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:
- sin7tcos2t
[ 1
2 (sin9t+sin5t)
]
- cos8xsin2x
[ 1
2 (sin10x−sin6x)
]
- 2sin7tsin3t [cos4t−cos 10t]
- 4cos3θcosθ [2(cos4θ+cos2θ)]
- 3sin
π
3
cos
π
6
[
3
2
(
sin
π
2
+sin
π
6
)]
- Determine
∫
2sin3tcos[tdt.
−
cos4t
4
−
cos2t
2
+c
]
- Evaluate
∫ π
2
0
4cos5xcos2xdx.
[
−
20
21
]
- 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