CHAPTER 44 COMPOUND ANGLES
EXERCISE 182 Page 488
1. Reduce the following to the sine of one angle:
(a) sin 37° cos 21° + cos 37° sin 21°
(b) sin 7t cos 3t – cos 7t sin 3t
(a) sin(A + B) = sin A cos B + cos A sin B
Hence, sin 37° cos 21° + cos 37° sin 21° = sin(37° + 21°) = sin 58°
(b) sin(A + B) = sin A cos B + cos A sin B
Hence, sin 7t cos 3t – cos 7t sin 3t = sin(7t – 3t) = sin 4t
2. Reduce the following to the cosine of one angle:
(a) cos 71° cos 33° – sin 71° sin 33°
(b) cos
π
3
cos
π
4
+ sin
π
3
sin
π
4
(a) cos(A + B) = cos A cos B – sin A sin B
Hence, cos 71° cos 33° – sin 71° sin 33° = sin(71° + 33°) = cos 104°
(b) cos(A – B) = cos A cos B + sin A sin B
Hence, cos
3. Show that:
and
π
3
cos
π
4
+ sin
(a) sin (x +
(b) – sin(
π
3
π
3
sin
π π 
π 
= cos  −  = cos  
4
3 4
 12 
π
) + sin (x +
2π
) = 3 cos x
3
3π
– φ) = cos φ
2
2π 
2π
2π
π
π
π


(a) L.H.S. = sin  x +  + sin  x +=
+ cos x sin
 sin x cos + cos x sin + sin x cos
3
3 
3
3
3
3


729
© 2014, John Bird
 3
 3
1
 1
= sin x   + cos x 
 + sin x  −  + cos x 


2
 2
 2 
 2 
 3
= 2
=
 2  cos x


3 cos x = R.H.S.
The diagram below shows an equilateral triangle ABC of side 2 with each angle 60°. Angle A is
bisected. By Pythagoras, AD = 22 − 12 =3 . Hence, sin
π
3
= sin 60° =
3
1
and cos 60° =
2
2
3π
 3π

 3π

(b) L.H.S. = − sin 
−φ  =
− sin
cos φ − cos sin φ 
2
2
 2



= − [ (−1) cos φ − (0) sin φ ] = cos φ = R.H.S.
4. Prove that: (a) sin(θ +
and
(b)
π
4
) – sin(θ –
3π
)=
4
2 (sin θ + cos θ)
cos(270° + θ )
= tan θ
cos(360° − θ )
π
3π  
π
π 
3π
3π 


(a) L.H.S. = sin  θ +  − sin  θ=
−
− cos θ sin
  sin θ cos + cos θ sin  −  sin θ cos

4
4  
4
4 
4
4 



 1 
 1  
 1 
 1 
= sin θ 
 + cos θ 
  − sin θ  −
 − cos θ 

2
 2
 2  

 2 

=
1
θ]
[sin θ + cos θ + sin θ + cos=
2
=
2 (sin θ + cos θ ) = R.H.S.
2
( sin θ + cos θ )
2
The diagram below shows an isosceles triangle where AB = BC = 1 and angles A and C are both
45°. By Pythagoras, AC = 12 + 12 =2 . Hence, sin
730
π
4
= sin 45° = cos 45°=
1
2
© 2014, John Bird
cos ( 270° + θ ) cos 270° cos θ − sin 270° sin θ 0 − (−1) sin θ
(b) L.H.S. =
=
=
cos ( 360° − θ ) cos 360° cos θ + sin 360° sin θ
(1) cos θ + 0
=
sin θ
= tan θ = R.H.S.
cos θ
5. Given cos A = 0.42 and sin B = 0.73 evaluate:
(a) sin (A – B), (b) cos (A – B), (c) tan (A + B), correct to 4 decimal places.
Since cos A = 0.42 then A = cos −1 0.42 = 65.17°
Thus sin A = sin 65.17° = 0.9075 and tan A = tan 65.17° = 2.1612
Since sin B = 0.73, B = sin −1 0.73 = 46.89°
Thus cos B = cos 46.89° = 0.6834 and tan B = tan 46.89° = 1.0682
(a) sin(A – B) = sin A cos B – cos A sin B
= (0.9075)(0.6834) – (0.42)(0.73) = 0.6202 – 0.3066 = 0.3136
(b) cos(A – B) = cos A cos B + sin A sin B
= (0.42)(0.6834) + (0.9075)(0.73) = 0.2870 + 0.6625 = 0.9495
(c) tan(A + B) =
(2.1612) + (1.0682)
tan A + tan B
3.2294
=
=
= –2.4678
−1.3086
1 − tan A tan B 1 − (2.1612)(1.0682)
6. Solve for values of θ between 0° and 360°: 3 sin(θ + 30°) = 7 cos θ
3 sin(θ + 30°) = 3[sin θ cos 30° + cos θ sin 30°]
from the formula for sin (A + B)
= 3[sin θ (0.8660) + cos θ (0.50)]
= 2.5980 sin θ + 1.50 cos θ
Since 3 sin(θ + 30°) = 7 cos θ then 2.5980 sin θ + 1.50 cos θ = 7 cos θ
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© 2014, John Bird
Rearranging gives: 2.5980 sin θ = 7 cos θ – 1.50 cos θ = 5.50 cos θ
sin θ
5.50
=
= 2.1170
cos θ
2.5980
and
tan θ = 2.1170 and θ = tan −1 2.1170 = 64.72° or 244.72° since tangent is positive in the 1st
i.e.
and 3rd quadrants
Hence, the solution of 3 sin(θ + 30°) = 7 cos θ for values of θ between 0° and 360° are:
θ = 64.72° or 244.72°
7. Solve for values of θ between 0° and 360°: 4 sin(θ – 40°) = 2 sin θ
4 sin(θ – 40°) = 2 sin θ
i.e.
4[sin θ cos 40° – cos θ sin 40°] = 2 sin θ
i.e.
3.064178 sin θ – 2.57115 cos θ = 2 sin θ
Hence,
1.064178 sin θ = 2.57115 cos θ
sin θ
2.57115
= = 2.4160901
cos θ 1.064178
i.e.
tan θ = 2.4160901
and
θ = tan −1 (2.4160901) = 67°31′ and 247°31′ (see diagram below)
or θ = 67.52° and 247.52°
732
© 2014, John Bird
EXERCISE 183 Page 492
1. Change 5 sin ωt + 8 cos ωt into the form R sin(ωt ± α)
Let 5 sin ωt + 8 cos ωt = R sin(ωt + α)
Then 5 sin ωt + 8 cos ωt = R[sin ωt cos α + cos ωt sin α]
= (R cos α) sin ωt + (R sin α) cos ωt
Equating coefficients of sin ωt gives: 5 = R cos α, from which, cos α =
5
R
Equating coefficients of cos ωt gives: 8 = R sin α, from which, sin α =
8
R
There is only one quadrant where both sin α and cos α are positive, and this is the first, as shown
below.
By Pythagoras’s theorem: R =
52 + 82 = 9.433
From trigonometric ratios: α = tan −1
Hence,
8
= 57.99° or 1.012 radians
5
5 sin ωt + 8 cos ωt = 9.433 sin(ωt + 1.012)
2. Change 4 sin ωt – 3 cos ωt into the form R sin(ωt ± α)
Let
4 sin ωt – 3 cos ωt = R sin( ωt + α)
= R[sin ωt cos α + cos ωt sin α]
= (R cos α) sin ωt + (R sin α) cos ωt
Hence,
4 = R cos α
from which, cos α =
4
R
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© 2014, John Bird
and
–3 = R sin α
from which, sin α = −
3
R
There is only one quadrant where both cosine is positive and sine is negative, i.e. the 4th quadrant,
as shown in the diagram below.
R=
( 42 + 32 )
and α = tan −1
=5
3
= 0.644 rad
4
(make sure your calculator is on radians)
4 sin ωt – 3 cos ωt = 5 sin(ωt – 0.644)
Hence,
3. Change –7 sin ωt + 4 cos ωt into the form R sin(ωt ± α)
Let
–7 sin ωt + 4 cos ωt = R sin( ωt + α)
= R[sin ωt cos α + cos ωt sin α]
= (R cos α) sin ωt + (R sin α) cos ωt
Hence,
and
–7 = R cos α from which, cos α = −
4 = R sin α
7
R
4
R
from which, sin α =
There is only one quadrant where both cosine is negative and sine is positive, i.e. the 2nd quadrant,
as shown in the diagram below.
R=
( 7 2 + 42 )
= 8.062
Thus, in the diagram,
Hence,
and φ = tan −1
4
= 0.519 rad
7
α = π – 0.519 = 2.622
–7 sin ωt + 4 cos ωt = 8.062 sin( ωt + 2.622)
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© 2014, John Bird
4. Change –3 sin ωt – 6 cos ωt into the form R sin(ωt ± α)
Let –3 sin ωt – 6 cos ωt = R sin(ωt + α)
= R {sin ωt cos α + cos ωt sin α]
= (R cos α) sin ωt + (R sin α) cos ωt
Equating coefficients gives: –3 = R cos α, from which, cos α =
– 6 = R sin α, from which, sin α =
and
−3
R
−6
R
There is only one quadrant in which both cosine and sine are negative, i.e. the third quadrant, as shown
below.
By Pythagoras, R =
32 + 62 = 6.708
θ = tan −1
and
6
= 63.435°
3
Hence, α = 180° + 63.435° = 243.435° or 4.249 radians
Thus,
–3 sin ωt – 6 cos ωt = 6.708 sin(ωt + 4.249)
An angle of 243.435° is the same as –116.565° or –2.034 radians
Hence,
–3 sin ωt – 6 cos ωt = 6.708 sin(ωt – 2.034)
5. Solve the following equations for values of θ between 0° and 360°:
(a) 2 sin θ + 4 cos θ = 3
(a) Let
(b) 12 sin θ – 9 cos θ = 7
2 sin θ + 4 cos θ = R sin(θ + α)
735
© 2014, John Bird
= R[sin θ cos α + cos θ sin α]
= (R cos α) sin θ + (R sin α) cos θ
Hence,
2 = R cos α
from which, cos α =
2
R
and
4 = R sin α
from which, sin α =
4
R
There is only one quadrant where both cosine is positive and sine is positive, i.e. the 1st
quadrant, as shown in the diagram below
R=
Hence,
( 42 + 22 )
= 4.472
and
4
= 63.43°
2
2 sin θ + 4 cos θ = 4.472 sin(θ + 63.43°)
Thus, since 2 sin θ + 4 cos θ = 3
i.e.
α = tan −1
sin(θ + 63.43°) =
then
4.472 sin(θ + 63.43°) = 3
3
= 0.67084
4.472
and
θ + 63.43° = sin −1 0.67084 = 42.13° or 180° – 42.13° = 137.87°
Thus,
θ = 42.13° – 63.43° = – 21.30° ≡ 360° – 21.30° = 338.70°
or
θ = 137.87° – 63.43° = 74.44°
i.e.
(b) Let
θ = 74.44° and 338.70° satisfies the equation 2 sin θ + 4 cos θ = 3
12 sin θ – 9 cos θ = R sin(θ + α)
= R[sin θ cos α + cos θ sin α]
= (R cos α) sin θ + (R sin α) cos θ
12
R
Hence,
12 = R cos α
from which, cos α =
and
–9 = R sin α
from which, sin α = −
9
R
There is only one quadrant where both cosine is positive and sine is negative, i.e. the 4th
quadrant, as shown in the diagram below.
736
© 2014, John Bird
R=
Hence,
(122 + 92 )
= 15
and
sin(θ – 36.87°) =
15 sin(θ – 36.87°) = 7
then
7
15
7
= 27.82° or
15
and
θ – 36.87° = sin −1
Thus,
θ = 27.82° + 36.87° = 64.69°
i.e.
9
= 36.87°
12
12 sin θ – 9 cos θ = 15 sin(θ – 36.87°)
Thus, since 12 sin θ – 9 cos θ = 7
i.e.
α = tan −1
or
180° – 27.82° = 152.18°
θ = 152.18° + 36.87° = 189.05°
θ = 64.69° and 189.05° satisfies the equation 12 sin θ – 9 cos θ = 7
6. Solve the following equations for 0° < A < 360°:
(a) 3 cos A + 2 sin A = 2.8
(a) Let
(b) 12 cos A – 4 sin A = 11
3 cos A + 2 sin A = R sin(A + α)
= R[sin A cos α + cos A sin α]
= (R cos α) sin A + (R sin α) cos A
Hence,
2 = R cos α
from which, cos α =
2
R
and
3 = R sin α
from which, sin α =
3
R
There is only one quadrant where both sine and cosine are positive, i.e. the 1st quadrant, as
shown in the diagram below
737
© 2014, John Bird
R=
( 22 + 32 )
Hence,
and α = tan −1
3
= 56.31°
2
3 cos A + 2 sin A = 3.606 sin(A + 56.31°)
Thus, since
i.e.
= 13 = 3.606
3 cos A + 2 sin A = 2.8
then
3.606 sin(A + 56.31°) = 2.8
2.8
= 0.776580...
13
sin(A + 56.31°) =
and
A + 56.31° = sin −1 0.776580... = 50.95° or 180° – 50.95° = 129.05°
Thus,
θ = 50.95° – 56.31° = –5.36° ≡ 360° – 5.36° = 354.64°
or
θ = 129.05° – 56.31° = 72.74°
θ = 72.74° and 354.64° satisfies the equation 3 cos A + 2 sin A = 2.8
i.e.
12 cos A – 4 sin A = R sin(A + α)
(b) Let
= R[sin A cos α + cos A sin α]
= (R cos α) sin A + (R sin α) cos A
Hence,
– 4 = R cos α
from which, cos α =
and
12 = R sin α
from which, sin α =
−4
R
12
R
There is only one quadrant where both sine is positive and cosine is negative, i.e. the 2nd
quadrant, as shown in the diagram below
R=
42 + 122 = 12.649 and A = tan −1
12
= 71.565°
4
Hence, α = 180° – 71.565° = 108.435°
Thus, 12 cos A – 4 sin A = 12.649 sin(A + 108.435°) = 11
Hence, sin(A + 108.435°) =
11
12.649
from which,
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© 2014, John Bird
(A + 108.435°) = sin −1
11
= 60.416° or 119.584°
12.649
A = 60.416° – 108.435° = – 48.019° ≡ (– 48.019° + 360°) = 311.98°
Thus
or
A = 119.584° – 108.435° = 11.15°
The solutions are thus A = 11.15° or 311.98°, which may be checked in the original equation
7. Solve the following equations for values of θ between 0° and 360°:
(a) 3sin θ + 4 cos θ =
3
(b) 2 cos θ + sin θ =
2
3sin θ + 4 cos θ = R sin(θ + α)
(a) Let
= R[sin θ cos α + cos θ sin α]
= (R cos α) sin θ + (R sin α) cos θ
Hence,
3 = R cos α
from which, cos α =
3
R
and
4 = R sin α
from which, sin α =
4
R
There is only one quadrant where both sine and cosine are positive, i.e. the 1st quadrant, as
shown in the diagram below
R=
( 32 + 42 )
Hence,
Thus, since
i.e.
=5
and α = tan −1
4
= 53.13°
3
3sin θ + 4 cos θ = 5 sin(A + 53.13°)
3sin θ + 4 cos θ =
3 then
sin(A + 53.13°) =
5 sin(A + 53.13°) = 3
3
= 0.60
5
and
θ + 53.13° = sin −1 0.60 = 36.87° or 180° – 36.87° = 143.13°
Thus,
θ = 36.87° – 53.13° = – 16.26° ≡ 360° – 16.26° = 343.74°
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© 2014, John Bird
θ = 143.13° – 53.13° = 90°
or
θ = 90° and 343.74° satisfies the equation 3sin θ + 4 cos θ =
3
i.e.
2 cos θ + sin θ = R sin(θ + α)
(b) Let
= R[sin θ cos α + cos θ sin α]
= (R cos α) sin θ + (R sin α) cos θ
Hence,
1 = R cos α
from which, cos α =
1
R
and
2 = R sin α
from which, sin α =
2
R
There is only one quadrant where both sine is positive and cosine is negative, i.e. the 2nd
quadrant, as shown in the diagram below.
R = 12 + 22 = 2.236 and α = tan −1
2
= 63.43°
1
Thus, 2 cos θ + sin θ = 2.236 sin(θ + 63.43°) = 2
Hence, sin(θ + 63.43°) =
(θ + 63.43°) = sin
Thus
or
2
2.236
−1
from which,
2
= 63.435° or 116.565°
2.236
θ = 63.435° – 63.43° = 0
θ = 116.565° – 63.43° = 53.14°
The solutions are thus θ = 0° or 53.14°, which may be checked in the original equation
8. Solve the following equations for values of θ between 0° and 360°:
(a) 6 cos θ + sin θ =
3
(a) Let
(b) 2sin 3θ + 8cos 3θ =
1
sin θ + 6 cos θ = R sin(θ + α)
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© 2014, John Bird
= R[sin θ cos α + cos θ sin α]
= (R cos α) sin θ + (R sin α) cos θ
Hence,
1 = R cos α
from which, cos α =
1
R
and
6 = R sin α
from which, sin α =
6
R
There is only one quadrant where both cosine is positive and sine is positive, i.e. the 1st
quadrant, as shown in the diagram below.
R=
= 6.083
α = tan −1
and
6
= 80.54°
1
sin θ + 6 cos θ = 6.083 sin(θ + 80.54°)
Hence,
Thus, since
i.e.
(12 + 62 )
sin θ + 6 cos θ =
3
sin(θ + 80.54°) =
6.083 sin(θ + 80.54°) =
then
3
3
= 0.284736...
6.083
and
θ + 80.54° = sin −1 0.284736... = 16.54° or 180° – 16.54° = 163.44°
Thus,
θ = 16.54° – 80.54° = –64° ≡ 360° –64° = 296°
or
θ = 163.44° – 80.54° = 82.90°
i.e.
(b) Let
θ = 82.90° and 296° satisfies the equation 2 sin θ + 4 cos θ = 3
2sin 3θ + 8cos 3θ = R sin(3θ + α)
= R[sin 3θ cos α + cos 3θ sin α]
= (R cos α) sin 3θ + (R sin α) cos 3θ
Hence,
2 = R cos α
from which, cos α =
and
8 = R sin α
from which, sin α =
2
R
8
R
There is only one quadrant where both cosine is positive and sine is positive, i.e. the 1st
quadrant, as shown in the diagram below
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© 2014, John Bird
R=
( 2 2 + 82 )
=
68
α = tan −1
and
2 sin 3θ + 8 cos 3θ =
Hence,
Thus, since 2 sin 3θ + 8 cos 3θ = 1
sin(3θ + 75.96°) =
i.e.
8
= 75.96°
2
68 sin(3θ + 75.96°)
68 sin(3θ + 75.96°) = 1
then
1
68
1
= 6.97° or
68
and
3θ + 75.96° = sin −1
Thus,
3θ = 6.97° – 75.96° = –68.99° = 291.01°
180° – 6.93° = 173.03°
or
3θ = 173.03° – 75.96° = 97.07°
i.e.
3θ = 291.01° or 291.01° + 360° = 651.01° or 651.01° + 360° = 1011.01°
and
3θ = 97.07° or 97.07° + 360° = 457.07° or 457.07° + 360° = 817.07°
Hence,
θ=
291.01°
= 97° or
3
and
θ=
97.07°
= 32.36° or
3
Thus,
651.01°
= 217° or
3
1011.01°
= 337°
3
457.07°
= 152.36° or
3
817.07°
= 272.36°
3
θ = 32.36°, 97°, 152.36°, 217°, 272.36° and 337° all satisfy the equation
2sin 3θ + 8cos 3θ =
1
9. The third harmonic of a wave motion is given by 4.3 cos 3θ – 6.9 sin 3θ.
Express this in the form R sin(3θ ± α).
Let
4.3 cos 3θ – 6.9 sin 3θ = R sin(3θ + α)
= R[sin 3θ cos α + cos 3θ sin α]
= (R cos α) sin 3θ + (R sin α) cos 3θ
Hence,
–6.9 = R cos α
from which, cos α = −
742
6.9
R
© 2014, John Bird
4.3 = R sin α
and
from which, sin α =
4.3
R
There is only one quadrant where both cosine is negative and sine is positive, i.e. the 2nd quadrant,
as shown in the diagram below.
R=
( 6.92 + 4.32 )
= 8.13
and
φ = tan −1
4.3
= 31°56′
6.9
α = 180° – 31°56′ = 148°4′ = 2.584 rad
and
4.3 cos 3θ – 6.9 sin 3θ = 8.13 sin(3θ + 2.584)
Hence,
10. The displacement x metres of a mass from a fixed point about which it is oscillating is given by:
x = 2.4 sin ωt + 3.2 cos ωt, where t is the time in seconds. Express x in the form R sin(ωt + α).
Let
x = 2.4 sin ωt + 3.2 cos ωt = R sin( ωt + α)
= R[sin ωt cos α + cos ωt sin α]
= (R cos α) sin ωt + (R sin α) cos ωt
Hence,
2.4 = R cos α
from which, cos α =
2.4
R
and
3.2 = R sin α
from which, sin α =
3.2
R
There is only one quadrant where both cosine is positive and sine is positive, i.e. the 1st quadrant, as
shown in the diagram below
743
© 2014, John Bird
R=
( 2.42 + 3.22 )
= 4.0
and
α = tan −1
3.2
= 53.13° or 0.927 rad
2.4
x = 2.4 sin ωt + 3.2 cos ωt = 4.0 sin( ωt + 0.927) m
Hence,
11. Two voltages, v 1 = 5 cos ωt and v2 = –8 sin ωt are inputs to an analogue circuit. Determine an
expression for the output voltage if this is given by (v 1 + v 2 ).
v 1 + v2 = 5 cos ωt + (–8 sin ωt) = 5 cos ωt – 8 sin ωt
5 cos ωt – 8 sin ωt = R sin(ωt + α) = R [sin ωt cos α + cos ωt sin α]
Let
Equating coefficients gives:
= (R cos α) sin ωt + (R sin α) cos ωt
5
5 = R sin α, from which, sin α =
R
– 8 = R cos α, from which, cos α =
and
−8
R
There is only one quadrant in which both sine is positive and cosine is negative, i.e. the second, as shown
below
By Pythagoras, R =
82 + 52 = 9.434 and θ = tan −1
5
= 32.01°
8
Hence, α = 180° – 32.01° = 147.99° = 2.583 rad
Thus,
v1 + v 2 = 5 cos ωt – 8 sin ωt = 9.434 sin(ωt + 2.583)
12. The motion of a piston moving in a cylinder can be described by: x = (5 cos 2t + 5 sin 2t) cm
Express x in the form R sin(ωt + α).
Let
5 cos 2t + 5 sin 2t = R sin(2t + α) = R [sin 2t cos α + cos 2t sin α]
= (R cos α) sin 2t + (R sin α) cos 2t
744
© 2014, John Bird
Equating coefficients gives:
and
5 = R sin α, from which, sin α =
5
R
5 = R cos α, from which, cos α =
5
R
There is only one quadrant in which both sine and cosine are positive, i.e. the 1st, as shown
below
By Pythagoras, R =
Thus,
52 + 52 = 7.07 and α = tan −1
5
π
= 45° =
rad
5
4
π

x = (5 cos 2t + 5 sin 2t) cm = 7.07 sin  2t +  cm
4

745
© 2014, John Bird
EXERCISE 184 Page 493
1. The power p in an electrical circuit is given by p =
v2
. Determine the power in terms of V, R and
R
cos 2t when v = V cos t
v2
Power
p=
=
R
cos t )
(V=
2
R
V 2 cos 2 t V 2
=
cos 2 t
R
R
One of the formulae for the cosine of the double angle is: cos 2t = 2 cos 2 t − 1
2t
cos=
from which,
Hence, power p =
1
(1 + cos 2t )
2
V2
V2 1
 V2
=
cos 2 t
(1 + cos 2t )
 (1 + cos 2t )  =
R
R 2
 2R
2. Prove the following identities:
(a) 1 –
(c)
cos 2φ
= tan2 ϕ
cos 2 φ
(b)
2
(tan 2 x)(1 + tan x)
=
tan x
1 − tan x
(a) L.H.S. = 1 −
1 + cos 2t
= 2 cot2 t
sin 2 t
(d) 2 cosec 2θ cos 2θ = cot θ – tan θ
 cos 2 φ − sin 2 φ 
 cos 2 φ sin 2 φ 
cos 2φ
1− 
1− 
=
=
−


cos 2 φ
cos 2 φ


 cos 2 φ cos 2 φ 
= 1– (1 − tan 2 φ ) = tan 2 φ = R.H.S.
(b) L.H.S. =
1 + cos 2t 1 + ( 2 cos 2 t − 1) 2 cos 2 t
=
= = 2 cot 2 t = R.H.S.
sin 2 t
sin 2 t
sin 2 t
 2 tan x 
(1 + tan x )
tan 2 x )(1 + tan x )  1 − tan 2 x 
(
(c) L.H.S. =
=
=
tan x
tan x
( 2 tan x )(1 + tan x )
(1 − tan x )(1 + tan x )
tan x
2 tan x
(1 − tan x ) =
2 tan x
2
=
= R.H.S.
=
tan x
tan x (1 − tan x ) 1 − tan x
746
© 2014, John Bird
 2
(d) L.H.S. = 2 cosec 2θ cos 2θ = 
 sin 2θ
=
2

=
=
2θ ) 2 cot
2θ =
 (cos
tan 2θ

2 (1 − tan 2 θ )
2 tan θ
=
2
2 tan θ
1 − tan 2 θ
1 − tan 2 θ
1
=
− tan θ = cot θ − tan θ = R.H.S.
tan θ
tan θ
3. If the third harmonic of a waveform is given by V3 cos 3θ, express the third harmonic in terms of the first
harmonic cos θ, when V3 = 1
When V3 = 1, V3 cos 3θ = cos 3θ = cos(2θ + θ) = cos 2θ cos θ – sin 2θ sin θ
= ( 2 cos 2 θ − 1) (cos θ ) − ( 2sin θ cos θ ) (sin θ )
= 2 cos3 θ − cos θ − 2 cos θ sin 2 θ
= 2 cos3 θ − cos θ − 2 cos θ (1 − cos 2 θ )
= 2 cos3 θ − cos θ − 2 cos θ + 2 cos3 θ
i.e.
cos 3θ = 4 cos3 θ − 3cos θ
4. Solve for θ in the range −180° ≤ θ ≤ 180° : cos 2θ = sin θ
then 1 − 2sin 2 θ =
sin θ
If cos 2θ = sin θ
Rearranging gives:
2sin 2 θ + sin θ − 1 =0
which is a quadratic equation in sin θ
(2 sin θ – 1)(sin θ + 1) = 0
Solving by quadratic formula or by factorizing:
1
2
from which,
sin θ =
or sin θ = –1 then
from which,
θ = 30° or 150° or 270°
Since the required range is −180° ≤ θ ≤ 180° then
θ = 30° or 150° or –90°
5. Solve for θ in the range −180° ≤ θ ≤ 180° : 3sin 2θ + 2 cos θ =
0
Since 3sin 2θ + 2 cos θ =
0 then 3(2 sin θ cos θ) + 2 cos θ = 0
747
© 2014, John Bird
6 sin θ cos θ + 2 cos θ = 0
i.e.
2 cos θ(3 sin θ + 1) = 0
Factorizing gives:
Hence,
either
2 cos θ = 0 or
cos θ = 0
i.e.
(3 sin θ + 1) = 0
sin θ = −
or
 1
or
=
θ sin −1  −  = 199.47° and 340.53°
 3
θ = cos −1 0 = 90° and 270°
Thus,
Hence, in the range −180° ≤ θ ≤ 180° :
1
3
θ = 90°, – 90°, – 19.47° and – 160.47°
6. Solve for θ in the range −180° ≤ θ ≤ 180° : sin 2θ + cos θ =
0
2sin θ cos θ + cos θ =
0
Since sin 2θ + cos θ =
0 then
cos θ(2 sin θ + 1) = 0
i.e.
from which,
cos θ = 0 or
2 sin θ + 1 = 0
i.e.
cos θ = 0 or
sin θ = −
Thus,
1
2
θ = 90° or 270° or 210° or 330°
However, in the required range of −180° ≤ θ ≤ 180° :
θ = 90° or – 90° or – 150° or – 30°
7. Solve for θ in the range −180° ≤ θ ≤ 180° : cos 2θ + 2sin θ =
−3
Since cos 2θ + 2sin θ =
−3 then
2sin 2 θ − 2sin θ − 4 =
0
Rearranging gives:
sin 2 θ − sin θ − 2 =
0
or
(sin θ – 2)(sin θ + 1) = 0
Factorizing gives:
from which,
i.e.
Hence,
−3
(1 − 2sin 2 θ ) + 2sin θ =
either
(sin θ – 2) = 0
sin θ = 2 (which has no solution)
or
(sin θ + 1) = 0
or sin θ = –1
θ = sin −1 (−1) = 270°
i.e. in the range −180° ≤ θ ≤ 180° :
θ = –90°
748
© 2014, John Bird
8. Solve for θ in the range −180° ≤ θ ≤ 180° : tan θ + cot θ =
2
Since tan θ + cot θ =
2 then
sin θ cos θ
+
=
2
cos θ sin θ
Multiplying each term by sin θ cos θ gives:
( sin θ cos θ )
Cancelling gives:
Now
sin θ
cos θ
+ ( sin θ cos θ )
=
2 ( sin θ cos θ )
cos θ
sin θ
sin 2 θ + cos 2 θ =
2sin θ cos θ
sin 2 θ + cos 2 θ =
1 and
Hence,
from which,
Hence,
In the range −180° ≤ θ ≤ 180° :
2sin θ cos θ = sin 2θ
1 = sin 2θ
2θ = sin −1 1 = 90° or 90° + 360° = 450°
θ=
90°
450°
= 45° or
= 225°
2
2
θ = 45° or –135°
749
© 2014, John Bird
EXERCISE 185 Page 495
1. Express sin 7t cos 2t as a sum or difference.
1
[sin(7t + 2t ) + sin(7t − 2t )]
2
sin 7t cos 2t =
from (1), page 494
1
[sin 9t + sin 5t ]
2
=
2. Express cos 8x sin 2x as a sum or difference.
1
[sin(8 x + 2 x) − sin(8 x − 2 x)]
2
cos 8x sin 2x =
=
from (2), page 494
1
[sin10 x − sin 6 x ]
2
3. Express 2 sin 7t sin 3t as a sum or difference.
 1
2 sin 7t sin 3t = (2)  −  [ cos(7t + 3t ) − cos(7t − 3t ) ]
 2
= − [ cos10t − cos 4t ]
or
from (4), page 494
cos 4t – cos 10t
4. Express 4 cos 3θ cos θ as a sum or difference.
1

4 cos 3θ cos θ = 4  [ cos(3θ + θ ) + cos(3θ − θ ) ]
2

from (3), page 494
= 2[cos 4θ + cos 2θ]
5. Express 3 sin
3 sin
π
3
cos
π
3
cos
π
6
as a sum or difference.
π
1  π π 
 π π 
= (3) sin  +  + sin  −  
6
2 3 6
 3 6 
=
from (1), page 494
3 π
π
sin
+
sin
2 
2
6 
750
© 2014, John Bird
∫ 2sin 3t cos t d t
6. Determine
1

2 sin 3t cost = 2  [sin(3t + t ) + sin(3t − t ) ]
2

from (1), page 494
= sin 4t + sin 2t
Hence,
=
t d t ∫ ( sin 4t + sin 2t ) d t
∫ 2sin 3t cos
7. Evaluate
∫
π /2
0
=−
cos 4t cos 2t
1
1
−
+ c or − cos 4t − cos 2t + c
4
2
4
2
4 cos 5 x cos 2 x d x
1

4 cos 5 x=
cos 2 x 4  [ cos(5 x + 2 x) + cos(5 x − 2 x) ]
2

= 2[cos 7x + cos 3x]
Hence,
∫
π /2
0
4 cos 5=
x cos 2 x d x
∫
π /2
0
2(cos 7 x + cos 3 x) d x
π /2
 2sin 7 x 2sin 3x 
=
+
3  0
 7
7π 2
3π
2
+ sin
=  sin
2 3
2
7
2
 2

 −  sin 0 + sin 0 
3
 7

20
 2 2
=  − −  − (0) = −
21
 7 3
8. Solve the equation: 2 sin 2φ sin φ = cos φ in the range φ = 0 to φ = 180°
2 sin 2φ sin φ = cos φ
i.e.
2(2 sin φ cos φ) sin φ = cos φ
i.e.
4sin 2 φ cos φ = cos φ
i.e.
and
Hence,
4sin 2 φ cos φ − cos φ =
0
cos φ ( 4sin 2 φ − 1) =
0
cos φ = 0
from which,
φ = cos −1 0 = 90°
751
© 2014, John Bird
1
4
and sin φ =
and
4sin 2 φ = 1
Hence,
φ = sin −1 0.5 = 30° and 150° (see diagram below)
and
φ = sin −1 (−0.5) = 210° and 330°
from which, sin 2 φ =
1
= ±0.5
4
Since the range is from φ = 0° to φ = 180°, then the only values of φ to satisfy: 2 sin 2φ sin φ = cos φ
are:
φ = 30°, 90° and 150°
752
© 2014, John Bird
EXERCISE 186 Page 496
1. Express sin 3x + sin x as products.
 3x + x 
 3x − x 
sin 3x + sin x = 2sin 
 cos 

 2 
 2 
from (5), page 495
= 2 sin 2x cos x
2. Express
1
(sin 9θ – sin 7θ) as products.
2
1
1
 9θ + 7θ
(sin 9θ – sin 7θ) = (2) cos 
2
2
 2
  9θ − 7θ 
 sin 

  2 
from (6), page 495
= cos 8θ sin θ
3. Express cos 5t + cos 3t as products.
 5t + 3t 
 5t − 3t 
cos 5t + cos 3t = 2 cos 
 cos 

 2 
 2 
from (7), page 495
= 2 cos 4t cos t
4. Express
1
(cos 5t – cos t) as products.
8
1
1
 5t + t   5t − t 
(cos 5t – cos t) =   (−2) sin 
 sin 

8
8
 2   2 
from (8), page 495
1
= − sin 3t sin 2t
4
5. Express
1
π
π
(cos
+ cos ) as products .
2
3
4

π π 
 π π 
+ 


 3 − 4  
1
π
π  1
3
4
2 cos 
 cos 

 cos + cos  =
2
3
4 2
 2 
 2 



 

753
from (7), page 495
© 2014, John Bird
 7π

= cos  12
 2

6. Show that:(a)
(b)

π 

 12 
7π
π
cos
 cos   = cos
24
24

 2 

 
sin 4 x − sin 2 x
= tan x
cos 4 x + cos 2 x
1
2
{
sin(5x – α) – sin(x + α)
} = cos 3x sin(2x – α)
 4x + 2x   4x − 2x 
2 cos 
 sin 

sin 4 x − sin 2 x
2   2 

(a) L.H.S. =
=
cos 4 x + cos 2 x
 4x + 2x 
 4x − 2x 
2 cos 
 cos 

 2 
 2 
=
(b) L.H.S. =
2 cos 3 x sin x sin x
= = tan x = R.H.S.
2 cos 3 x cos x cos x
1
{ sin(5 x − α ) − sin( x + α ) }
2
=
1
[(sin 5 x cos α − cos 5 x sin α ) − (sin x cos α + cos x sin α )]
2
=
1
[cos α (sin 5 x − sin x) − sin α (cos 5 x + cos x)]
2
=


1
 5x + x   5x − x 
 5x + x 
 5 x − x  
cos α 2 cos 
 sin 
  − sin α 2 cos 
 cos 
 
2
 2   2 
 2 
 2  


=
1
[ 2 cos α (cos 3x sin 2 x) − 2sin α (cos 3x cos 2 x)]
2
= cos 3x (cos α sin 2x – sin α cos 2x)
= cos 3x (sin 2x cos α – cos 2x sin α)
= cos 3x sin(2x – α) = R.H.S.
7. Solve for θ in the range 0° ≤ θ ≤ 180° : cos 6θ + cos 2θ =
0
cos 6θ + cos 2θ =
0 hence
 6θ + 2θ
2 cos 
 2
i.e.
2 cos 4θ cos 2θ = 0
i.e.

 6θ − 2θ
 cos 

 2
cos 4θ cos 2θ = 0
754

=0

from (7), page 495
© 2014, John Bird
Hence, either
i.e.
i.e.
i.e.
cos 4θ = 0
4θ = cos −1 0
or
cos 2θ = 0
or
2θ = cos −1 0
4θ = 90° or 270° or 450° or 630°
or
θ = 22.5° or 67.5° or 112.5° or 157.5°
2θ = 90° or 270°
or θ = 45° or 135°
Hence, in the range 0° ≤ θ ≤ 180° : θ = 22.5° or 45° or 67.5° or 112.5° or 135° or 157.5°
8. Solve for θ in the range 0° ≤ θ ≤ 180° : sin 3θ − sin θ =
0
 3θ + θ
sin 3θ – sin θ = 2 cos 
 2
  3θ − θ 
 sin 

  2 
from (6), page 495
= 2 cos 2θ sin θ
Since sin 3θ − sin θ =
0 then 2 cos 2θ sin θ = 0
Hence,
either cos 2θ = 0
i.e.
either 2θ = cos −1 0 = 90° and 270° from which, θ = 45° and 135°
sin θ = 0
θ = sin −1 0 = 0° and 180°
or
Hence,
or
θ = 0°, 45°, 135° and 180° all satisfy the equation sin 3θ − sin θ =
0
9. Solve in the range 0° ≤ θ ≤ 360° : cos 2x = 2 sin x
Since cos 2x = 2 sin x
then
1 − 2sin 2 x =
2sin x
2sin 2 x + 2sin x − 1 =0
Rearranging gives:
Using the quadratic formula gives: sin x =
Hence,
−2 ± 22 − 4(2)(−1) −2 ± 12
=
2(2)
4
= 0.3660254… or –1.3660254… (which has no solution)
x = sin 0.3660254... = 21.47° or 158.53°
−1
10. Solve in the range 0° ≤ θ ≤ 360° : sin 4t + sin 2t = 0
sin 4t + sin 2t = 0
 4t + 2t 
 4t − 2t 
hence, 2sin 
 cos 
=0
 2 
 2 
755
from (5), page 495
© 2014, John Bird
i.e.
2 sin 3t cos t = 0
i.e.
sin 3t cos t = 0
Hence, either
i.e.
i.e.
i.e.
sin 3t = 0
3t = sin −1 0
or
or
cos t = 0
t = cos −1 0
3t = 0° or 180° or 360° or 540° or 720° or 900° or 1080°
or
t = 90° or 270°
t = 0° or 60° or 120° or 180° or 240° or 300° or 360° or t = 90° or 270°
Hence, in the range 0° ≤ θ ≤ 360° :
θ = 0° or 60° or 90° or 120° or 180° or 240° or 270° or 300° or 360°
756
© 2014, John Bird