172 Higher Engineering Mathematics
Solving the simultaneous equations gives:
A =
X + Y
2
and B =
X − Y
2
Thus sin(A + B) + sin(A − B) = 2 sin A cos B becomes,
sin X + sin Y = 2 sin
X + Y
2
cos
X − Y
2
(5)
Similarly,
sin X − sin Y = 2cos
X + Y
2
sin
X − Y
2
(6)
cos X + cos Y = 2cos
X + Y
2
cos
X − Y
2
(7)
cos X − cos Y = −2 sin
X + Y
2
sin
X − Y
2
(8)
Problem 19. Express sin 5θ + sin 3θ as a product.
From equation (5),
sin 5θ + sin 3θ = 2 sin
5θ + 3θ
2
cos
5θ − 3θ
2
= 2 sin 4θ cos θ
Problem 20. Express sin 7x − sin x as a product.
From equation (6),
sin 7x − sin x = 2 cos
7x + x
2
sin
7x − x
2
= 2 cos 4x sin 3x
Problem 21. Express cos 2t − cos 5t as a
product.
From equation (8),
cos 2t − cos 5t = −2 sin
2t + 5t
2
sin
2t − 5t
2
= −2 sin
7
2
t sin
−
3
2
t
= 2 sin
7
2
t sin
3
2
t
since sin
−
3
2
t
= −sin
3
2
t
Problem 22. Show that
cos 6x + cos 2x
sin 6x + sin 2x
= cot 4x.
From equation (7),
cos 6x + cos 2x = 2 cos 4x cos 2x
From equation (5),
sin 6x + sin 2x = 2 sin4x cos 2x
Hence
cos 6x + cos 2x
sin 6x + sin 2x
=
2 cos4x cos 2x
2 sin4x cos 2x
=
cos 4x
sin 4x
= cot 4 x
Problem 23. Solve the equation
cos 4θ + cos 2θ = 0 for θ in the range 0 ◦ ≤ θ ≤ 360 ◦ .
From equation (7),
cos 4θ + cos 2θ = 2 cos
4θ + 2θ
2
cos
4θ − 2θ
2
Hence,
2 cos3θ cos θ = 0
Dividing by 2 gives:
cos 3θ cos θ = 0
Hence, either
cos 3θ = 0 or cosθ = 0
Thus,
3θ = cos
−1 0 or θ = cos
−1 0
from which, 3θ = 90 ◦ or 270 ◦ or 450 ◦ or 630 ◦ or
810 ◦ or 990 ◦
and θ = 30
◦ ,90
◦ , 150
◦ ,210
◦ , 270
◦ or 330
◦
Now try the following exercise
Exercise 76 Further problems on changing
sums or differences of sines and cosines into
products
In Problems 1 to 5, express as products:
1. sin 3x + sin x
[2 sin 2x cos x]
2.
1
2 (sin 9θ − sin 7θ)
[cos 8θ sin θ]
3. cos 5t + cos 3t
[2 cos 4t cos t ]
4.
1
8 (cos 5t − cos t )
−
1
4 sin 3t sin 2t
5.
1
2
cos
π
3
+ cos
π
4
cos
7π
24
cos
π
24
6. Show that:
(a)
sin 4x − sin 2x
cos 4x + cos 2x
= tan x
(b)
1
2 {sin(5x − α) − sin(x + α)}
= cos 3x sin(2x − α)