## Sums of Powers of Trigonometric Functions

Sums of powers of trigonometric functions - sin, cos, tan, cot - often simply. Often the series is a geometric series and we can use the formula for the sum of an infinite series,
$S= \frac{a}{1-r} , \: -1 < r < 1$
where
$a$
is the first term and
$r$
is the common ratio.
$1+ cos^2 x + cos^4 x + ...+ cos^{2n} x + ...=\frac{1}{1-cos^2 x}$

Use the identity
$cos^2 x + sin^2 x =1$
written as
$1- cos^2 x = sin^2 x$
to give
$1+ cos^2 x + cos^4 x + ...+ cos^{2n} x + ...=\frac{1}{sin^2 x} =cosec^2 x$

$1+ sin^2 x + sin^4 x + ...+ sin^{2n} x + ...=\frac{1}{1-sin^2 x}$

Use the identity
$cos^2 x + sin^2 x =1$
written as
$1- sin^2 x = cos^2 x$
to give
$1+ sin^2 x + sin^4 x + ...+ sin^{2n} x + ...=\frac{1}{cos^2 x} =sec^2 x$

$cos^2 x + cos^4 x + ...+ cos^{2n} x + ...=\frac{cos^2 x}{1-cos^2 x}$

Use the identity
$cos^2 x + sin^2 x =1$
written as
$1- cos^2 x = sin^2 x$
to give
$cos^2 x + cos^4 x + ...+ cos^{2n} x + ...=\frac{cos^2 x}{sin^2 x} =\frac{1}{tan^2 x} =cot^2 x$

$sin^2 x + sin^4 x + ...+ sin^{2n} x + ...=\frac{sin^2 x}{1-sin^2 x}$

Use the identity
$cos^2 x + sin^2 x =1$
written as
$1- sin^2 x = cos^2 x$
to give
$sin^2 x + sin^4 x + ...+ sin^{2n} x + ...=\frac{sin^2 x}{cos^2 x} =tan^2 x$

$1- cos x + cos^2 x + ...+ (-1)^{n-1} cos^{n-1} x + ...=\frac{1}{1-(-cos x)} =\frac{1}{1+ cos x}$

Use the identity
$cos x =2 cos^2 \frac{x}{2} -1$
written as
$1+ cos x = 2cos^2 \frac{x}{2}$
to give
$1- cos x + cos^2 x + ...+ (-1)^{n-1} cos^{n-1} x + ...=\frac{1}{2 cos^2 \frac{x}{2}} =\frac{sec^2 \frac{x}{2}}{2}$

$1+ cos x + cos x + ...+ cos^{n-1} x + ...=\frac{1}{1-cos x}$

Use the identity
$cos x =1-2 sin^2 \frac{x}{2}$
written as
$1- cos x = 2sin^2 \frac{x}{2}$
to give
$1+ cos x + cos x + ...+ cos^{n-1} x + ...=\frac{1}{2 sin^2 \frac{x}{2}} =\frac{cosec^2 \frac{x}{2}}{2}$

$1-cot^2 x + cot^4 x - ...+ (-1)^{n-1} cot^{2n-2} x + ...=\frac{1}{1-(-cot^2 x)}$

Use the identity
$cot^2 x + 1 =cosec^2 x$
to give
$1-cot^2 x + cot^4 x - ...+ (-1)^{n-1} cot^{2n-2} x + ...=\frac{1}{cosec^2 x} =sin^2 x$

$1-tan^2 x + tan^4 x - ...+(-1)^{n-1}tan^{2n-2} x + ...=\frac{1}{1-(-tan^2 x)} =\frac{1}{1+tan^2 x}$

Use the identity
$tan^2 x + 1 =sec^2 x$
to give
$1-tan^2 x + tan^4 x - ...+(-1)^{n-1}tan^{2n-2} x + ...=\frac{1}{sec^2 x} =cos^2 x$

$cot^2 x - cot^4 x + ...+ (-1)^{n-1} cot^{2n} x + ...=\frac{cot^2 x}{1-(-cot^2 x)} =\frac{cot^2 x}{1+cot^2 x}$

Use the identity
$cot^2 x + 1 =cosec^2 x$
to give
$cot^2 x - cot^4 x + ...+ + (-1)^{n-1} cot^{2n} x + ...=\frac{cot^2 x}{cosec^2 x} = \frac{(cos^2 x)/(sin^2)}{1/(sin^2 x)} =cos^2 x$

$tan^2 x - tan^4 x + ...+(-1)^{n-1} tan^{2n} x + ...=\frac{tan^2 x}{1-(-tan^2 x)} =\frac{tan^2 x}{1+tan^2 x}$

Use the identity
$tan^2 x + 1 =sec^2 x$
to give
$tan^2 x - tan^4 x + ...+(-1)^{n-1} tan^{2n} x + ...=\frac{tan^2 x}{sec^2 x} =\frac{(sin^2 x)/(cos^2 x)}{1/(cos^2 x)} = sin^2 x$

Aby summation of cosines is only valid if
$x \neq m \pi$
and any summation of sines is only valid if
$x \neq m \pi + \frac{\pi}{2}$
.
Any summation of
$tan x$
is only valid for
$- \frac{\pi}{4} + m \pi < x < \frac{\pi}{4} + m \pi$
and any summation of
$cot$
is only valid for
$\frac{\pi}{4} + m \pi < x < \frac{3 \pi}{4}+ m \pi$
where
$m$
is any integer. positive or negative.