Trig addition formulas

sin(a + b) = sin a cos b + cos a sin b

cos(a + b) = cos a cos b - sin a sin b

Proof of the addition formulas, in the case when a, b, and a + b are acute angles:

Construct right triangles OEP and OGQ so that a is the angle measure of EOP, b is the angle measure of GOQ, G lies on OP, OP = 1, and OQ = 1. Let GD be the perpendicular from G to OE, let QC be the perpendicular from Q to OE, and let GF be the perpendicular from G to QC. Let x = OE, y = PE, u = OG, v = QG, s = QF, t = FC. Thus we have x = cos a, y = sin a, u = cos b, v = sin b, and s + t = sin(a + b).

Let a' be the angle measure of QGF. Now angle FGO has measure a, since FG is parallel to OE. Thus a + a' = 90, so cos a = sin a'. From right triangle FGQ we have sin a' = s/v. Thus s = v cos a = cos a sin b. From right triangle GDO we have sin a = t/u, so t = u sin a = sin a cos b.

Since sin(a + b) = s + t, we conclude that sin(a + b) = sin a cos b + cos a sin b.

To prove the cosine addition formula we use the same two right triangles FGQ and GDO. We have sin a = cos a', and from the right triangle FGQ we have cos a' = (CD)/v. Thus CD = v sin a, and so CD = sin a sin b. From the right triangle GDO we have cos a = (OD)/u, so OD = u cos a, and so OD = cos a cos b. Since cos(a + b) = OC, and OC = OD - CD, we conclude that cos (a + b) = cos a cos b - sin a sin b.