Question

proof the formula for cos(A+B)=cosAcosB-sinAsinB

Answer 1

http://www.khanacademy.org/math/trigonometry/v/proof--cos-a-b-----cos-a--cos-b---sin-a--sin-b

Answer 2

We know that sin(a + b) = sin(a) cos(b) + sin(b) cos(a)
and that cos(a) = sin(90-a)

cos (A+B)
= cos (A-(-B))
= sin (90-(A-(-B))
= sin (90-A-B)
= sin ((90-A)-B)
=sin(90-A) cos(B)- sin(B) cos(90-A)
=cos(A) cos(B) - sin(B) sin(A)

Answer 3

there's a geometric one, i like this one personally
\[e^{i A} = \cos A + i \sin A\]

\[e^{i A + B} = \cos(A+B) + i \sin(A+B) = (\cos A + i \sin A)(\cos B + i \sin B)\\ \]

equating real parts

cos(A+B) = cosAcosB - sinAsinB

Answer 4

and equating imaginary parts we have
sin(A+B) = sinAcosB + cosAsinB