Cosine of Sum

Theorem

 * $\cos \left({a + b}\right) = \cos a \cos b - \sin a \sin b$


 * $\sin \left({a + b}\right) = \sin a \cos b + \cos a \sin b$

where $\sin$ and $\cos$ are sine and cosine.

Corollary 1

 * $\cos \left({a - b}\right) = \cos a \cos b + \sin a \sin b$


 * $\sin \left({a - b}\right) = \sin a \cos b - \cos a \sin b$

Corollary 2

 * $\displaystyle \tan \left({a + b}\right) = \frac {\tan a + \tan b} {1 - \tan a \tan b}$


 * $\displaystyle \tan \left({a - b}\right) = \frac {\tan a - \tan b} {1 + \tan a \tan b}$

where $\tan$ is tangent.

Proof of Corollary 1

 * From Cosine Function is Even we have $\cos \left({- b}\right) = \cos b$


 * From Sine Function is Odd we have $\sin \left({- b}\right) = - \sin b$.

Thus:

Similarly:

So:

Proof of Corollary 2

 * $\displaystyle \tan \left({a + b}\right) = \frac {\tan a + \tan b} {1 - \tan a \tan b}$:


 * $\displaystyle \tan \left({a - b}\right) = \frac {\tan a - \tan b} {1 + \tan a \tan b}$:

As above, we have:

Thus:
 * $\tan \left({- b}\right) = - \tan b$

Therefore:

Historical Note
These formulas were proved by François Viète in about 1579.