# Proof of Compound Angles?

• November 23rd 2007, 07:05 AM
qspeechc
Proof of Compound Angles?
Could someone please give me the proofs, or point me to a website, of the compound angles formulae?
That is:
cos(A+B) = cosA.cosB - sinA.sinB

And the other ones, cos(A-B), sin(A+B), sin(A-B), tan(A+B), tan(A-B).

Are they just geometrical proofs, or are there algebraic ones? Thank-you.
• November 23rd 2007, 07:14 AM
topsquark
Quote:

Originally Posted by qspeechc
Could someone please give me the proofs, or point me to a website, of the compound angles formulae?
That is:
cos(A+B) = cosA.cosB - sinA.sinB

And the other ones, cos(A-B), sin(A+B), sin(A-B), tan(A+B), tan(A-B).

Are they just geometrical proofs, or are there algebraic ones? Thank-you.

The only proofs I have seen are geometric.

-Dan
• November 23rd 2007, 07:33 AM
ThePerfectHacker
The formal mathematical proof uses power series and the Cauchy product formula. But I doubt that is how you want it done.
• November 23rd 2007, 07:51 AM
qspeechc
Thank-you for the link topsquark, and TPH, I would like to see the power series proof, and you are correct in assuming that I do not know about Cauchy product formulae.
• November 23rd 2007, 08:47 AM
ThePerfectHacker
Quote:

Originally Posted by qspeechc
Thank-you for the link topsquark, and TPH, I would like to see the power series proof, and you are correct in assuming that I do not know about Cauchy product formulae.

The Power series proof is to know that,
$\sin x = \sum_{n=0}^{\infty} \frac{(-1)^nx^{2n+1}}{(2n+1)!}$
$\cos x = \sum_{n=0}^{\infty} \frac{(-1)^nx^{2n}}{(2n)!}$
You can to show,
$\sin(x+y) = \sin x \cos y + \cos x \sin y$
Now substitute those formulas and use the Cauchy product:
$\sum_{n=0}^{\infty}a_n \cdot \sum_{n=0}^{\infty}b_n = \sum_{n=0}^{\infty} c_n \mbox{ where }c_n = a_nb_0+a_{n-1}b_1+...+a_1b_{n-1}+a_0b_n$.