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
We have:
 * From the definition of sine:
 * $\displaystyle \sin x = \sum_{n=0}^\infty \left({-1}\right)^n \frac {x^{2n+1}}{\left({2n+1}\right)!} = x - \frac {x^3} {3!} + \frac {x^5} {5!} - \cdots$


 * From the definition of cosine:
 * $\displaystyle \cos x = \sum_{n=0}^\infty \left({-1}\right)^n \frac {x^{2n}}{\left({2n}\right)!} = 1 - \frac {x^2} {2!} + \frac {x^4} {4!} - \cdots$.

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 Differentiation of a Constant, $\forall a: g \left({a}\right)^2 + h \left({a}\right)^2 = c$.

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

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

Hence the result.