1
$\begingroup$

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

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

$\endgroup$
  • 5
    $\begingroup$ Move the cosines around a bit. $\endgroup$ – Daniel Fischer Sep 7 '15 at 11:25
  • 1
    $\begingroup$ Basic trig identities should help here. Just look for a trig identity that matches the problem somewhat. $\endgroup$ – jdods Sep 7 '15 at 11:27
  • $\begingroup$ @DanielFischer How I haven't seen this.... thanks any way :) $\endgroup$ – LiziPizi Sep 7 '15 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 '15 at 13:06
2
$\begingroup$

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

$\endgroup$
2
$\begingroup$

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

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.