trig identity

A trig identity is an equation that involves trig functions.

Remembering trig identities can help you simplify some expressions.

For visualization of these identities, see unit circle. For a better explanation of these identities, see here

And I have a mindmap that is awesome if you can use imagination.

reciprocal and quotient identities

From the definitions of the trig functions, we can express all trig functions with just sine and cosine.

$\tan (a)=\frac{\sin (a)}{\cos (a)}$

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==$\tan(a)$== = ==$\dfrac{\sin(a)}{\cos(a)}$==

$\csc (a)=\frac{1}{\sin (a)}$

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==$\csc(a)$== = ==$\dfrac{1}{\sin(a)}$==

$\sec (a)=\frac{1}{\cos (a)}$

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==$\sec(a)$== = ==$\dfrac{1}{\cos(a)}$==

$\cot (a)=\frac{1}{\tan (a)}=\frac{\cos (a)}{\sin (a)}$

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==$\cot(a)$== = ==$\dfrac{1}{\tan(a)}=\dfrac{\cos(a)}{\sin(a)}$==

Pythagorean identities

Pythagorean identities can be found by applying the Pythagorean theorem to the stuff in the unit circle.

${\sin }^2(a)+{\cos }^2(a)=1$

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==$\sin^2(a)+\cos^2(a)$== = ==$1$==

${\sin }^2(a)=\frac{1-\cos (2a)}{2}$

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==$\sin^2(a)$== = ==$\dfrac{1-\cos(2a)}{2}$==

${\cos }^2(a)=\frac{1+\cos (2a)}{2}$

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==$\cos^2(a)$== = ==$\dfrac{1+\cos(2a)}{2}$==

cosecant and secant squared

${\csc }^2(a)={\cot }^2(a)+1$

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==$\csc^2(a)$== = ==$\cot^2(a)+1$==

${\sec }^2(a)={\tan }^2(a)+1$

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==$\sec^2(a)$== = ==$\tan^2(a)+1$==

angle sum and difference identities

$\sin (a+b)=\sin (a)\cdot \cos (b)+\sin (b)\cdot \cos (a)$

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==$\sin(a+b)$== = ==$\sin(a)\cdot\cos(b)+\sin(b)\cdot\cos(a)$==

$\sin (a-b)=\sin (a)\cdot \cos (b)-\sin (b)\cdot \cos (a)$

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==$\sin(a-b)$== = ==$\sin(a)\cdot\cos(b)-\sin(b)\cdot\cos(a)$==

$\cos (a+b)=\cos (a)\cdot \cos (b)-\sin (b)\cdot \sin (a)$

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==$\cos(a+b)$== = ==$\cos(a)\cdot\cos(b)-\sin(b)\cdot\sin(a)$==

$\cos (a-b)=\cos (a)\cdot \cos (b)+\sin (b)\cdot \sin (a)$

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==$\cos(a-b)$== = ==$\cos(a)\cdot\cos(b)+\sin(b)\cdot\sin(a)$==

$\tan (a+b)=\frac{\tan (a)+\tan (b)}{1-\tan (a)\tan (b)}$

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==$\tan(a+b)$== = ==$\dfrac{\tan(a)+\tan(b)}{1-\tan(a)\tan(b)}$==

$\tan (a-b)=\frac{\tan (a)-\tan (b)}{1+\tan (a)\tan (b)}$

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==$\tan(a-b)$== = ==$\dfrac{\tan(a)-\tan(b)}{1+\tan(a)\tan(b)}$==

double angle identities

these are derived from the angle sum identities and Pythagorean identities

$\sin (2a)=2\cdot \sin (a)\cdot \cos (a)$

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==$\sin(2a)$== = ==$2\cdot\sin(a)\cdot\cos(a)$==

$\cos (2a)={\cos }^2(a)-{\sin }^2(a)$

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==$\cos(2a)$== = ==$\cos^2(a)-\sin^2(a)$==

$\cos (2a)=2\cdot {\cos }^2(a)-1$

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==$\cos(2a)$== = ==$2\cdot\cos^2(a)-1$==

$\cos (2a)=1-2\cdot {\sin }^2(a)$

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==$\cos(2a)$== = ==$1-2\cdot\sin^2(a)$==

$\tan (2a)=\frac{2\tan (a)}{1-{\tan }^2(a)}$

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==$\tan(2a)$== = ==$\dfrac{2\tan(a)}{1-\tan^2(a)}$==

half angle identities

these are derived from the double angle identities

$\sin (\frac{a}{2})=\pm \sqrt{\frac{1-\cos (a)}{2}}$

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==$\sin(\dfrac{a}{2})$== = ==$\pm\sqrt{\dfrac{1-\cos(a)}{2}}$==

$\cos (\frac{a}{2})=\pm \sqrt{\frac{1+\cos (a)}{2}}$

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==$\cos(\dfrac{a}{2})$== = ==$\pm\sqrt{\dfrac{1+\cos(a)}{2}}$==

$\tan (\frac{a}{2})=\frac{\sin (a)}{1+\cos (a)}$

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==$\tan(\dfrac{a}{2})$== = ==$\dfrac{\sin(a)}{1+\cos(a)}$==

symmetry and periodicity identities

$\sin (-a)=-\sin (a)$

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==$\sin(-a)$== = ==$-\sin(a)$==

$\cos (-a)=\cos (a)$

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==$\cos(-a)$== = ==$\cos(a)$==

$\tan (-a)=-\tan (a)$

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==$\tan(-a)$== = ==$-\tan(a)$==

$\sin (a+2\pi )=\sin (a)$

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==$\sin(a+2\pi)$== = ==$\sin(a)$==

$\cos (a+2\pi )=\cos (a)$

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==$\cos(a+2\pi)$== = ==$\cos(a)$==

$\tan (a+\pi )=\tan (a)$

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==$\tan(a+\pi)$== = ==$\tan(a)$==

other identities

$\tan (\theta +\frac{\pi }{2})=\frac{1}{\tan (\theta )}$

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==$\tan(\theta+\dfrac{\pi}{2})$== = ==$\dfrac{1}{\tan(\theta)}$==