Pythagorean identity for sin and cos

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Theorem

The following formula holds for all $z \in \mathbb{C}$: $$\sin^2(z)+\cos^2(z)=1,$$ where $\sin$ denotes the sine function and $\cos$ denotes the cosine function.

Proof

From the definitions $$\sin(z)=\dfrac{e^{iz}-e^{-iz}}{2i}$$ and $$\cos(z)=\dfrac{e^{iz}+e^{-iz}}{2},$$ using the square of i in the denominator of the first term, we see $$\begin{array}{ll} \sin^2(z)+\cos^2(z)&=\left( \dfrac{e^{iz}-e^{-iz}}{2i} \right)^2 + \left( \dfrac{e^{iz}+e^{-iz}}{2} \right)^2 \\ &= -\dfrac{1}{4} (e^{2iz}-2+e^{-2iz})+ \dfrac{1}{4} (e^{2iz}+2+e^{-2iz}) \\ &= 1, \end{array}$$ as was to be shown. █