2
$\begingroup$

How to solve the following question?

If $n$ is an integer, show that

\begin{eqnarray} \left(\frac{1+\sin x+i\cos x}{1+\sin x-i\cos x}\right)^n=\cos n\left(\frac{\pi }{2}-x\right)+i\sin n\left(\frac{\pi}{2}-x\right) \end{eqnarray}

where x is a real number such that $\left(1+\sin x\right)^2+\cos^2x>0$.

My work enter image description here enter image description here

I have no idea how to continue...

Thank you for your help.

$\endgroup$
2
$\begingroup$

$$\frac{1+\sin x+i\cos x}{1+\sin x-i\cos x}=\frac{(1+\sin x+i\cos x)^2}{(1+\sin x)^2+\cos^2 x}$$

$$=\frac{2(1+\sin x)(\sin x+i\cos x)}{2(1+\sin x)}=\cos\left(\frac\pi2-x\right)+i\sin\left(\frac\pi2-x\right)$$

Apply de Movire's Theorem assuming $1+\sin x\ne0$

as $\sin x+1=0\implies 1+\sin x-i\cos x=0$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.