Sine of Sum/Proof 2

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


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

Let:

Let us differentiate 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 so:
 * $g \left({0}\right)^2 + h \left({0}\right)^2 = 0$

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

But from Square of Real Number is Non-Negative:
 * $g \left({a}\right)^2 \ge 0$ and $h \left({a}\right)^2 \ge 0$

So it follows that:
 * $g \left({a}\right) = 0$

and:
 * $h \left({a}\right) = 0$

Hence the result.