# Trigonometry/Identities

Let us take a right angled triangle with hypotenuse length 1. If we mark one of the acute angles as ${\displaystyle \theta }$, then using the definition of the sine ratio, we have

${\displaystyle \sin \theta ={\cfrac {opposite}{hypotenuse}}}$

As the hypotenuse is 1,

${\displaystyle \sin \theta ={\cfrac {opposite}{1}}=opposite}$

Repeating the same process using the definition of the cosine ratio, we have

${\displaystyle \cos \theta ={\cfrac {adjacent}{hypotenuse}}={\cfrac {adjacent}{1}}=adjacent}$

## Pythagorean identities

Since this is a right triangle, we can use the Pythagorean Theorem:

${\displaystyle x^{2}+y^{2}=r^{2}}$

${\displaystyle {\frac {x^{2}}{r^{2}}}+{\frac {y^{2}}{r^{2}}}={\frac {r^{2}}{r^{2}}}}$

${\displaystyle \operatorname {cos} ^{2}\theta +\operatorname {sin} ^{2}\theta =1}$

This is the most fundamental identity in trigonometry.

${\displaystyle {\frac {x^{2}}{y^{2}}}+{\frac {y^{2}}{y^{2}}}={\frac {r^{2}}{y^{2}}}}$

${\displaystyle \operatorname {cot} ^{2}x+1=\operatorname {csc} ^{2}}$

${\displaystyle {\frac {x^{2}}{x^{2}}}+{\frac {y^{2}}{x^{2}}}={\frac {r^{2}}{x^{2}}}}$

${\displaystyle \operatorname {1} +\operatorname {tan} ^{2}\theta =\operatorname {sec} ^{2}\theta }$

From this identity, if we divide through by squared cosine, we are left with:

${\displaystyle {\cfrac {\operatorname {sin} ^{2}\theta +\operatorname {cos} ^{2}\theta }{\operatorname {cos} ^{2}\theta }}={\cfrac {1}{\operatorname {cos} ^{2}\theta }}}$

${\displaystyle \operatorname {tan} ^{2}\theta +1=\operatorname {sec} ^{2}\theta }$

${\displaystyle \operatorname {sec} ^{2}\theta -\operatorname {tan} ^{2}\theta =1}$

If instead we divide the original identity by squared sine, we are left with:

${\displaystyle {\cfrac {\operatorname {sin} ^{2}\theta +\operatorname {cos} ^{2}\theta }{\operatorname {sin} ^{2}\theta }}={\cfrac {1}{\operatorname {sin} ^{2}\theta }}}$

${\displaystyle \operatorname {cot} ^{2}\theta +1=\operatorname {csc} ^{2}\theta }$

${\displaystyle \operatorname {csc} ^{2}\theta -\operatorname {cot} ^{2}\theta =1}$

There are basically 3 main trigonometric identities. The proofs come directly from the definitions of these functions and the application of the Pythagorean theorem:

${\displaystyle \operatorname {sin} ^{2}\theta +\operatorname {cos} ^{2}\theta =1}$
${\displaystyle \operatorname {sec} ^{2}\theta -\operatorname {tan} ^{2}\theta =1}$
${\displaystyle \operatorname {csc} ^{2}\theta -\operatorname {cot} ^{2}\theta =1}$

## Angle sum-difference identities

${\displaystyle \sin(\alpha \pm \beta )=\sin \alpha \cos \beta \pm \cos \alpha \sin \beta }$
${\displaystyle \cos(\alpha \pm \beta )=\cos \alpha \cos \beta \mp \sin \alpha \sin \beta }$

## Cofunction identities

 ${\displaystyle \cos(90-\theta )=\sin \theta }$ ${\displaystyle \sec(90-\theta )=\csc \theta }$ ${\displaystyle \tan(90-\theta )=\cot \theta }$ ${\displaystyle \sin(90-\theta )=\cos \theta }$ ${\displaystyle \csc(90-\theta )=\sec \theta }$ ${\displaystyle \cot(90-\theta )=\tan \theta }$

## Multiple angle identities

${\displaystyle \cos 2A=\cos ^{2}A-\sin ^{2}A}$
${\displaystyle \sin 2A=2\sin A\cos A}$
${\displaystyle \sin 2\theta ={\frac {tan2\theta *tan\theta }{tan2\theta -tan\theta }}}$
${\displaystyle \cos 2\theta ={\frac {tan\theta }{tan2\theta -tan\theta }}}$
${\displaystyle \tan 2\theta =tan\theta ({\frac {1}{cos2\theta }}+1)}$
${\displaystyle \tan 2\theta ={\frac {2sin^{2}\theta }{sin2\theta -tan\theta }}}$