Derivative of Cosine Function/Proof 1

Theorem

$\map {\dfrac \d {\d x} } {\cos x} = -\sin x$

From the definition of the cosine function, we have:

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

Then:

$\ds \map {\frac \d {\d x} } {\cos x}$

$=$

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

$\ds $

$=$

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

$\ds $

$=$

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

changing summation index

$\ds $

$=$

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

The result follows from the definition of the sine function.

$\blacksquare$

1977: K.G. Binmore: Mathematical Analysis: A Straightforward Approach ... (previous) ... (next): $\S 16.3 \ (1) \ \text{(v)}$