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I need to prove this identity, any idea on how I do it?

$$\sin^2\alpha-\cos^2\beta = \sin^2\beta-\cos^2\alpha$$

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    $\begingroup$ Move the cosines around a bit. $\endgroup$ Sep 7, 2015 at 11:25
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    $\begingroup$ Basic trig identities should help here. Just look for a trig identity that matches the problem somewhat. $\endgroup$
    – jdods
    Sep 7, 2015 at 11:27
  • $\begingroup$ @DanielFischer How I haven't seen this.... thanks any way :) $\endgroup$
    – LiziPizi
    Sep 7, 2015 at 11:30
  • $\begingroup$ Too bad the equation wasn't $\sin^2\alpha + 1 -\cos^2\beta = \sin^2\beta +1- \cos^2\alpha$. Opps! Did I just give away the answer? $\endgroup$
    – John Joy
    Sep 7, 2015 at 13:06

2 Answers 2

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$$1= \sin^2\alpha+\cos^2\alpha = \sin^2\beta+\cos^2\beta$$ Then by moving cosine's to other sides, we have $$\sin^2\alpha-\cos^2\beta = \sin^2\beta-\cos^2\alpha$$

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Simply think to use the identity $\sin^2 x+\cos^2 x=1$. Using that for $x=a$ and $x=b$ you get:

$$\sin^2 \alpha+\cos^2 \alpha=1$$ $$\sin^2 \beta+\cos^2 \beta=1$$

Equating LHS, because RHS=1, you get: $$\sin^2 \alpha+\cos^2 \alpha=\sin^2 \beta+\cos^2 \beta$$

Now, rearrange and prove the final identity: $$\sin^2\alpha-\cos^2\beta = \sin^2\beta-\cos^2\alpha$$

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