Question

The real part of $\sin (\alpha +i\beta )$ is

• A. $\sin \alpha .\frac{e^{\beta }+e^{-\beta }}{2}$
• B. $\cos \alpha .\frac{e^{\beta }+e^{-\beta }}{2}$
• C. $\sin \alpha$
• D. No real part

Answer 1

The correct option is A $\sin \alpha .\frac{e^{\beta }+e^{-\beta }}{2}$

$\sin (\alpha +i\beta )=\sin \alpha .\cos \left(i\beta \right)+\cos \alpha .\sin \left(i\beta \right)$
$\cos \left(i\beta \right)=1-\frac{{\left(i\beta \right)}^2}{2!}+\frac{{\left(i\beta \right)}^4}{4!}-...$
$=1+\frac{{\beta }^2}{2!}+\frac{{\beta }^4}{4!}+...$
$\frac{e^{\beta }+e^{-\beta }}{2}=\frac{1}{2}[(1+\frac{\beta }{1!}+\frac{{\beta }^2}{2!}+...)+(1-\frac{\beta }{1!}+\frac{{\beta }^2}{2!}-...)]$
$=1+\frac{{\beta }^2}{2!}+\frac{{\beta }^4}{4!}+...=\cos \left(i\beta \right)$
$\sin \left(i\beta \right)=i\beta -\frac{{\left(i\beta \right)}^3}{3!}+\frac{{\left(i\beta \right)}^5}{5!}-...$
$=i[\beta +\frac{{\beta }^3}{3!}+\frac{{\beta }^5}{5!}-...]=\frac{i[e^{\beta }+e^{-\beta }]}{2}$
$\therefore \sin (\alpha +i\beta )=\sin \alpha .\frac{e^{\beta }+e^{-\beta }}{2}+\cos \alpha .i\left(\frac{e^{\beta }+e^{-\beta }}{2}\right)$
$\therefore$ Real part of $\sin (\alpha +i\beta )=\sin \alpha .\frac{e^{\beta }+e^{-\beta }}{2}$