Question

The sum value of the series \( \sin p+x \sin (p+q)+ \) \( \left[\left(x^{2}\right) / 2 !\right] \sin (p+2 q)+\ldots \ldots . . \) is

A) \( e^{(x \cos q)}[\sin (p+x \sin q)] \)

B) \( e^{(x \cos q)} \)

C) 0

D) \( e^{(x \cos q)}[\cos (p+x \sin q)] \)

Answer 1

Given series is 

sin(p + x) sin(p + q) + x²/2! sin(p + 2q) + ....

We know that e^iα = cos α + i sin α.

We develop new series

cos p + x cos(p + q) + x²/2! cos(p + 2q) + ... + i(sin p + x sin(p + q) + x²/2! sin(p + 2q)+....)

= e^ip + xe^{i(p + q)} + x²/2! e^{i(p + 2q)}+.....

= e^ip(1 + x^iq + x²/2! e^i2q + ....)

= e^ip(1 + (x e^iq) + \(\frac{(xe^{i2q})^2}{2!}\)+....)

= e^ip. e^{\(xe^{iq}\)} (\(\because\) e^x = 1 + x + x²/2! + .......)

= e^ip + xe^iq

= e^{ip + x(cos q + i sin q)} (\(\therefore \) e^iα = cos α + i sin α)

= e^{xcosq + i (p + x sin q)}

 = e^xcosq. e^{i(p +x sin q)} (\(\because\) e^{a + b} = e^a. e^b)

= e^xcosq (cos(p +sin q) + i sin (p + x sin q))

By comparing imaginary part we get

sin p + x sin (p + q) + x²/2! sin(p + 2q) = e^xcosq sin(p + x sin q).

Hence, option (A) is correct.