
A right-angled isosceles triangle has two acute angles of 45°.
Suppose the equal sides are of 1
unit length.
45°
2
1
Using Pythagoras’ theorem,
The hypotenuse =  1² + 1²
45°
= 2
1
We can use this triangle to write exact values for sin, cos and
tan 45°:
1
sin 45° =
2
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1
cos 45° =
2
tan 45° = 1
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Sin, cos and tan of 30°
Suppose we have an equilateral triangle of side length 2.
60°
30°
2
3
60°
1
2
2
If we cut the triangle in half then we have
a right-angled triangle with acute angles
of 30° and 60°.
Using Pythagoras’ theorem,
60° The height of the triangle =  2² – 1²
= 3
We can use this triangle to write exact values for sin, cos and
tan 30°:
1
sin 30° =
2
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3
cos 30° =
2
1
tan 30° =
3
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Sin, cos and tan of 60°
Suppose we have an equilateral triangle of side length 2.
If we cut the triangle in half then we have
a right-angled triangle with acute angles
of 30° and 60°.
30°
2
3
60°
1
Using Pythagoras’ theorem,
The height of the triangle =  2² – 1²
= 3
We can also use this triangle to write exact values for sin, cos
and tan 60°:
3
sin 60° =
2
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1
cos 60° =
2
tan 60° = 3
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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:
sin
cos
tan
30°
45°
60°
1
2
3
2
1
3
1
2
1
2
3
2
1
2
1
3
Use this table to write the exact value of sin 150°:
1
sin 150° = 2
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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:
sin
cos
tan
30°
45°
60°
1
2
3
2
1
3
1
2
1
2
3
2
1
2
1
3
Use this table to write the exact value of cos 135°:
–1
cos 135° = 2
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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:
sin
cos
tan
30°
45°
60°
1
2
3
2
1
3
1
2
1
2
3
2
1
2
1
3
Use this table to write the exact value of tan 120°
tan 120° = –3
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Sin, cos and tan of 30°, 45° and 60°
Write the following ratios exactly:
1) cos 300° =
1
2
2) tan 315° =
3) tan 240° =
3
4) sin –330° =
1
2
5) cos –30° =
3
2
6) tan –135° =
1
7) sin 210° =
–1
2
8) cos 315° =
1
2
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–1
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Contents
S4 Further trigonometry
A S4.1 Sin, cos and tan of any angle
A S4.2 Sin, cos and tan of 30°, 45° and 60°
A S4.3 Graphs of trigonometric functions
A S3.4 Area of a triangle using ½ab sin C
A S3.5 The sine rule
A S4.6 The cosine rule
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The graph of sin θ
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The graph of cos θ
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The graph of tan θ
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Transforming trigonometric graphs
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Contents
S4 Further trigonometry
A S4.1 Sin, cos and tan of any angle
A S4.2 Sin, cos and tan of 30°, 45° and 60°
A S4.3 Graphs of trigonometric functions
A S3.4 Area of a triangle using ½ab sin C
A S3.5 The sine rule
A S4.6 The cosine rule
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The area of a triangle
Remember,
h
b
Area of a triangle =
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1
bh
2
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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
h
B
7 cm
Let’s call the height of the
triangle h.
4 cm
We can find h using the sine ratio.
h
= sin 47°
47° C
4
h = 4 sin 47°
Area of triangle ABC = ½ × base × height
= ½ × 7 × 4 sin 47°
= 10.2 cm2 (to 1 d.p.)
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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.
A
c
B
b
a
Area of triangle ABC =
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C
1
ab sin C
2
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The area of a triangle using ½ ab sin C
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Contents
S4 Further trigonometry
A S4.1 Sin, cos and tan of any angle
A S4.2 Sin, cos and tan of 30°, 45° and 60°
A S4.3 Graphs of trigonometric functions
A S3.4 Area of a triangle using ½ab sin C
A S3.5 The sine rule
A S4.6 The cosine rule
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The sine rule
Consider any triangle ABC,
If we drop a perpendicular
line, h from C to AB, we
can divide the triangle into
two right-angled triangles,
ACD and BDC.
C
b
A
h
D
h
sin A =
b
h = b sin A
So,
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a
B
a is the side opposite A and
b is the side opposite B.
h
a
h = a sin B
sin B =
b sin A = a sin B
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The sine rule
b sin A = a sin B
Dividing both sides of the equation by sin A and then by sin B
we have:
a
b
=
sin A
sin B
If we had dropped a perpendicular from A to BC we would have
found that:
b sin C = c sin B
Rearranging:
c
b
=
sin C
sin B
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The sine rule
For any triangle ABC,
C
b
A
b
c
a
=
=
sin B
sin C
sin A
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a
c
or
B
sin A
sin B
sin C
=
=
a
b
c
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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
B
39°
a
Using the sine rule,
7
a
=
sin 39°
sin 118°
118°
C
7 cm
A
7 sin 118°
a =
sin 39°
a = 9.82 (to 2 d.p.)
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Using the sine rule to find side lengths
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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
C
8 cm
A
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46°
6 cm
B
Using the sine rule,
sin 46°
sin B
=
8
6
8 sin 46°
sin B =
6
8 sin 46°
–1
B = sin
6
B = 73.56° (to 2 d.p.)
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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°.
There is a second possible value for the angle at B.
Instead of this triangle …
… we could have this triangle.
C
Remember, sin θ = sin (180° – θ)
So for every acute solution, there
is a corresponding obtuse solution.
8 cm
6 cm
B = 73.56° (to 2 d.p.)
6 cm
or
46°
B = 180° – 73.56°
A
B
B
= 106.44° (to 2 d.p.)
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Using the sine rule to find angles
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Contents
S4 Further trigonometry
A S4.1 Sin, cos and tan of any angle
A S4.2 Sin, cos and tan of 30°, 45° and 60°
A S4.3 Graphs of trigonometric functions
A S3.4 Area of a triangle using ½ab sin C
A S3.5 The sine rule
A S4.6 The cosine rule
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The cosine rule
Consider any triangle ABC.
If we drop a perpendicular
line, h from C to AB, we
can divide the triangle into
two right-angled triangles,
ACD and BDC.
C
b
A
h
x
a
D c–x
B
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.
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The cosine rule
Using Pythagoras’ theorem
in triangle ACD,
C
b
A
a
h
D c–x
x
In triangle BCD,
Substituting
1
and
b2 = x2 + h2
1
Also,
B
x
cos A =
b
x = b cos A
2
a2 = (c – x)2 + h2
a2 = c2 – 2cx + x2 + h2
,
a2 = c2 – 2cb cos A + b2
2
This is the
cosine rule.
a2 = b2 + c2 – 2bc cos A
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The cosine rule
For any triangle ABC,
a2 = b2 + c2 – 2bc cos A
A
c
B
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or
b
a
C
cos A =
b2 + c2 – a2
2bc
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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
a
C
a2 = b2 + c2 – 2bc cos A
4 cm
48°
7 cm
a2 = 72 + 42 – 2 × 7 × 4 × cos 48°
A
a2 = 27.53 (to 2 d.p.)
a = 5.25 cm (to 2 d.p.)
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Using the cosine rule to find side lengths
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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.
B
8 cm
6 cm
cos A =
b2 + c2 – a2
2bc
cos A =
42 + 62 – 82
2×4×6
cos A = –0.25
C
4 cm
A
This is negative so
A must be obtuse.
A = cos–1 –0.25
A = 104.48° (to 2 d.p.)
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Using the cosine rule to find angles
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