﻿ Table of Trigonometric Identities

# Table of Trigonometric Identities

Collection of trigonometric identities for math solutions

## Symmetry Identities

\def\arraystretch{1.5} \begin{array}{cc} \sin (-\alpha )=-\sin (\alpha )&\cos (-\alpha )=\cos (\alpha )& \\ \tan (-\alpha )=-\tan (\alpha )&\cot (-\alpha )=\cot (\alpha )& \\ \csc (-\alpha )=-\csc (\alpha )&\sec (-\alpha )=\sec (\alpha )& \\ \end{array}

## Reciprocal identities

\def\arraystretch{2.5} \begin{array}{cc} \sin \alpha = \dfrac{1}{\csc \alpha} &\csc \alpha = \dfrac{1}{\sin \alpha} & \\ \cos \alpha = \dfrac{1}{\sec \alpha} &\sec \alpha = \dfrac{1}{\cos \alpha} & \\ \tan \alpha = \dfrac{1}{\cot \alpha} &\cot \alpha = \dfrac{1}{\tan \alpha} & \\ \end{array}

## Quotient Identities

\def\arraystretch{2.2} \begin{array}{cc} \tan \alpha = \dfrac{\sin\alpha }{\cos \alpha } & \\ \cot \alpha = \dfrac{\cos \alpha }{\sin \alpha } & \\ \end{array}

## Cofunction Identities

Two trigonometric functions are cofunctions if they are equal on complementary angles. There 3 pairs of cofunctions, Sine and cosine, tangent and cotangent, secant and cosecant.

\def\arraystretch{2} \begin{array}{cc} \cos(\dfrac{\pi}{2}-\alpha)=\sin \alpha&\cos(90\degree-\alpha)=\sin \alpha& \\ \sin(\dfrac{\pi}{2}-\alpha)=\cos \alpha&\sin(90\degree-\alpha)=\cos \alpha& \\ \tan(\dfrac{\pi}{2}-\alpha)=\cot \alpha&\tan(90\degree-\alpha)=\cot \alpha& \\ \cot(\dfrac{\pi}{2}-\alpha)=\tan \alpha&\cot(90\degree-\alpha)=\tan \alpha& \\ \sec(\dfrac{\pi}{2}-\alpha)=\csc \alpha&\sec(90\degree-\alpha)=\csc \alpha& \\ \csc(\dfrac{\pi}{2}-\alpha)=\sec \alpha&\csc(90\degree-\alpha)=\sec \alpha& \\ \end{array}

## Pythagorean Identities

\def\arraystretch{1.5} \begin{array}{cc} \sin^2\alpha +\cos^2\alpha =1 & \\ 1+\tan^2\alpha = \sec^2\alpha & \\ 1+\cot^2\alpha =\csc^2\alpha & \\ \end{array}

## Power-Reducing Identities

\def\arraystretch{2.2} \begin{array}{cc} \sin^2\alpha = \dfrac{1-\cos (2\alpha)}{2} & \\ \cos^2\alpha = \dfrac{1+\cos (2\alpha)}{2} & \\ \tan^2\alpha = \dfrac{1-\cos (2\alpha)}{1+\cos (2\alpha)} & \\ \end{array}

## Sum and difference identities

\def\arraystretch{2.2} \begin{array}{ll} \cos(\alpha +\beta ) = \cos \alpha\cos \beta -\sin \alpha \sin \beta & \\ \cos(\alpha -\beta ) = \cos \alpha\cos \beta +\sin \alpha \sin \beta & \\ \sin(\alpha +\beta ) = \sin \alpha\cos \beta +\cos \alpha \sin \beta & \\ \sin(\alpha -\beta ) = \sin \alpha\cos \beta -\cos \alpha \sin \beta & \\ \tan(\alpha+\beta) = \dfrac{\tan \alpha + \tan \beta }{1-\tan \alpha\tan \beta} & \\ \tan(\alpha -\beta) = \dfrac{\tan \alpha - \tan \beta }{1+\tan \alpha\tan \beta} & \\ \cot(\alpha+\beta) = \dfrac{\cos(\alpha +\beta)}{\sin(\alpha +\beta)} = \dfrac{ \cos \alpha\cos \beta -\sin \alpha \sin \beta}{\sin \alpha\cos \beta +\cos \alpha \sin \beta} = \dfrac{\dfrac{ \cos \alpha\cos \beta -\sin \alpha \sin \beta}{\sin \alpha \sin \beta } }{\dfrac{\sin \alpha\cos \beta +\cos \alpha \sin \beta}{\sin \alpha \sin \beta } } =\dfrac{\cot \alpha \cot \beta -1 }{\cot \alpha+ \cot \beta } & \\ \cot(\alpha-\beta) =\dfrac{\cot \alpha \cot \beta +1 }{\cot \alpha- \cot \beta } \end{array}

## Double angle identities

\def\arraystretch{2.5} \begin{array}{ll} \sin2 \alpha = 2\sin \alpha \cos \alpha & \\ \cos2 \alpha =\cos^2\alpha - \sin^2 \alpha & \\ \cos2 \alpha = 2\cos^2\alpha -1 & \\ \cos2 \alpha = 1-2 \sin^2 \alpha & \\ \tan 2 \alpha = \dfrac{2\tan \alpha }{1-\tan^2\alpha } & \\ \cot 2 \alpha = \dfrac{1}{\tan 2\alpha } = \dfrac{ 1-\tan^2\alpha }{ 2\tan \alpha }= \dfrac{ 1-\dfrac{1}{\cot^2\alpha } }{ \dfrac{2}{\cot \alpha } } =\dfrac{\cot^2\alpha -1 }{2\cot \alpha } \\ \sin 2\alpha = 2\sin \alpha \cos \alpha=\dfrac{ 2\sin \alpha \cos \alpha}{\sin^2 \alpha +\cos^2 \alpha } = \dfrac{2\tan \alpha }{1+\tan^2 \alpha } & \\ \cos 2\alpha = \cos^2\alpha - \sin^2 \alpha = \dfrac{\cos^2\alpha - \sin^2 \alpha }{\cos^2 \alpha+\sin^2\alpha } =\dfrac{1-\tan^2 \alpha }{1+\tan^2\alpha } & \\ \end{array}

## Half angle identities

\def\arraystretch{3} \begin{array}{ll} \sin \dfrac{\alpha }{2}=\pm\sqrt{\dfrac{1-\cos \alpha }{2} } & \\ \cos \dfrac{\alpha }{2}=\pm\sqrt{\dfrac{1+\cos \alpha }{2} } & \\ \tan \dfrac{\alpha }{2}=\pm\sqrt{\dfrac{1-\cos \alpha }{1+\cos \alpha} } & \\ \tan \dfrac{\alpha }{2}= \dfrac{\sin \dfrac{\alpha }{2}}{\cos \dfrac{\alpha }{2}} = \dfrac{2\sin \dfrac{\alpha }{2}\cos \dfrac{\alpha }{2}}{2\cos^2 \dfrac{\alpha }{2}} = \dfrac{\sin \alpha }{1+\cos \alpha } & \\ \tan \dfrac{\alpha }{2}= \dfrac{\sin \dfrac{\alpha }{2}}{\cos \dfrac{\alpha }{2}} = \dfrac{2\sin^2 \dfrac{\alpha }{2}}{2\sin \dfrac{\alpha }{2}\cos \dfrac{\alpha }{2}} =\dfrac{1-\cos \alpha }{\sin \alpha } \\ \end{array}

## Triple angle identities

\def\arraystretch{1.5} \begin{array}{ll} \sin 3\alpha =3\sin \alpha -4\sin^3\alpha &\\ \cos 3\alpha =4\cos^3\alpha -3\cos \alpha &\\ \tan 3\alpha =\dfrac{3\tan \alpha -\tan^3\alpha }{1-3\tan^2\alpha } &\\ \end{array}

## Sum to product identities

\def\arraystretch{2.2} \begin{array}{ll} \sin \alpha + \sin \beta = 2\sin\dfrac{\alpha +\beta }{2} \cos\dfrac{\alpha -\beta }{2} &\\ \sin \alpha -\sin \beta = 2\cos\dfrac{\alpha +\beta }{2} \sin\dfrac{\alpha -\beta }{2} &\\ \cos \alpha - \cos \beta = -2\sin\dfrac{\alpha +\beta }{2} \sin\dfrac{\alpha -\beta }{2} &\\ \cos \alpha + \cos \beta = 2\cos\dfrac{\alpha +\beta }{2} \cos\dfrac{\alpha -\beta }{2} &\\ \tan \alpha +\tan \beta =\dfrac{\sin(\alpha +\beta )}{\cos\alpha \cos \beta } &\\ \tan \alpha -\tan \beta =\dfrac{\sin(\alpha -\beta )}{\cos\alpha \cos \beta } &\\ \cot \alpha +\cot \beta =\dfrac{\sin(\alpha +\beta )}{\sin\alpha \sin \beta } &\\ \cot \alpha -\cot \beta =\dfrac{\sin(\alpha -\beta )}{\sin\alpha \sin \beta } &\\ \end{array}

## Product to sum identities

\def\arraystretch{2.2} \begin{array}{ll} \cos \alpha\sin \beta = \dfrac{1}{2} [\sin(\alpha+\beta ) - \sin(\alpha - \beta) ] &\\ \sin \alpha\cos \beta = \dfrac{1}{2} [\sin(\alpha+\beta ) +\sin(\alpha - \beta) ] &\\ \sin \alpha\sin \beta = \dfrac{1}{2} [\cos(\alpha-\beta ) - \cos(\alpha + \beta) ] &\\ \cos \alpha\cos \beta = \dfrac{1}{2} [\cos(\alpha+\beta ) +\cos(\alpha - \beta) ] &\\ \end{array}

Collected in the board: Trigonometry

Steven Zheng posted 4 months ago

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