Higher Engineering Mathematics, Sixth Edition

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:

1. sin3x+sinx [2sin2xcosx]
   2.^12 (sin 9θ−sin7θ) [cos8θsinθ]


2. 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

]

6. Show that:
   (a)

sin4x−sin2x

cos4x+cos2x

=tanx

(b)^12 {sin( 5 x−α)−sin(x+α)}

=cos3xsin( 2 x−α)