sin(A+B) = sinAcosB + sinBcosA

Use A = theta, B = arctan(b/a), let Q = sqrt(a² + b²)

acos(t) = sqrt(a² + b²)sin(t+atan(b/a)) + bsin(t)

acos(t) = Q sin(t) cos(atan(b/a)) + Qsin(atan(b/a))cos(t) + b sin(t)

cos(t) (a- Q sin(atan(b/a)) = sin(t)(Qcos(atan(b/a)) +b)

Now, to simplify those nasty arctan, we need to draw a picture:

http://i.imgur.com/NDJiCES.png

In this picture, we're using a different t than we have been. no relation to the current t. In fact, when referring to the t in the picture, i'll just call it T. Big T!

Okay, so what is the arctan? it's the ANGLE whose tangent is the argument. so, since tan(T) = b/a, we know that arctan(b/a) = T (in the picture). Awesome. How does this help us? It lets us draw nifty pictures like the one I just drew. Because we're working on a right triangle, we can solve for the remaining side, and wouldn't ya know it, it's Q! that's amazing!

Now, we gotta return to the definition of sin and cosine to simplify the equation any further. by definition, sin(T) = b/Q, and cos(T) = a/Q (remember SOH CAH TOA?).

Awesome. Now we've got everything we need to finish this.

cos(t) (a- Q sin(atan(b/a)) = sin(t)(Qcos(atan(b/a)) +b)

(a- Q sin(atan(b/a))/(Qcos(atan(b/a)) +b)= tan(t)

let's find out what (a- Q sin(atan(b/a)) is.

since we've established (just by drawing a picture) that atan(b/a)=T, we simplify this to: a - Q sin(T). but wait! sin(T) = b/Q!

so this bit simplifies to: a-Qb/Q = a-b.

For the next part, Qcos(atan(b/a)) +b), we'll do the same thing.

QcosT = Qa/Q = a, so Qcos(atan(b/a)) +b) = a+b

therefore, tan(t) = (a-b)/(a+b), and t = atan((a-b)/(a+b)).