Difference between revisions of "Pythagorean identity for sin and cos"
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| − | <strong>[[Pythagorean identity for sin and cos|Theorem]]: (Pythagorean identity)</strong> The following formula holds for all $ | + | <strong>[[Pythagorean identity for sin and cos|Theorem]]: (Pythagorean identity)</strong> The following formula holds for all $z \in \mathbb{C}$: |
| − | $$\sin^2( | + | $$\sin^2(z)+\cos^2(z)=1,$$ |
where $\sin$ denotes the [[sine]] function and $\cos$ denotes the [[cosine]] function. | where $\sin$ denotes the [[sine]] function and $\cos$ denotes the [[cosine]] function. | ||
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Revision as of 20:42, 15 May 2016
Theorem: (Pythagorean identity) The following formula holds for all $z \in \mathbb{C}$: $$\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},$$ we see $$\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. █
