Question

Find the value of $\cosh \left( ix \right)$.
(a) $\cos \left( x \right)$
(b) $i\times \cos \left( x \right)$
(c) $-\cos \left( x \right)$
(d) $i\times \cos \left( x \right)$

Answer 1

Consider $i$ as the imaginary number $\sqrt{-1}$ and use the formula for the exponential form of the hyperbolic function $\cosh \left( x \right)$ given as $\cosh \left( x \right)=\dfrac{{{e}^{x}}+{{e}^{-x}}}{2}$. Substitute $ix$ in place of x to get the expression for $\cosh \left( ix \right)$. Now, use the Euler’s formulas in complex number given as ${{e}^{ix}}=\cos x+i\sin x$ and substitute –x in place of x to find the value of ${{e}^{-ix}}$. Simplify the obtained expression to get the answer.

Complete step by step answer:
Here we have been provided with the hyperbolic function $\cosh \left( ix \right)$ and we are asked to find its simplified form. Let us use the exponential form of the hyperbolic function and some properties of the complex numbers to get the answer.
Now, here $i$ is the imaginary number $\sqrt{-1}$, so using the formula for the hyperbolic function $\cosh \left( x \right)$ given as $\cosh \left( x \right)=\dfrac{{{e}^{x}}+{{e}^{-x}}}{2}$ in exponential form we get on replacing x with $ix$,
$\Rightarrow \cosh \left( ix \right)=\dfrac{{{e}^{ix}}+{{e}^{-ix}}}{2}$
Now, in complex numbers we have Euler’s formula relating the exponential function having complex exponential and trigonometric functions (sine and cosine function) given as ${{e}^{ix}}=\cos x+i\sin x$, so using this formulas and also replacing x with –x to get the value of ${{e}^{-ix}}$ we get,
$\Rightarrow \cosh \left( ix \right)=\dfrac{\left[ \cos \left( x \right)+i\sin \left( x \right) \right]+\left[ \cos \left( -x \right)+i\sin \left( -x \right) \right]}{2}$
Using the properties $\cos \left( -x \right)=\cos x$ and $\sin \left( -x \right)=-\sin x$ we get,
$\Rightarrow \cosh \left( ix \right)=\dfrac{\cos \left( x \right)+i\sin \left( x \right)+\cos \left( x \right)-i\sin \left( x \right)}{2}$
Cancelling the like terms and simplifying we get,
$\begin{aligned} & \Rightarrow \cosh \left( ix \right)=\dfrac{2\cos \left( x \right)}{2} \\\ & \therefore \cosh \left( ix \right)=\cos \left( x \right) \\\ \end{aligned}$

So, the correct answer is “Option a”.

Note: Note that you must not consider $i$ as any variable otherwise you will not be able to use the Euler’s formula to simplify the expression. Always remember that $i$ is the solution of the quadratic equation ${{x}^{2}}+1=0$ and it is an imaginary number. You must remember the above used trigonometric identities so that you can cancel the like terms wherever possible.