Difference between revisions of "Pythagorean identity for sin and cos"
From specialfunctionswiki
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where $\sin$ denotes the [[sine]] function and $\cos$ denotes the [[cosine]] function. | where $\sin$ denotes the [[sine]] function and $\cos$ denotes the [[cosine]] function. | ||
<div class="mw-collapsible-content"> | <div class="mw-collapsible-content"> | ||
| − | <strong>Proof:</strong> █ | + | <strong>Proof:</strong> From the definitions |
| + | $$\sin(z)=\dfrac{e^{iz}-e^{-iz}}{2i}$$ | ||
| + | and | ||
| + | $$\cos(z)=\dfrac{e^{-iz}+e^{-iz}}{2},$$ | ||
| + | we see | ||
| + | $$\begin{array}{ll} | ||
| + | \sin^2(z)&=\left( \dfrac{e^{iz}-e^{-iz}}{2i} \right)^2 + \left( \dfrac{e^{iz}+e^{-iz}}{2} \right)^2 \\ | ||
| + | &= -\dfrac{1}{4} (e^{2iz}-2+e^{-2iz})+ \dfrac{1}{4} (e^{2iz}+2+e^{-2iz}) \\ | ||
| + | &= 1, | ||
| + | \end{array}$$ | ||
| + | as was to be shown. █ | ||
</div> | </div> | ||
</div> | </div> | ||
Revision as of 04:13, 25 March 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 $$\begin{array}{ll} \sin^2(z)&=\left( \dfrac{e^{iz}-e^{-iz}}{2i} \right)^2 + \left( \dfrac{e^{iz}+e^{-iz}}{2} \right)^2 \\ &= -\dfrac{1}{4} (e^{2iz}-2+e^{-2iz})+ \dfrac{1}{4} (e^{2iz}+2+e^{-2iz}) \\ &= 1, \end{array}$$ as was to be shown. █
