\[S= \frac{a}{1-r} , \: -1 < r < 1\]
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.