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.