Difference of Fourth Powers of Cosine and Sine

From ProofWiki

Jump to: navigation, search

Theorem

$\sin^4 x - \cos^4 x = \sin^2 x - \cos^2 x$

$\displaystyle $	$\displaystyle $	$\displaystyle $	$\displaystyle \sin^4 x - \cos^4 x$	$=$	$\displaystyle \sin^2 x \left({1 - \cos^2 x}\right) - \cos^2 x \left({1 - \sin^2 x}\right)$	$\displaystyle $	$\displaystyle $	$\displaystyle $	Sum of Squares of Sine and Cosine
$\displaystyle $	$\displaystyle $	$\displaystyle $	$\displaystyle $	$=$	$\displaystyle \sin^2 x - \sin^2 x \ \cos^2 x - \cos^2 x + \sin^2 x \ \cos^2 x$	$\displaystyle $	$\displaystyle $	$\displaystyle $
$\displaystyle $	$\displaystyle $	$\displaystyle $	$\displaystyle $	$=$	$\displaystyle \sin^2 x - \cos^2 x$	$\displaystyle $	$\displaystyle $	$\displaystyle $

$\blacksquare$

\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \sin^4 x - \cos^4 x\)	\(=\)	\(\displaystyle \sin^2 x \left({1 - \cos^2 x}\right) - \cos^2 x \left({1 - \sin^2 x}\right)\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	Sum of Squares of Sine and Cosine
\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(=\)	\(\displaystyle \sin^2 x - \sin^2 x \ \cos^2 x - \cos^2 x + \sin^2 x \ \cos^2 x\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)
\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(=\)	\(\displaystyle \sin^2 x - \cos^2 x\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)