The half angle identities come from the power reduction formulas using the key substitution alpha=theta/2 twice, once on the left and right sides of the equation. With half angle identies, on the left side, this yields (after a square root) cos(theta/2) or sin(theta/2); on the right side cos(2theta) becomes cos( ) because 2(1/2)=1. For a problem like sin(pi/12), remember that theta/2=pi/12, or theta=pi/6, when substituting into the identity.
I want to use the power reduction formulas to derive the half angle formulas. Recall the power reduction formulas cosine squared alpha equals 1 plus cosine 2 alpha over 2 and sine squared alpha equals 1 minus cosine 2 alpha over 2. Now the way to derive the half angle formulas is to make the substitution alpha equals theta over 2. Now if I make that substitution, this becomes cosine squared theta over 2 equals 1 plus cosine theta over 2. All I have to do is extract square roots and I get cosine of theta over 2 is plus or minus the square root of 1 plus cosine theta over 2. Now with the plus or minus the plus or minus stays in the identity. The way we determine the sine of the final answer is we have to draw a unit circle and determine what quadrant we're in, so there's little bit of work involved but we can't say in the formula whether or not we're going to get a positive or negative answer. Now similarly we make the same substitution alpha becomes theta over 2 in the sine squared identity and we get 1 minus cosine theta over 2 and here again we extract the square root and so we get sine of theta over 2 is plus or minus the square root of 1 minus cosine theta over 2. These are the half angle identities sine of theta over 2 plus or minus the square root of 1 minus cosine theta over 2 and the cosine half angle identity which is identical except the minus becomes a plus.