$\sin h i x$ is equal to
$\sin h i x$ is equal to
- $i \sin x$
- $\sin i x$
- $-i \sin x$
- $i \sin i x$
Solution
We have
$
\begin{aligned}
\sin h x & =\frac{e^x-e^{-x}}{2} \\
\sin h i x & =\frac{e^{i x}-e^{-i x}}{2}=i \sin x
\end{aligned}
$
Asked in: AP EAMCET 2002
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