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Question: The value of \({\sin h}(x + 2\pi ni)\) is....

The value of sin⁡h(x+2πni){\sin h}(x + 2\pi ni) is.

A

ex+e−x2\frac{e^{x} + e^{- x}}{2}

B

ex−e−x2\frac{e^{x} - e^{- x}}{2}

C

ex−e−x2i\frac{e^{x} - e^{- x}}{2i}

D

ex+e−x2i\frac{e^{x} + e^{- x}}{2i}

Answer

ex−e−x2\frac{e^{x} - e^{- x}}{2}

Explanation

Solution

Period of sin⁡hx=2πi{\sin h}x = 2\pi i

Hence value of sinh⁡(x+2nπi)=ex−e−x2\sinh(x + 2n\pi i) = \frac{e^{x} - e^{- x}}{2}