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
- 1977: K.G. Binmore: Mathematical Analysis: A Straightforward Approach ... (previous) ... (next): $\S 16.3 \ (2)$