Question

The value of $\sinh \left(x+\mathit{2}\pi ni\right)$ is

• A. $\frac{e^x+e^{-x}}{2}$
• B. $\frac{e^x-e^{-x}}{2}$
• C. $\frac{e^x-e^{-x}}{2i}$
• D. $\frac{e^x+e^{-x}}{2i}$

Answer 1

The correct option is B

$\frac{e^x-e^{-x}}{2}$

The explanation for the correct option

The given trigonometric expression: $\sinh \left(x+\mathit{2}\pi ni\right)$.

$$\begin{gathered} \sinh \left(x+\mathit{2}\pi ni\right)=\frac{e^{\left(x+\mathit{2}\pi ni\right)}-e^{-\left(x+\mathit{2}\pi ni\right)}}{2} \\ \Rightarrow \sinh \left(x+\mathit{2}\pi ni\right)=\frac{\left[e^x\times e^{\mathit{2}\pi ni}\right]-\left[e^{-x}\times e^{-\mathit{2}\pi ni}\right]}{2} \\ \Rightarrow \sinh \left(x+\mathit{2}\pi ni\right)=\frac{e^x\left[\cos \left(2\mathrm{\pi }n\right)+i\sin \left(2\mathrm{\pi }n\right)\right]-e^{-x}\left[\cos \left(-2\mathrm{\pi }n\right)+i\sin \left(-2\mathrm{\pi }n\right)\right]}{2} \\ \Rightarrow \sinh \left(x+\mathit{2}\pi ni\right)=\frac{e^x\left[1+i·0\right]-e^{-x}\left[1+i·0\right]}{2} \\ \Rightarrow \sinh \left(x+\mathit{2}\pi ni\right)=\frac{e^x-e^{-x}}{2} \end{gathered}$$

Therefore, the value of $\sinh \left(x+\mathit{2}\pi ni\right)$ is $\frac{e^x-e^{-x}}{2}$.

Hence, the correct option is (B).