Question

The sum of the series $\sin \alpha +x\sin (\alpha +\beta )+\frac{x^2}{2!}\sin (\alpha +2\beta )+\dots \dots ..\infty$ is

• A. $e^{x\cos \beta }\sin (\alpha +x\sin \beta )$
• B. $e^{x\sin \beta }\sin (\alpha +x\sin \beta )$
• C. $e^{xs\dot{m}\beta }\cos (\alpha +x\sin \beta )$
• D. $e^{xs\dot{m}\beta }\cos (\alpha -x\sin \beta )$

Answer 1

The correct option is A $e^{x\cos \beta }\sin (\alpha +x\sin \beta )$

We know,
$cos\alpha +isin\alpha =e^{i\alpha }$
Multiplying given series by i and adding terms of $cos\alpha +xcos(\alpha +\beta )+x^{2}cos(\alpha +2\beta )$
We develop new series
$\Rightarrow cos\alpha +xcos(\alpha +\beta )......+i(sin\alpha +xsin(\alpha +\beta )+\frac{x^2}{2!}........)$
$=e^{i\alpha }+x{e}^{i(\alpha +\beta )}+\frac{x^2}{2!}{e}^{i(\alpha +2\beta )}+\frac{x^3}{3!}{e}^{i(\alpha +3\beta )}+.....soon$
$e^{i\alpha }(1+(x{e}^{i\beta })+\frac{(x{e}^{i\beta }{)}^2}{2!}+\frac{(x{e}^{e\beta }{)}^3}{3!})$ $\to$
$e^{i\alpha }\underset{}{(1+(x{e}^{i\beta })+\frac{(x{e}^{i\beta }{)}^2}{2!}+\frac{(x{e}^{e\beta }{)}^3}{3!})}$
Expansion of $e^p$ where $p=x{e}^{i\beta }$
$=e^{i\alpha }\left(e^{x{e}^{i\beta }}\right)$
The required summation is actually the imaginary part of our new series, thus comparing imaginary parts
$S=Im$ {$e^{i\alpha }\left(e^{xcos\beta +ixsin\beta }\right)$}$=Im${$e^{xcos\beta }\cdot e^{i(\alpha +xsin\beta )}$}
$=e^{xcos\beta }sin(\alpha +xsin\beta )$