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

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(Proof)
 
Line 9: Line 9:
 
and
 
and
 
$$\cos(z)=\dfrac{e^{iz}+e^{-iz}}{2},$$
 
$$\cos(z)=\dfrac{e^{iz}+e^{-iz}}{2},$$
we see
+
using the [[square of i]] in the denominator of the first term, we see
 
$$\begin{array}{ll}
 
$$\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 \\
 
\sin^2(z)+\cos^2(z)&=\left( \dfrac{e^{iz}-e^{-iz}}{2i} \right)^2 + \left( \dfrac{e^{iz}+e^{-iz}}{2} \right)^2 \\
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&= 1,
 
&= 1,
 
\end{array}$$  
 
\end{array}$$  
as was to be shown. █  
+
as was to be shown. █
  
 
==References==
 
==References==

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. █

References