Cosine of Sum/Proof 2

Theorem

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

where $\cos$ is the cosine.

Proof
Recall the analytic definitions of sine and cosine:
 * $\displaystyle \sin x = \sum_{n \mathop = 0}^\infty \left({-1}\right)^n \frac {x^{2n+1}}{\left({2n+1}\right)!}$


 * $\displaystyle \cos x = \sum_{n \mathop = 0}^\infty \left({-1}\right)^n \frac {x^{2n}}{\left({2n}\right)!}$

Let:

Let us derive these with respect to $a$, keeping $b$ constant.

Then from Derivative of Sine Function and Derivative of Cosine Function, we have:

Hence:

Thus from Derivative of Constant, $\forall a \in \R: g \left({a}\right)^2 + h \left({a}\right)^2 = c$.

In particular, it is true for $a = 0$, and $g \left({0}\right)^2 + h \left({0}\right)^2 = 0$.

So $g \left({a}\right)^2 + h \left({a}\right)^2 = 0$

But $g \left({a}\right)^2 \ge 0$ and $h \left({a}\right)^2 \ge 0$ from Even Powers are Positive.

So it follows that $g \left({a}\right) = 0$ and $h \left({a}\right) = 0$.

Hence the result.