Higher Engineering Mathematics

COMPOUND ANGLES 183

B

Now try the following exercise.

Exercise 82 Further problems on double

angles

1. The powerpin an electrical circuit is given

byp=

v^2

R

. Determine the power in terms of
V,Rand cos 2twhenv=[Vcost.
V^2
2 R

(1+cos 2t)

]

2. Prove the following identities:

(a) 1−

cos 2φ

cos^2 φ

=tan^2 φ

(b)

1 +cos 2t

sin^2 t

=2 cot^2 t

(c)

(tan 2x)(1+tanx)

tanx

=

2

1 −tanx

(d) 2 cosec 2θcos 2θ=cotθ−tanθ

3. If the third harmonic of a waveform is given
   by V 3 cos 3θ, express the third harmonic
   in terms of the first harmonic cosθ, when
   V 3 =1.
   [cos 3θ=4 cos^3 θ−3 cosθ]

18.4 Changing products of sines and

cosines into sums or differences

(i) sin(A+B)+sin(A−B)=2 sinAcosB (from

the formulae in Section 18.1)

i.e. sinAcosB

=^12 [sin(A+B)+sin(A−B)] (1)

(ii) sin(A+B)−sin(A−B)=2 cosAsinB

i.e. cosAsinB

=^12 [sin(A+B)−sin(A−B)] (2)

(iii) cos(A+B)+cos(A−B)=2 cosAcosB

i.e. cosAcosB

=^12 [cos(A+B)+cos(A−B)] (3)

(iv) cos(A+B)−cos(A−B)=−2 sinAsinB

i.e. sinAsinB

=−^12 [cos(A+B)−cos(A−B)] (4)

Problem 14. Express sin 4xcos 3xas a sum or

difference of sines and cosines.

From equation (1),

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

=^12 (sin 7x+sinx)

Problem 15. Express 2 cos 5θsin 2θas a sum

or difference of sines or cosines.

From equation (2),

2 cos 5θsin 2θ= 2

{

1

2

[sin(5θ+ 2 θ)−sin(5θ− 2 θ)]

}

=sin 7θ−sin 3θ

Problem 16. Express 3 cos 4tcostas a sum or

difference of sines or cosines.

From equation (3),

3 cos 4tcost= 3

{

1

2

[cos(4t+t)+cos(4t−t)]

}

=

3

2

(cos 5t+cos 3t)

Thus, if the integral

∫

3 cos 4tcostdtwas required

(for integration see Chapter 37), then

∫

3 cos 4tcostdt=

∫

3

2

(cos 5t+cos 3t)dt

=

3

2

[

sin 5t

5

+

sin 3t

3

]

+c

Problem 17. In an alternating current circuit,

voltagev=5 sinωtand currenti=10 sin(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=(5 sinωt)

[

10 sin(ωt−π/ 6 )

]

=50 sinωtsin(ωt−π/6)

From equation (4),

50 sinωtsin(ωt−π/6)

=(50)

[

−^12

{

cos (ωt+ωt−π/6)

−cos

[

ωt−(ωt−π/6)

]}]

=− 25 {cos(2ωt−π/6)−cosπ/ 6 }