Trigonometric identities are used to manipulate trigonometry equations in certain ways. Here is a list of them:

Basic Definitions

The six basic trigonometric functions can be defined using a right triangle:

The six trig functions are sine, cosine, tangent, cosecant, secant, and cotangent. They are abbreviated by using the first three letters of their name (except for cosecant which uses $\csc$ ). They are defined as follows:

Basic Definitions
$\sin A = \frac ac$	$\csc A = \frac ca$	$\cos A = \frac bc$	$\sec A = \frac cb$	$\tan A = \frac ab$	$\cot A = \frac ba$

Even-Odd Identities

$\sin (-\theta) = -\sin (\theta)$

$\cos (-\theta) = \cos (\theta)$

$\tan (-\theta) = -\tan (\theta)$

$\sec (-\theta) = \sec (\theta)$

$\csc (-\theta) = -\csc (\theta)$

$\cot (-\theta) = -\cot (\theta)$

Further Conclusions

Based on the above identities, we can also claim that

$\sin(\cos(-\theta)) = \sin(\cos(\theta))$

$\cos(\sin(-\theta)) = \cos(-\sin(\theta)) = \cos(\sin(\theta))$

This is only true when $\sin(\theta)$ is in the domain of $\cos(\theta)$ .

Reciprocal Relations

From the first section, it is easy to see that the following hold:

$\sin A = \frac 1{\csc A}$

$\cos A = \frac 1{\sec A}$

$\tan A = \frac 1{\cot A}$

Another useful identity that isn't a reciprocal relation is that $\tan A =\frac{\sin A}{\cos A}$ .

Note that $\sin^{-1} A \neq \csc A$ ; the former refers to the inverse trigonometric functions.

Pythagorean Identities

Using the Pythagorean Theorem on our triangle above, we know that $a^2 + b^2 = c^2$ . If we divide by $c^2$ we get $\left(\frac{a}{c}\right)^2 + \left(\frac{b}{c}\right)^2 = 1$ , which is just $\sin^2 A + \cos^2 A =1$ . Dividing by $a^2$ or $b^2$ instead produces two other similar identities. The Pythagorean Identities are listed below:

$\sin^2x + \cos^2x = 1$
$1 + \cot^2x = \csc^2x$
$\tan^2x + 1 = \sec^2x$

(Note that the last two are easily derived by dividing the first by $\sin^2x$ and $\cos^2x$ , respectively.)

Angle Addition/Subtraction Identities

Once we have formulas for angle addition, angle subtraction is rather easy to derive. For example, we just look at $\sin(\alpha+(-\beta))$ and we can derive the sine angle subtraction formula using the sine angle addition formula.

$\sin(\alpha \pm \beta) = \sin \alpha\cos \beta \pm\sin \beta \cos \alpha$
$\cos(\alpha \pm \beta) = \cos \alpha \cos \beta \mp \sin \alpha \sin \beta$
$\tan(\alpha \pm \beta) = \frac{\tan \alpha \pm \tan \beta}{1\mp\tan \alpha \tan \beta}$

We can prove $\cos(\alpha + \beta) = \cos \alpha \cos \beta - \sin \alpha \sin \beta$ easily by using $\sin(\alpha + \beta) = \sin \alpha\cos \beta +\sin \beta \cos \alpha$ and $\sin(x)=\cos(90-x)$ .

$\cos (\alpha + \beta)$

$= \sin((90 -\alpha) - \beta)$ $= \sin (90- \alpha) \cos (\beta) - \sin ( \beta) \cos (90- \alpha)$

$=\cos \alpha \cos \beta - \sin \beta \sin \alpha$

Double Angle Identities

Double angle identities are easily derived from the angle addition formulas by just letting $\alpha = \beta$ . Doing so yields:

$\begin{eqnarray*} \sin 2\alpha &=& 2\sin \alpha \cos \alpha\\ \cos 2\alpha &=& \cos^2 \alpha - \sin^2 \alpha\\ &=& 2\cos^2 \alpha - 1\\ &=& 1-2\sin^2 \alpha\\ \tan 2\alpha &=& \frac{2\tan \alpha}{1-\tan^2\alpha} \end{eqnarray*}$

Further Conclusions

We can see from the above that

$\csc(2a) = \frac{\csc(a)\sec(a)}{2}$
$\sec(2a) = \frac{1}{2\cos^2(a) - 1} = \frac{1}{\cos^2(a) - \sin^2(a)} = \frac{1}{1 - 2\sin^2(a)}$
$\cot(2a) = \frac{1 - \tan^2(a)}{2\tan(a)}$

Half Angle Identities

Using the double angle identities, we can derive half angle identities. The double angle formula for cosine tells us $\cos 2\alpha = 2\cos^2 \alpha - 1$ . Solving for $\cos \alpha$ we get $\cos \alpha =\pm \sqrt{\frac{1 + \cos 2\alpha}2}$ where we look at the quadrant of $\alpha$ to decide if it's positive or negative. Likewise, we can use the fact that $\cos 2\alpha = 1 - 2\sin^2 \alpha$ to find a half angle identity for sine. Then, to find a half angle identity for tangent, we just use the fact that $\tan \frac x2 =\frac{\sin \frac x2}{\cos \frac x2}$ and plug in the half angle identities for sine and cosine.

To summarize:

$\sin \frac{\theta}2 = \pm \sqrt{\frac{1-\cos \theta}2}$
$\cos \frac{\theta}2 = \pm \sqrt{\frac{1+\cos \theta}2}$
$\tan \frac{\theta}2 = \pm \sqrt{\frac{1-\cos \theta}{1+\cos \theta}}=\frac{\sin \theta}{1+\cos\theta}=\frac{1-\cos\theta}{\sin \theta}$

Prosthaphaeresis Identities

(Otherwise known as sum-to-product identities)

$\displaystyle{\sin \theta + \sin \gamma = 2 \sin \frac{\theta + \gamma}2 \cos \frac{\theta - \gamma}2}$
$\displaystyle{\sin \theta - \sin \gamma = 2 \sin \frac{\theta - \gamma}2 \cos \frac{\theta + \gamma}2}$
$\displaystyle{\cos \theta + \cos \gamma = 2 \cos \frac{\theta+\gamma}2 \cos \frac{\theta-\gamma}2}$
$\displaystyle{\cos \theta - \cos \gamma = -2 \sin \frac{\theta+\gamma}2 \sin \frac{\theta-\gamma}2}$

Law of Sines

Main article: Law of Sines

The extended Law of Sines states

$\frac a{\sin A} = \frac b{\s

Law of Cosines

Main article: Law of Cosines

The Law of Cosines states

$a^2 = b^2 + c^2 - 2bc\cos A.$

Law of Tangents

Main article: Law of Tangents

The Law of Tangents states that if $A$ and $B$ are angles in a triangle opposite sides $a$ and $b$ respectively, then

$\frac{\tan{\left(\frac{A-B}{2}\right)}}{\tan{\left(\frac{A+B}{2}\right)}}=\frac{a-b}{a+b} .$

A further extension of the Law of Tangents states that if $A$ , $B$ , and $C$ are angles in a triangle, then $\tan(A)\cdot\tan(B)\cdot\tan(C)=\tan(A)+\tan(B)+\tan(C)$

Other Identities

$\sin(90-\theta) = \cos(\theta)$
$\cos(90-\theta)=\sin(\theta)$
$\tan(90-\theta)=\cot(\theta)$
$\sin(180-\theta) = \sin(\theta)$
$\cos(180-\theta) = -\cos(\theta)$
$\tan(180-\theta) = -\tan(\theta)$
$e^{i\theta} = \cos \theta + i\sin \theta$ (This is also written as $\text{cis }\theta$ )
$|1-e^{i\theta}|=2\sin\frac{\theta}{2}$
$\left(\tan\theta + \sec\theta\right)^2 = \frac{1 + \sin\theta}{1 - \sin\theta}$
$\sin(\theta) = \cos(\theta)\tan(\theta)$
$\cos(\theta) = \frac{\sin(\theta)}{\tan(\theta)}$
$\sec(\theta) = \frac{\tan(\theta)}{\sin(\theta)}$
$\arctan(x) + \arctan(y) = \arctan \left( \dfrac{x+y}{1-xy} \right)$
$\sin^2(\theta) + \cos^2(\theta) + \tan^2(\theta) = \sec^2(\theta)$
$\sin^2(\theta) + \cos^2(\theta) + \cot^2(\theta) = \csc^2(\theta)$