Proof of half angle identity

Question

Remember one of the step in proving it is to replace theta with 1/2 A. I am wondering how could you replace a variable to prove the identity. A is not the same as theta. Plus aren't we trying to find half theta?

Answer 1

Presumably, you are asking to prove the half-angle identities for cosine and sine given the double-angle formulas for cosine: $$\begin{align} \cos 2 \theta &= 2 \cos^2 \theta - 1 &(1) \\ &= 1 - 2 \sin^2\theta &(2) \end{align}$$

The identities show that trig functions of "some angle" (shown as "$\theta$") on the right are equal to a trig function of "twice that angle" ("$2\theta$") on the left; as identities, these relations hold no matter what "some angle" is.

Being clever, we can replace "some angle" with "half some other angle", as in $\theta \to \alpha/2$, whereupon "twice that angle" becomes simply "the other angle" ($2\theta \to 2(\alpha/2) = \alpha$), and we have:

$$\begin{align} \cos\alpha &= 2 \cos^2\frac{\alpha}{2} - 1 &(1^\prime) \\ &= 1 - 2 \sin^2\frac{\alpha}{2} &(2^\prime) \end{align}$$

You can then solve these equations for $\cos^2\frac{\alpha}{2}$ and $\sin^2\frac{\alpha}{2}$ in terms of $\cos\alpha$.

(And then, if having the identities in terms of "$\alpha$" bothers you, you can change the name of the angle to "$\theta$", so that you have $\cos^2\frac{\theta}{2}$ and $\sin^2\frac{\theta}{2}$ in terms of $\cos\theta$. I suspect that the instructions tell you to use $\alpha$ to avoid some confusion with replacing $\theta \to \theta/2$ in the earlier step.)