Difference between revisions of "Pythagorean identity for sin and cos"
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and | and | ||
$$\cos(z)=\dfrac{e^{iz}+e^{-iz}}{2},$$ | $$\cos(z)=\dfrac{e^{iz}+e^{-iz}}{2},$$ | ||
| − | we see | + | using the [[square of i]] in the denominator of the first term, we see |
$$\begin{array}{ll} | $$\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 \\ | \sin^2(z)+\cos^2(z)&=\left( \dfrac{e^{iz}-e^{-iz}}{2i} \right)^2 + \left( \dfrac{e^{iz}+e^{-iz}}{2} \right)^2 \\ | ||
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&= 1, | &= 1, | ||
\end{array}$$ | \end{array}$$ | ||
| − | as was to be shown. █ | + | as was to be shown. █ |
==References== | ==References== | ||
Latest revision as of 18:51, 15 December 2016
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. █
