Difference between revisions of "Pythagorean identity for sin and cos"
From specialfunctionswiki
(→Proof) |
|||
| (7 intermediate revisions by the same user not shown) | |||
| Line 1: | Line 1: | ||
| − | + | ==Theorem== | |
| − | + | The following formula holds for all $z \in \mathbb{C}$: | |
| − | $$\sin^2( | + | $$\sin^2(z)+\cos^2(z)=1,$$ |
| − | where $\sin$ denotes the [[sine | + | where $\sin$ denotes the [[sine]] function and $\cos$ denotes the [[cosine]] function. |
| − | + | ||
| − | + | ==Proof== | |
| − | + | From the definitions | |
| − | + | $$\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}$$ | ||
| + | as was to be shown. █ | ||
| + | |||
| + | ==References== | ||
| + | |||
| + | [[Category:Theorem]] | ||
| + | [[Category:Proven]] | ||
Latest revision as of 18:51, 15 December 2016
Theorem
The following formula holds for all $z \in \mathbb{C}$: $$\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. █
