Cosine of Sum/Proof 2

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Theorem

$\map \cos {a + b} = \cos a \cos b - \sin a \sin b$


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:

\(\ds \map g a\) \(=\) \(\ds \map \sin {a + b} - \sin a \cos b - \cos a \sin b\)
\(\ds \map h a\) \(=\) \(\ds \map \cos {a + b} - \cos a \cos b + \sin a \sin b\)

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

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

\(\ds \map {g'} a\) \(=\) \(\ds \map \cos {a + b} - \cos a \cos b + \sin a \sin b = \map h a\)
\(\ds \map {h'} a\) \(=\) \(\ds -\map \sin {a + b} + \sin a \cos b + \cos a \sin b = -\map g a\)

Hence:

\(\ds \map {D_a} {\paren {\map g a}^2 + \paren {\map h a}^2}\) \(=\) \(\ds 2 \map g a \map {g'} a + 2 \map h a \map {h'} a\)
\(\ds \) \(=\) \(\ds 0\)

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.

$\blacksquare$


Sources