# I need to prove this identity, any idea on how I do it? sin 2 </msup> &#x2061

I need to prove this identity, any idea on how I do it?
${\mathrm{sin}}^{2}\alpha -{\mathrm{cos}}^{2}\beta ={\mathrm{sin}}^{2}\beta -{\mathrm{cos}}^{2}\alpha$
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Tristin Case
$1={\mathrm{sin}}^{2}\alpha +{\mathrm{cos}}^{2}\alpha ={\mathrm{sin}}^{2}\beta +{\mathrm{cos}}^{2}\beta$
Then by moving cosine's to other sides, we have
${\mathrm{sin}}^{2}\alpha -{\mathrm{cos}}^{2}\beta ={\mathrm{sin}}^{2}\beta -{\mathrm{cos}}^{2}\alpha$
###### Not exactly what you’re looking for?
sweetymoeyz
Simply think to use the identity ${\mathrm{sin}}^{2}x+{\mathrm{cos}}^{2}x=1$. Using that for x=a and x=b you get:
${\mathrm{sin}}^{2}\alpha +{\mathrm{cos}}^{2}\alpha =1$
${\mathrm{sin}}^{2}\beta +{\mathrm{cos}}^{2}\beta =1$
Equating LHS, because RHS=1, you get:
${\mathrm{sin}}^{2}\alpha +{\mathrm{cos}}^{2}\alpha ={\mathrm{sin}}^{2}\beta +{\mathrm{cos}}^{2}\beta$
Now, rearrange and prove the final identity:
${\mathrm{sin}}^{2}\alpha -{\mathrm{cos}}^{2}\beta ={\mathrm{sin}}^{2}\beta -{\mathrm{cos}}^{2}\alpha$