Verifying the identity

gnlele - please note that you can't cut and paste math formulas from other applications into this web site. I thought perhaps you meant this:


\frac {\sin^2 \theta} 2 = \frac {(sin^2 \theta + \cos \theta -1)} {2 \cos \theta


but that's not correct. Please retype the formula so we can see what you are asking.

OK, I think I've got it. The identity you are looking for is this:


sin^2(\frac {a} 2) = \frac {\sin^2 a + \cos a -1}{2 \cos a}


** I have edited the original subject title to reflect this correction **

The trick here is to substitute \omega = a/2. This turns the problem into:


sin^2(\omega) = \frac {\sin^2 (2 \omega)+ \cos(2 \omega) -1}{2 \cos(2 \omega)}


Then use the usual identities for \sin (2 \omega) and \cos (2 \omega) :


\sin(2 \omega) = 2 \sin \omega \cos \omega\\
\cos ( 2 \omega ) = 1 - 2 \sin^2 \omega


The rest is just algebra.