Difference between revisions of "Pythagorean identity for sin and cos"
From specialfunctionswiki
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$$\sin(z)=\dfrac{e^{iz}-e^{-iz}}{2i}$$ | $$\sin(z)=\dfrac{e^{iz}-e^{-iz}}{2i}$$ | ||
and | and | ||
| − | $$\cos(z)=\dfrac{e^{ | + | $$\cos(z)=\dfrac{e^{iz}+e^{-iz}}{2},$$ |
we see | we see | ||
$$\begin{array}{ll} | $$\begin{array}{ll} | ||
| − | \sin^2(z)&=\left( \dfrac{e^{iz}-e^{-iz}}{2i} \right)^2 + \left( \dfrac{e^{iz}+e^{-iz}}{2} \right)^2 \\ | + | \sin^2(z)+\cos^2(z)&=\left( \dfrac{e^{iz}-e^{-iz}}{2i} \right)^2 + \left( \dfrac{e^{iz}+e^{-iz}}{2} \right)^2 \\ |
&= -\dfrac{1}{4} (e^{2iz}-2+e^{-2iz})+ \dfrac{1}{4} (e^{2iz}+2+e^{-2iz}) \\ | &= -\dfrac{1}{4} (e^{2iz}-2+e^{-2iz})+ \dfrac{1}{4} (e^{2iz}+2+e^{-2iz}) \\ | ||
&= 1, | &= 1, | ||
Revision as of 20:41, 15 May 2016
Theorem: (Pythagorean identity) The following formula holds for all $x$: $$\sin^2(x)+\cos^2(x)=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},$$ we see $$\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. █
