The sum value of the series \( \sin p+x \sin (p+q)+ \) \( \left[\left(

1.	The sum value of the series \( \sin p+x \sin (p+q)+ \) \( \left[\left(x^{2}\right) / 2 !\right] \sin (p+2 q)+\ldots \ldots . . \) isA) \( e^{(x \cos q)}[\sin (p+x \sin q)] \)B) \( e^{(x \cos q)} \)C) 0D) \( e^{(x \cos q)}[\cos (p+x \sin q)] \)
Answer» 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.

The sum value of the series \( \sin p+x \sin (p+q)+ \) \( \left[\left(x^{2}\right) / 2 !\right] \sin (p+2 q)+\ldots \ldots . . \) isA) \( e^{(x \cos q)}[\sin (p+x \sin q)] \)B) \( e^{(x \cos q)} \)C) 0D) \( e^{(x \cos q)}[\cos (p+x \sin q)] \)

Answer»

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.