Show that $\sin\left(\frac{\pi}{2}-x\right)=\cos x$ using complex exponentials.
This is where I've got to:
$$\sin\left(\frac{\pi}{2}-x\right)=\frac{i}{2}\left(e^{xi-\frac{\pi i}{2}}-e^{\frac{π}{2}-xi}\right)$$
What should I do next?
$\endgroup$ 11 Answer
$\begingroup$$$\begin{align}\sin{\left(\frac\pi2-x\right)} &=\frac1{2i}\left(e^{i(\frac\pi2-x)}-e^{-i(\frac\pi2-x)}\right)\\ &=\frac1{2i}\left(e^{i\frac\pi2-ix}-e^{-i\frac\pi2+ix}\right)\\ &=\frac1{2i}\left(e^{i\frac\pi2}(e^{-ix})-e^{-i\frac\pi2}(e^{ix})\right)\\ &=\frac1{2i}\left(i(e^{-ix})+i(e^{ix})\right)\\ &=\frac1{2}\left(e^{-ix}+e^{ix}\right)\\ &=\frac1{2}\left(e^{ix}+e^{-ix}\right)\\ &=\cos{(x)}\\ \end{align}$$Where I have used Euler's formula to find$$e^{i\frac\pi2}=\cos{\left(\frac\pi2\right)}+i\sin{\left(\frac\pi2\right)}=i$$$$e^{-i\frac\pi2}=\cos{\left(-\frac\pi2\right)}+i\sin{\left(-\frac\pi2\right)}=-i$$
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