Question

Imaginary part of $\cosh \left(\alpha +i\beta \right)-\cosh \left(\alpha -i\beta \right)$

• A. $2\sinh \alpha \sinh \beta$
• B. $2\sinh \alpha \sin \beta$
• C. $2\cos \text{h}\alpha \cos \text{h}\beta$
• D. $2\cos \alpha \cos \beta$

Answer 1

The correct option is B

$2\sinh \alpha \sin \beta$

Explanation for the correct option.

Find the imaginary part.

Expand the expression $\cosh \left(\alpha +i\beta \right)-\cosh \left(\alpha -i\beta \right)$ using the formula $\cosh (A+B)-\cosh (A-B)=2\sinh A\sinh B$.

$\begin{array}{rcl}\cosh \left(\alpha +i\beta \right)-\cosh \left(\alpha -i\beta \right)& =& 2\sinh \left(\alpha \right)\sinh \left(i\beta \right)\\ & =& 2\sinh \left(\alpha \right)·i\sin \left(\beta \right)\left[\sinh \left(iz\right)=i\sin \left(z\right)\right]\\ & =& \left(2\sinh \alpha \sin \beta \right)i\end{array}$

So the imaginary part of $\cosh \left(\alpha +i\beta \right)-\cosh \left(\alpha -i\beta \right)$ is $2\sinh \alpha \sin \beta$.

Hence, the correct option is B.