Proving identity $\cos^2 a= \sinh^2 b$ if $\sin(a+ib)=\cos b + i\sin b$

Question

This is my first question on StackExchange. I have some trouble with proving this identity with the following condition:

If: $\sin(a+ib)= \cos b + i \sin b$

Prove: $\cos^2 (a) = \sinh^2 (b)$

I tried using different methods, but the furthest I've come to is this:

$\sinh^2 (b) = \sin^2 (b)/\cos^2 (a)$

I expanded $\sin(a+ib)$ and rearranged the equation to isolate $\sinh(b)$. I then squared it (to get $\sinh^2 (b)$) and got the above statement.

I would like to know whether I'm on the right track, or if there are any mistakes. Thanks a lot.

Answer 1

\begin{align} \sin(a+ib) &=\sin a\cos ib+\cos a\sin ib\\ &=\sin a\cosh b+i\cos a\sinh b\\ &= \cos b + i \sin b \end{align} so take real part and imaginary of sides give us $$\sin a\cosh b= \cos b$$ $$\cos a\sinh b= \sin b$$ then squaring two equations and adding concludes $$\sin^2 a\cosh^2 b+\cos^2 a\sinh^2 b=1$$ $$(1-\cos^2 a)(1+\sinh^2 b)+\cos^2 a\sinh^2 b=1$$ which gives $\color{blue}{\cos^2 a = \sinh^2 b}$.

Answer 2

Use that $$\sin(a+bi)=\sin(a)\cosh(b)+i\cos(a)\sinh(b)$$