---
disqus: abhyas29
---
# Trigonometry
[TOC]

###### tags: `Mathematics`

[All Mathematics Formula by Abhyas here](/@abhyas/maths_formula)

Please see [README](https://hackmd.io/@abhyas/maths_formula#README) if this is the first time you are here.

[Basic Geometry notes here](/@abhyas/geometry)
## From Pythgorous Theorem
$H^2=B^2+P^2$
$\sin^2\theta+\cos^2\theta=1$
$\csc^2\theta-\cot^2\theta=1$
$\sec^2\theta-\tan^2\theta=1$

$\sin^4\theta+\cos^4\theta=1-2\sin^2\theta\cos^2\theta$
$\sin^4\theta-\cos^4\theta=\sin^2\theta-\cos^2\theta$
$\sin^6\theta+\cos^6\theta=1-3\sin^2\theta\cos^2\theta$

$(\sin\theta\pm\cos\theta)^2=1\pm\sin2\theta$

## Circular Function

360, 720, 1080, 1440, 1800

### nPI+-X
<small>_Because both $\sin$ and $\cos$ negaive in IV quadrent, so same behaviour. $\tan$ will be positive._</small>
$\sin(n\pi+\theta)=(-1)^n\sin\theta$
$\cos(n\pi+\theta)=(-1)^ncos\theta$
$\tan(n\pi+\theta)=\tan\theta$

<small>_Because $\sin$ positive in I so need odd multiples of $\pi$
$\tan$ is negative in II and IV.
$\cos$ only positive in IV so need even multiple of $\pi$_</small>
$\sin(n\pi-\theta)=(-1)^{n-1}\sin\theta$
$\cos(n\pi-\theta)=(-1)^ncos\theta$
$\tan(n\pi-\theta)=-\tan\theta$

### nPI
$\sin(n\pi)=0$ <small>_because perpendicular is 0_</small>
$\cos(n\pi)=(-1)^n$ <small>_because base eighter left or right of y axis_</small>
$\tan(n\pi)=0$ <small>_because perpendicular is 0_</small>

### -x
<small>_Because only $\cos$ is positive in IV_</small>
$sin(-\theta)=-\sin\theta$
$cos(-\theta)=\cos\theta$
$tan(-\theta)=-\tan\theta$

### Sum and difference Identities (A+B)

$\sin(A+B)=\sin A\cos B+\cos A\sin B$
$\sin(A-B)=\sin A\cos B - \cos A \sin B$

$\cos(A+B) = \cos A \cos B - \sin A \sin B$
$\cos(A-B)=\cos A \cos B + \sin A \sin B$

$\tan(A+B) = \frac{\tan A + \tan B}{1 - \tan A \tan B}$
$\tan(A-B)= \frac{\tan A - \tan B}{1 + \tan A \tan B}$

$\cot(A+B) = \frac{\cot B \cot A - 1}{\cot B + cot A}$
$\cot(A - B) = \frac{\cot B \cot A +1}{\cot B - \cot A}$

### Product of (A+B) and (A-B)

$\sin(A+B)\sin(A-B) = \sin^2A-\sin^2B$
$\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad=\cos^2B-cos^2A$
$cos(A+B)cos(A-B)=\cos^2A-sin^2B$
$\tan(A+B)\tan(A-B)=\frac{\tan^2A-\tan^2B}{1-\tan^2A\tan^2B}$

### Double Angle Identities 2A

$\sin2\theta=2\sin\theta\cos\theta$
$\quad\quad\,\,\,\,=\frac{2\tan A}{1+\tan^2A}$
$\cos2\theta=\cos^2\theta-\sin^2\theta$
$\quad\quad\,\,\,\,=2\cos^2\theta-1$
$\quad\quad\,\,\,\,=1-2\sin^2\theta$
$\quad\quad\,\,\,\,=\frac{1-\tan^2 A}{1+\tan^2A}$
$\tan2\theta=\frac{2\tan\theta}{1-\tan^2\theta}$

$\tan A + cotA = \frac{1}{\sin A \cos A}$
$\quad\quad\quad\quad\quad\,\,\,\,\,=\frac{2}{\sin2A}$
$\quad\quad\quad\quad\quad\,\,\,\,\,=2\csc2A$

### Triple Angle Identities 3A

$\sin3A=3\sin A -4\sin^3A$
$\cos3A=4\cos^3A - 3 \cos A$
$\tan3A = \frac{3\tan A - tan^3 A}{1 - 3 \tan^2A}$

### Half Angle Identities X/2

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

### C-D Formula

$\sin C + \sin D = 2\sin\frac{C+D}{2}\cos\frac{C-D}{2}$
$\sin C - \sin D = 2\cos\frac{C+D}{2}\sin\frac{C-D}{2}$
$\cos C + \cos D = 2\cos\frac{C+D}{2}\cos\frac{C-D}{2}$
$\cos C - \cos D = -2\sin\frac{C+D}{2}\sin\frac{C-D}{2}$

### Value of common angles

$\sin15^\circ=\frac{\sqrt3-1}{2\sqrt2}=\cos75^\circ$
$\cos15^\circ=\frac{\sqrt3+1}{2\sqrt2}=\sin75^\circ$

$\sin18^\circ=\frac{\sqrt5-1}{4}=\cos72^\circ$
$\cos18=\frac{\sqrt{10+2\sqrt5}}{4}=\sin72^\circ$
$\sin36^\circ=\frac{\sqrt{10-2\sqrt5}}{4}=\cos54^\circ$
$\cos36^\circ=\frac{\sqrt5+1}{4}=\sin54^\circ$

### A, A+60, A-60 form

$\sin A\sin(A+60^\circ)\sin(A-60^\circ)=\frac{1}{4}\sin3A$
$\cos A\cos(A+60^\circ)\cos(A-60^\circ)=\frac{1}{4}\cos3A$
$\tan A+\tan(A+60^\circ)-\tan(A-60^\circ)=\tan3A$

### cosA cos2A cos4A...

$\cos A\cos2A\cos4A=\frac{sin8A}{2^3\sin A}$

*Numerator: $\sin(2\times$ largest angle$)$
Denominator: $2$^{number of cos terms}$\times \sin($ smallest angle$)$*

### tan Or cot and x/2 Or (Pi/2 + x)

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

$\frac{1+\sin\theta}{\cos\theta}=\tan(\frac{\pi}{4}+\frac{\theta}{2})$
$\frac{1-\sin\theta}{\cos\theta}=\tan(\frac{\pi}{4}-\frac{\theta}{2})$

$\frac{1+\sin\theta}{1-\sin\theta}=\tan(\frac{\pi}{4}+\frac{\theta}{2})$
$\frac{1-\sin\theta}{1+\sin\theta}=\cot(\frac{\pi}{4}+\frac{\theta}{2})$

$\frac{1+\cos\theta}{1-\cos\theta}=\cot^2(\frac{\pi}{4}+\frac{\theta}{2})$
$\frac{1-\cos\theta}{1+\cos\theta}=\tan^2(\frac{\pi}{4}+\frac{\theta}{2})$

$\frac{\cos\theta +sin\theta}{\cos\theta-\sin\theta}=\tan(\frac{\pi}{4}+\theta)$
$\frac{\cos\theta -sin\theta}{\cos\theta+\sin\theta}=\tan(\frac{\pi}{4}-\theta)$

### Range of Trigonometric expressoins

$-1\le\sin\theta\le1$
$-1\le\cos\theta\le1$
$-\infty\le\tan\theta\le\infty$

$\sec\theta\in(-\infty,-1]\cup[1, \infty)$
$\csc\theta\in(-\infty,-1]\cup[1, \infty)$

$0\le\sin^2\theta\le1$
$0\le\cos^2\theta\le1$
$0\le\tan^2\theta\le1$

$-\sqrt{a^2+b^2} \le (a\cos \theta + b \sin \theta) \le \sqrt{a^2+b^2}$

## Period of Trigonometic funtion

Period: $\frac{2\pi}{|a|}$
$\quad\sin(ax+b)$
$\quad\cos(ax+b)$
$\quad\sec(ax+b)$
$\quad\csc(ax+b)$

Period: $\frac{\pi}{|a|}$
$\quad\tan(ax+b)$
$\quad\cot(ax+b)$

## Trigonometric Equation

### Type 1 x = 0

$\sin x = 0 \implies x = n\pi$
$\tan x = 0 \implies x = n\pi$

$\cos x = 0 \implies x = (2n+1)\frac{\pi}{2}$
$\cot x = 0 \implies x = (2n+1)\frac{\pi}{2}$

$\sec x = 0 \implies x$ is not defined
$\csc x = 0 \implies x$ is not defined

### Type 2 theta = alpha

$sin\theta=\sin\alpha \implies x = n\pi+(-1)^n\alpha$
$cos\theta=\cos\alpha \implies x = 2n\pi\pm\alpha$
$tan\theta=\tan\alpha \implies x = n\pi+\alpha$

### Type 3 squared

$\sin^2\theta=\sin^2\alpha\implies x=n\pi\pm\alpha$
$\cos^2\theta=\cos^2\alpha\implies x=n\pi\pm\alpha$
$\tan^2\theta=\tan^2\alpha\implies x=n\pi\pm\alpha$

## Properties of Triangles


### Basic Traingle Formulas

$\angle A+ \angle B + \angle C=180^\circ$
$A+B+C=180^\circ$

Perimeter of a circle
$2S=a+b+c$

Area of Triangle from Heron's Formuls
$\Delta=\sqrt{S(S-a)(S-b)(S-c)}$
where $S=\frac{a+b+c}{2}$

### Trigonometric ratios of sum of angles

#### A+B = Pi-C
$\because A+B=\pi-C$

$\sin(A+B)=\sin C$
$\sin(B+C)=\sin A$
$\sin(A+C)=\sin B$

$\cos(A+B)=-\cos C$
$\cos(B+C)=-\cos A$
$\cos(A+C)=-\cos B$

$\tan(A+B)=-\tan C$
$\tan(B+C)=-\tan A$
$\tan(A+C)=-\tan B$

#### (A+B)/2 = Pi/2 - C/2

$\because \frac{A+B}{2}=\frac{\pi}{2}-\frac{C}{2}$

$\sin(\frac{A+B}{2})=\cos\frac{C}{2}$
$\sin(\frac{B+C}{2})=\cos\frac{A}{2}$
$\sin(\frac{A+C}{2})=\cos\frac{B}{2}$

$\cos(\frac{A+B}{2})=\sin\frac{C}{2}$
$\cos(\frac{B+C}{2})=\sin\frac{A}{2}$
$\cos(\frac{A+C}{2})=\sin\frac{B}{2}$

$\tan(\frac{A+B}{2})=\cot\frac{C}{2}$
$\tan(\frac{B+C}{2})=\cot\frac{A}{2}$
$\tan(\frac{A+C}{2})=\cot\frac{B}{2}$

#### tan(A+B+C)

In $\triangle ABC$
$\tan(A+B+C)=\frac{\tan A+\tan B + \tan C-\tan A\tan B \tan C}{1-\tan A\tan B - \tan B\tan C - \tan C \tan A}$
Now,
$\tan(A+B+C) = \tan(\pi) = 0 = \frac{a}{b}$
$\implies a = 0$

We get the formula
$\implies \tan A+\tan B +\tan C=\tan A\tan B\tan C$
in any $\triangle ABC$

Now, Dividing both sides by $\tan A\tan B \tan C$
$\cot B\cot C + \cot A \cot C + \cot A \cot B = 1$

##### Similar Formula, not related to properties of triangle

Now, if $A+B+C=\frac{\pi}{2}$ then,
$\tan(A+B+C)=\tan(\frac{\pi}{2})=\frac{1}{0}=\frac{a}{b}$
$\implies b = 0$
$\implies 1-\tan A\tan B - \tan B\tan C - \tan C \tan A = 0$
$\implies \tan A\tan B + \tan B\tan C + \tan C \tan A = 1$

### Identities in any Triangle

#### 2A 2B 2C

<small>In sin</small>
$\sin2A+\sin2B+\sin2C=4\sin A\sin B\sin C$
<small>sin goes to sin</small>

<small>If one is negative</small>
$\sin2A+\sin2B-\sin2C=4\cos A\cos B\sin C$
<small>then negative will be sin.</small>

<small>If more then one negative. Then take common -1.</small>
$\sin2A-\sin2B-\sin2C$
$-1(-\sin2A+\sin2B+\sin2C)$
$=-4\sin A\cos B\cos C$

<small>In cos</small>
$\cos2A+\cos2B+cos2C=-1-4\cos A\cos B\cos C$
<small>cos goes to cos with -1</small>

<small>If one is negative</small>
$\cos2A+\cos2B-cos2C=1-4\sin A\sin B\cos C$
then negative will be cos

#### A+B+C

<small>In sin</small>
$\sin A+\sin B + \sin C=4\cos\frac{A}{2}cos\frac{B}{2}cos\frac{C}{2}$
<small>$\because \frac{\pi}{2} - \theta$ so change ratio </small>

<small>If one is negative</small>
$\sin A+\sin B - \sin C=4\sin\frac{A}{2}sin\frac{B}{2}cos\frac{C}{2}$
<small>Invert which are positive in +ve form</small>

<small>In cos</small>
$\cos A+\cos B + \cos C=1+4\sin\frac{A}{2}sin\frac{B}{2}sin\frac{C}{2}$
<small>$\because \frac{\pi}{2} - \theta$ so all positive and change ratio</small>

<small>If one is negative</small>
$\cos A+\cos B - \cos C=-1+4\cos\frac{A}{2}cos\frac{B}{2}sin\frac{C}{2}$
<small>-1 + Invert which are positive in +ve form</small>

<small>Notice: In case of negative invert from +ve from whichever are not negative.</small>

### Relation B/w side and angle (sine formula)

$\sin A \propto a$
$\sin B \propto b$
$\sin C \propto c$

Formula 1
$\frac{\sin A}{a}=\frac{\sin B}{b}=\frac{\sin C}{c}$
$\therefore \sin A = ka$
$\therefore \sin B = kb$
$\therefore \sin C = kc$

Formula 2
$\frac{a}{\sin A} = \frac{b}{\sin B}=\frac{c}{\sin C}$
$\therefore a = k\sin A$
$\therefore b = k\sin B$
$\therefore c = k\sin C$

<small>Notice: Same angle and same side relation
NOTE: Can only use either formula 1 or 2 in one question</small>

### cosine formula

$\cos A=\frac{b^2+c^2-a^2}{2bc}$
$\cos B=\frac{a^2+c^2-b^2}{2ac}$
$\cos C=\frac{a^2+b^2-c^2}{2ab}$
<small>Notice: One angle and all three sides relation</small>

### Projection formula

$a=b\cos C+ c\cos B$
$b=a\cos C+c\cos A$
$c=a\cos B+b\cos A$
<small>Two angles and all three sides relation</small>

### Area of Triangle

Area $=\frac{1}{2}ac\sin B=\frac{1}{2}ab\sin C=\frac{1}{2}bc\sin A$

### Half angle formula

$\sin\frac{A}{2}=\sqrt\frac{(S-b)(S-c)}{bc}$
$\sin\frac{B}{2}=\sqrt\frac{(S-a)(S-c)}{ac}$
$\sin\frac{C}{2}=\sqrt\frac{(S-a)(S-b)}{ab}$

$\cos\frac{A}{2}=\sqrt\frac{S(S-a)}{bc}$
$\cos\frac{B}{2}=\sqrt\frac{S(S-b)}{ac}$
$\cos\frac{C}{2}=\sqrt\frac{S(S-c)}{ab}$

$\tan\frac{A}{2}=\sqrt\frac{(S-b)(S-c)}{S(S-a)}$
$\tan\frac{B}{2}=\sqrt\frac{(S-a)(S-c)}{S(S-b)}$
$\tan\frac{C}{2}=\sqrt\frac{(S-a)(S-b)}{S(S-c)}$

### Trick

| | | |
| ------------ | ------------ | ------------ |
| $A=30^\circ$ | $B=60^\circ$ | $C=90^\circ$ |
| $a=1$ | $b=\sqrt3$ | $2$ |

||||
|-|-|-|
|$A=30^\circ$|$B=60^\circ$|$C=90^\circ$|
|$a=1$|$b=\sqrt3$|$2$|

||||
|-|-|-|
|$A=30^\circ$|$B=60^\circ$|$C=90^\circ$|
|$a=1$|$b=\sqrt3$|$2$|

### Circumcircle of a triangle


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

### Incenter of a triangle


$r=\frac{\Delta}{S}$
$\Delta=\frac{1}{2}ab\sin C=\frac{1}{2}bc\sin A=\frac{1}{2}ac\sin B$

## Inverse Trionometric Functions

$\sin^{-1}x=\csc^{-1}\frac{1}{x}$
$\cos^{-1}x=\sec^{-1}\frac{1}{x}$
$\tan^{-1}x=\cot^{-1}\frac{1}{x}$
$\csc^{-1}x=\sin^{-1}\frac{1}{x}$
$\sec^{-1}x=\cos^{-1}\frac{1}{x}$
$\cot^{-1}x=\tan^{-1}\frac{1}{x}$

$\sin^{-1}x+\cos^{-1}x=\frac{\pi}{2}$
$\tan^{-1}x+\cot^{-1}x=\frac{\pi}{2}$
$\sec^{-1}x+\csc^{-1}x=\frac{\pi}{2}$

$\sin^{-1}(\sin x)=x$, $x\in[-\frac{\pi}{2},\frac{\pi}{2}]$
$\cos^{-1}(\cos x)=x$, $x\in[0,\pi]$
$\sec^{-1}(\sec x)=x$, $x\in(-\frac{\pi}{2},\frac{\pi}{2})$

$\sin(\sin^{-1}x)=x$, $x\in[-1,1]$
$\cos(\cos^{-1}x)=x$, $x\in[-1,1]$
$\tan(\tan^{-1}x)=x$, $x\in[-1,1]$

$\sin^{-1}x+\sin^{-1}y=\sin^{-1}[x\sqrt{1-y^2}+y\sqrt{1-x^2}]$
$\sin^{-1}x-\sin^{-1}y=\sin^{-1}[x\sqrt{1-y^2}-y\sqrt{1-x^2}]$
$\cos^{-1}x+\cos^{-1}y=\sin^{-1}[xy-\sqrt{1-y^2}\sqrt{1-x^2}]$
$\cos^{-1}x-\cos^{-1}y=\sin^{-1}[xy+\sqrt{1-y^2}\sqrt{1-x^2}]$
$\tan^{-1}x-\tan^{-1}y=\tan^{-1}[\frac{x+y}{1-xy}]$
$\tan^{-1}x+\tan^{-1}y=\tan^{-1}[\frac{x-y}{1+xy}]$

$2\sin^{-1}x=\sin^{-1}(2x\sqrt{1-x^2})$
$2\cos^{-1}x=\cos^{-1}(2x^2-1)$
$2\tan^{-1}x=\tan^{-1}(\frac{2x}{1-x^2})$
$\quad\quad\quad\quad=\cos^{-1}(\frac{1-x^2}{1+x^2})$
$\quad\quad\quad\quad=\sin^{-1}(\frac{2x}{1+x^2})$

$3\sin^{-1}x=\sin^{-1}(3x-4x^3)$
$3\cos^{-1}x=\cos^{-1}(4x^3-3x)$
$3\tan^{-1}x=\tan^{-1}(\frac{3x-x^3}{1-3x^2})$

| | Domain | Range |
| -------------- |:---------------------------------:|:--------------------------------------:|
| $y=\sin x$ | $\mathbb R$ | $[-1,1]$ |
| $y=\sin^{-1}x$ | $[-1,1]$ | $[-\frac{\pi}{2},\frac{\pi}{2}]$ |
| $y=\csc x$ | $\mathbb R-n\pi$ | $(-\infty, -1]\cup[1,\infty)$ |
| $y=\csc^{-1}x$ | $(-\infty, -1]\cup[1,\infty)$ | $[-\frac{\pi}{2},\frac{\pi}{2}]-\{0\}$ |
| $y=\cos x$ | $\mathbb R$ | $[-1,1]$ |
| $y=\cos^{-1}x$ | $[-1,1]$ | $[0,\pi]$ |
| $y=\sec x$ | $\mathbb R$ | $(-\infty, -1]\cup[1,\infty)$ |
| $y=\sec^{-1}x$ | $(-\infty, -1]\cup[1,\infty)$ | $[0,\pi]-\{\frac{\pi}{2}\}$ |
| $y=\tan x$ | $\mathbb R - (2n+1)\frac{\pi}{2}$ | $(-\infty, \infty)$ |
| $y=\tan^{-1}x$ | $(-\infty, \infty)$ | $(-\frac{\pi}{2},\frac{\pi}{2})$ |
| $y=\cot x$ | $\mathbb R-\{n\pi\}$ | $(-\infty, \infty)$ |
| $y=\cot^{-1}x$ | $(-\infty, \infty)$ | $[0,\pi]$ |

## Licensing and Links

[All Mathematics Formula by Abhays here](/@abhyas/maths_formula)

<a rel="license" href="http://creativecommons.org/licenses/by-nc/4.0/"><img alt="Creative Commons License" style="border-width:0" src="https://i.creativecommons.org/l/by-nc/4.0/88x31.png" /></a><br />This work is licensed under a <a rel="license" href="http://creativecommons.org/licenses/by-nc/4.0/">Creative Commons Attribution-NonCommercial 4.0 International License</a>.

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