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gnlele · Jan 2, 2012, 09:14 PM

Verifying the identity

ebaines · Jan 3, 2012, 07:32 AM

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.

ebaines · Jan 4, 2012, 07:34 AM

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.