Question

The imaginary part of is ${\sin }^2(x+iy)$ is

• A. $\frac{1}{2}\cos h\left(2x\right)\cos 2y$
• B. $\frac{1}{2}\cos 2x\cos h\left(2y\right)$
• C. $\frac{1}{2}\sin h\left(2x\right)\sin 2y$
• D. $\frac{1}{2}\sin 2x\sin h\left(2y\right)$

Answer 1

The correct option is D

$\frac{1}{2}\sin 2x\sin h\left(2y\right)$

Find the imaginary part of the given expression

Given expression, ${\sin }^2(x+iy)$ can be simplified as,

$2{\sin }^2\left(x+iy\right)=1-\cos 2\left(x+iy\right)$

$=1-\left(\cos 2x\cos 2iy-\sin 2x\sin 2iy\right)$

$=1-\cos 2x\cos h\left(2y\right)+i\sin 2x\sin h\left(2y\right)$

$\therefore$${\sin }^2(x+iy)=$$\frac{1-\cos 2x\cos h\left(2y\right)}{2}+i\frac{\sin 2x\sin h\left(2y\right)}{2}$

So, the imaginary part of ${\sin }^2(x+iy)$ is $\frac{1}{2}\sin 2x\sin h\left(2y\right)$

Hence, option $\left(D\right)$ is the correct answer.