Difference between revisions of "Pythagorean identity for sin and cos"

From specialfunctionswiki
Jump to: navigation, search
Line 1: Line 1:
 
<div class="toccolours mw-collapsible mw-collapsed">
 
<div class="toccolours mw-collapsible mw-collapsed">
<strong>[[Pythagorean identity for sin and cos|Theorem]]: (Pythagorean identity)</strong> The following formula holds for all $x$:
+
<strong>[[Pythagorean identity for sin and cos|Theorem]]: (Pythagorean identity)</strong> The following formula holds for all $z \in \mathbb{C}$:
$$\sin^2(x)+\cos^2(x)=1,$$
+
$$\sin^2(z)+\cos^2(z)=1,$$
 
where $\sin$ denotes the [[sine]] function and $\cos$ denotes the [[cosine]] function.
 
where $\sin$ denotes the [[sine]] function and $\cos$ denotes the [[cosine]] function.
 
<div class="mw-collapsible-content">
 
<div class="mw-collapsible-content">

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. █