Editing Trigonometric functions (section)

===Partial fraction expansion===

There is a series representation as [[partial fraction expansion]] where just translated [[Multiplicative inverse|reciprocal function]]s are summed up, such that the [[Pole (complex analysis)|pole]]s of the cotangent function and the reciprocal functions match:<ref name="Aigner_2000"/>
: <math>
\pi \cot \pi x = \lim_{N\to\infty}\sum_{n=-N}^N \frac{1}{x+n}.
</math>
This identity can be proved with the [[Gustav Herglotz|Herglotz]] trick.<ref name="Remmert_1991"/>
Combining the {{math|(–''n'')}}th with the {{math|''n''}}th term lead to [[absolute convergence|absolutely convergent]] series:
:<math>
\pi \cot \pi x = \frac{1}{x} + 2x\sum_{n=1}^\infty \frac{1}{x^2-n^2}.
</math>
Similarly, one can find a partial fraction expansion for the secant, cosecant and tangent functions:
:<math>
\pi\csc\pi x = \sum_{n=-\infty}^\infty \frac{(-1)^n}{x+n}=\frac{1}{x} + 2x\sum_{n=1}^\infty \frac{(-1)^n}{x^2-n^2},
</math>
:<math>\pi^2\csc^2\pi x=\sum_{n=-\infty}^\infty \frac{1}{(x+n)^2},</math>
:<math>
\pi\sec\pi x = \sum_{n=0}^\infty (-1)^n \frac{(2n+1)}{(n+\tfrac12)^2 - x^2},
</math>
:<math>
\pi \tan \pi x = 2x\sum_{n=0}^\infty \frac{1}{(n+\tfrac12)^2 - x^2}.
</math>