chapter 44 compound angles AWS
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
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© 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 731
© 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°
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© 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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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) 734
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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
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= 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.
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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
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© 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° 739
© 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) 740
© 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 741
© 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 = −
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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
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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
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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
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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
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 731
© 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
733
© 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) 734
© 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,
738
© 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° 739
© 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) 740
© 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 741
© 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
