Higher Engineering Mathematics, Sixth Edition

Compound angles 171

(iv) cos(A+B)−cos(A−B)=−2sinAsinB

i.e. sinAsinB

=−^12 [cos(A+B)−cos(A−B)] (4)

Problem 15. Express sin4xcos3xas a sum or

difference of sines and cosines.

From equation (1),

sin4xcos3x=^12 [sin( 4 x+ 3 x)+sin( 4 x− 3 x)]

=^12 (sin7x+sinx)

Problem 16. Express 2cos5θsin2θas a sum or

difference of sines or cosines.

From equation (2),

2cos5θsin2θ= 2

{

1

2

[sin( 5 θ+ 2 θ)−sin( 5 θ− 2 θ)]

}

=sin7θ−sin3θ

Problem 17. Express 3cos4tcostas a sum or

difference of sines or cosines.

From equation (3),

3cos4tcost= 3

{

1

2

[cos( 4 t+t)+cos( 4 t−t)]

}

=

3

2

(cos5t+cos3t)

Thus, if the integral

∫
3cos4tcostdtwas required (for
integration see Chapter 37), then
∫
3cos4tcostdt=

∫

3

2

(cos 5t+cos3t)dt

=

3

2

[

sin5t

5

+

sin3t

3

]

+c

Problem 18. In an alternating current circuit,

voltagev=5sinωtand currenti=10sin(ωt−

π/ 6 ). Find an expression for the instantaneous

powerpat timetgiven thatp=vi, expressing the

answer as a sum or difference of sines and cosines.

p=vi=(5sinωt)[10sin(ωt−π/ 6 )]

=50sinωtsin(ωt−π/ 6 )

From equation (4),

50sinωtsin(ωt−π/ 6 )

=( 50 )

[

−^12

{

cos(ωt+ωt−π/ 6 )

−cos

[

ωt−(ωt−π/ 6 )

]}]

=− 25 {cos( 2 ωt−π/ 6 )−cosπ/ 6 }

i.e. instantaneous power,

p=25[cosπ/ 6 −cos(2ωt−π/6)]

Now try the following exercise

Exercise 75 Further problems on changing

products of sines and cosines into sums or

differences

In Problems 1 to 5, express as sums or differences:

1. sin7tcos2t

[ 1

2 (sin9t+sin5t)

]

2. cos8xsin2x

[ 1

2 (sin10x−sin6x)

]

3. 2sin7tsin3t [cos4t−cos 10t]


4. 4cos3θcosθ [2(cos4θ+cos2θ)]


5. 3sin

π

3

cos

π

6

[

3

2

(

sin

π

2

+sin

π

6

)]

6. Determine

∫

2sin3tcos[tdt.

−

cos4t

4

−

cos2t

2

+c

]

7. Evaluate

∫ π

2

0

4cos5xcos2xdx.

[

−

20

21

]

8. Solve the equation: 2sin2φsinφ=cosφin
   the rangeφ=0toφ= 180 ◦.
   [30◦,90◦or 150◦]

17.5 Changing sums or differences of

sines and cosines into products

In the compound-angle formula let,

(A+B)=X

and

(A−B)=Y