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Verify the identity by simplifying the left side.

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

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    $\begingroup$ p.s. trigonometric identity this is the way but I wouldn't call it "simplifying". $\endgroup$ – Kuba Dec 7 '13 at 14:00
  • $\begingroup$ Sum $(\sin (y))^2+(\cos(x))^2$to both sides of the equality. $\endgroup$ – Git Gud Dec 7 '13 at 14:15
  • $\begingroup$ Please, add some of your thoughts about the problem. $\endgroup$ – egreg Dec 7 '13 at 14:15
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We have $$\displaystyle \sin^2x+\cos^2x=\sin^2y+\cos^2y$$ as both are equal to $1$

Now change the sides of $\displaystyle \sin^2y,\cos^2x$

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This can be verified by using $\cos^2(x)=1-\sin^2(x)$ and $\cos^2(y)=1-\sin^2(y)$.

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$$\sin^2 x-\sin^2 y=1-\sin^2 y-(1-\sin^2 x)=\sin^2 x-\sin^2 y,$$ because $$\cos^2 x=1-\sin^2 x,$$ and $$\cos^2 y=1-\sin^2 y$$

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