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

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

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

• Move the cosines around a bit. Sep 7, 2015 at 11:25
• Basic trig identities should help here. Just look for a trig identity that matches the problem somewhat. Sep 7, 2015 at 11:27
• @DanielFischer How I haven't seen this.... thanks any way :) Sep 7, 2015 at 11:30
• 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? Sep 7, 2015 at 13:06

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