172 Higher Engineering Mathematics
Solving the simultaneous equations gives:
A=
X+Y
2
andB=
X−Y
2
Thus sin(A+B)+sin(A−B)=2sinAcosBbecomes,
sinX+sinY=2sin
(
X+Y
2
)
cos
(
X−Y
2
)
(5)
Similarly,
sinX−sinY=2cos
(
X+Y
2
)
sin
(
X−Y
2
)
(6)
cosX+cosY=2cos
(
X+Y
2
)
cos
(
X−Y
2
)
(7)
cosX−cosY=−2sin
(
X+Y
2
)
sin
(
X−Y
2
)
(8)
Problem 19. Express sin5θ+sin3θas a product.
From equation (5),
sin5θ+sin3θ=2sin
(
5 θ+ 3 θ
2
)
cos
(
5 θ− 3 θ
2
)
=2sin4θcosθ
Problem 20. Express sin7x−sinxas a product.
From equation (6),
sin7x−sinx=2cos
(
7 x+x
2
)
sin
(
7 x−x
2
)
=2cos4xsin3x
Problem 21. Express cos2t−cos5tas a
product.
From equation (8),
cos2t−cos5t=−2sin
(
2 t+ 5 t
2
)
sin
(
2 t− 5 t
2
)
=−2sin
7
2
tsin
(
−
3
2
t
)
=2sin
7
2
tsin
3
2
t
(
since sin
(
−
3
2
t
)
=−sin
3
2
t
)
Problem 22. Show that
cos6x+cos2x
sin6x+sin2x
=cot 4x.
From equation (7),
cos6x+cos2x=2cos4xcos2x
From equation (5),
sin6x+sin2x=2sin4xcos2x
Hence
cos6x+cos 2x
sin6x+sin2x
=
2cos4xcos2x
2sin4xcos2x
=
cos4x
sin4x
=cot 4x
Problem 23. Solve the equation
cos4θ+cos2θ=0forθin the range 0◦≤θ≤ 360 ◦.
From equation (7),
cos4θ+cos2θ=2cos
(
4 θ+ 2 θ
2
)
cos
(
4 θ− 2 θ
2
)
Hence, 2cos3θcosθ= 0
Dividing by 2 gives: cos3θcosθ= 0
Hence, either cos3θ=0orcosθ= 0
Thus, 3 θ=cos−^1 0orθ=cos−^10
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:
- sin3x+sinx [2sin2xcosx]
2.^12 (sin 9θ−sin7θ) [cos8θsinθ] - cos5t+cos3t [2cos4tcost]
4.^18 (cos 5t−cost)
[
−^14 sin3tsin2t
]
5.^12
(
cos
π
3
+cos
π
4
) [
cos
7 π
24
cos
π
24
]
- Show that:
(a)
sin4x−sin2x
cos4x+cos2x
=tanx
(b)^12 {sin( 5 x−α)−sin(x+α)}
=cos3xsin( 2 x−α)