Cosine of Integer Multiple of Argument/Formulation 2

Theorem

For $n \in \Z_{>0}$:

$\ds \cos n \theta$

$=$

$\ds \cos^n \theta \paren {1 - \dbinom n 2 \paren {\tan \theta}^2 + \dbinom n 4 \paren {\tan \theta}^4 - \cdots}$

$\ds $

$=$

$\ds \cos^n \theta \sum_{k \mathop \ge 0} \paren {-1}^k \dbinom n {2 k } \paren {\tan^{2 k } \theta}$

$\cos n \theta + i \sin n \theta = \paren {\cos \theta + i \sin \theta}^n$

As $n \in \Z_{>0}$, we use the Binomial Theorem on the right hand side, resulting in:

$\ds \cos n \theta + i \sin n \theta = \sum_{k \mathop \ge 0} \binom n k \paren {\cos^{n - k} \theta} \paren {i \sin \theta}^k$

When $k$ is even, the expression being summed is real.

Equating the real parts of both sides of the equation, replacing $k$ with $2 k$ to make $k$ even, gives:

$\ds \cos n \theta$

$=$

$\ds \sum_{k \mathop \ge 0} \paren {-1}^k \dbinom n {2 k } \paren {\cos^{n - \paren {2 k } } \theta} \paren {\sin^{2 k } \theta}$

$\ds $

$=$

$\ds \cos^n \theta \sum_{k \mathop \ge 0} \paren {-1}^k \dbinom n {2 k } \paren {\tan^{2 k } \theta}$

factor out $\cos^n \theta$

$\blacksquare$