Question

Prove that cosh iA=cos A where i is the square root of -1 and A is angle in radians.

Answer 1

\[\cosh (x)=\frac{e^x+e^{-x}}{2}\qquad\quad\qquad\frac{e^{ix}+e^{-ix}}{2}=\cos (x)\]
\[\cosh (iA)=\frac{e^{iA}+e^{-iA}}{2}\qquad\qquad\frac{e^{iA}+e^{-iA}}{2}=\cos (A)\]