Sine of Sum/Proof 2

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


 * $\ds \cos x = \sum_{n \mathop = 0}^\infty \paren {-1}^n \frac {x^{2 n} } {\paren {2 n}!}$

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:
 * $\forall a \in \R: \map g a^2 + \map h a^2 = c$

In particular, it is true for $a = 0$, and so:
 * $\map g 0^2 + \map h 0^2 = 0$

So:
 * $\map g a^2 + \map h a^2 = 0$

But from Square of Real Number is Non-Negative:
 * $\map g a^2 \ge 0$ and $\map h a^2 \ge 0$

So it follows that:
 * $\map g a = 0$

and:
 * $\map h a = 0$

Hence the result.