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The Problem Solver's Resource
Introduction Other Tips and Tricks Methods of Proof You are currently viewing page 1.

Trigonometric Formulas

Note that all measurements are in degrees, not radians.

Basic Facts

$\sin (-A)=-\sin A$

$\cos (-A)=\cos A$

$\tan (-A)=-\tan A$

$\sin (180-A) = \sin A$

$\cos (180-A) = -\cos A$

$\cos (360-A) = \cos A$

$\tan (180+A) = \tan A$

$\cos (90-A)=\sin A$

$\tan (90-A)=\cot A$

$\sec{90-A}=\csc A$

$\cos (90-A) = \sin A$

$\cot (90-A)=\tan A$

$\csc (90-A)=\sec A$

$\sin^2 A+\cos^2 A=1$

$\sec^2 A-\tan^2 A=1$

$\csc^2 A-\cot^2 A=1$

$\tan A=\frac{\sin A}{\cos A}$

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

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

$\tan \frac{A}{2}=\frac{1-\cos A}{\sin A}=\frac{\sin A}{1+\cos A}$

Terminology

$\cot A=\frac{1}{\tan A}$, but $\cot A\ne\tan^{-1} A}$ (Error compiling LaTeX. Unknown error_msg).

$\csc A=\frac{1}{\sin A}$, but $\csc A\ne\sin^{-1} A}$ (Error compiling LaTeX. Unknown error_msg).

$\sec A=\frac{1}{\sin A}$, but $\sec A\ne\cos^{-1} A}$ (Error compiling LaTeX. Unknown error_msg).

Also:

$\tan^{-1} A=\atan A=\arctan A$ (Error compiling LaTeX. Unknown error_msg)

$\tan^{-1} A=\asin A=\arcsin A$ (Error compiling LaTeX. Unknown error_msg)

$\tan^{-1} A=\asin A=\arcsin A$ (Error compiling LaTeX. Unknown error_msg)

Sum of Angle Formulas

$\sin (A \pm B)=\sin A \cos B \pm \cos A \sin B$

$\cos (A \pm B)=\cos A \cos B \mp \sin A \sin B$

$\tan (A \pm B)=\frac{\tan A \pm \tan B}{1 \mp \tan A \tan B}$

$\sin2A=2\sin A \cos A$

$\cos2A=\cos^2 A - \sin^2 A$ or $\cos2A=2\cos^2 A -1$ or $\cos2A=1- 2 \sin^2 A$

$\tan2A=\frac{2\tan A}{1-\tan^2 A}$

Pythagorean identities

$\sin^2 A+\cos^2 A=1$

$1 + \tan^2 A = \sec^2 A$

$1 + \cot^2 A = \csc^2 A$

for all $A$.

Other Formulas

==Law of Cosines

In a triangle with sides $a$, $b$, and $c$ opposite angles $A$, $B$, and $C$, respectively,

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

and:

Law of Sines

$\frac{a}{\sin A}=\frac{b}{\sin B}=\frac{c}{\sin C}=2R$

=Law of Tangents

For any $a$ and $b$ such that $\tan a,\tan b \subset \mathbb{R}$, $\frac{a-b}{a+b}=\frac{\tan(a-b)}{\tan(a+b)}$

Area of a Triangle

The area of a triangle can be found by

$\frac 12ab\sin C$

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