Trigonometry

• Trigonometry
  • From Pythgorous Theorem
  • Circular Function
    • nPI±X
    • nPI
    • -x
    • Sum and difference Identities (A+B)
    • Product of (A+B) and (A-B)
    • Double Angle Identities 2A
    • Triple Angle Identities 3A
    • Half Angle Identities X/2
    • C-D Formula
    • Value of common angles
    • A, A+60, A-60 form
    • cosA cos2A cos4A…
    • tan Or cot and x/2 Or (Pi/2 + x)
    • Range of Trigonometric expressoins
  • Period of Trigonometic funtion
  • Trigonometric Equation
    • Type 1 x = 0
    • Type 2 theta = alpha
    • Type 3 squared
  • Properties of Triangles
    • Basic Traingle Formulas
    • Trigonometric ratios of sum of angles
    • Identities in any Triangle
    • Relation B/w side and angle (sine formula)
    • cosine formula
    • Projection formula
    • Area of Triangle
    • Half angle formula
    • Trick
    • Circumcircle of a triangle
    • Incenter of a triangle
  • Inverse Trionometric Functions
  • Licensing and Links

tags: Mathematics

All Mathematics Formula by Abhyas here

Please see README if this is the first time you are here.

Basic Geometry notes here

From Pythgorous Theorem

$H^2=B^2+P^2$

${\sin }^2\theta +{\cos }^2\theta =1$

${\csc }^2\theta -{\cot }^2\theta =1$

${\sec }^2\theta -{\tan }^2\theta =1$

${\sin }^4\theta +{\cos }^4\theta =1-2{\sin }^2\theta {\cos }^2\theta$

${\sin }^4\theta -{\cos }^4\theta ={\sin }^2\theta -{\cos }^2\theta$

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

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

Circular Function

360, 720, 1080, 1440, 1800

nPI±X

Because both

$\sin$ and

$\cos$ negaive in IV quadrent, so same behaviour.

$\tan$ will be positive.

$\sin (n\pi +\theta )=(-1{)}^n\sin \theta$

$\cos (n\pi +\theta )=(-1{)}^{n}cos\theta$

$\tan (n\pi +\theta )=\tan \theta$

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

$\pi$

$\sin (n\pi -\theta )=(-1{)}^{n-1}\sin \theta$

$\cos (n\pi -\theta )=(-1{)}^{n}cos\theta$

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

nPI

$\sin (n\pi )=0$ because perpendicular is 0

$\cos (n\pi )=(-1{)}^n$ because base eighter left or right of y axis

$\tan (n\pi )=0$ because perpendicular is 0

-x

Because only

$\cos$ is positive in IV

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

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

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

Sum and difference Identities (A+B)

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

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

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

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

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

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

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

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

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

$\sin (A+B)\sin (A-B)={\sin }^{2}A-{\sin }^{2}B$

$={\cos }^{2}B-co{s}^{2}A$

$cos(A+B)cos(A-B)={\cos }^{2}A-si{n}^{2}B$

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

Double Angle Identities 2A

$\sin 2\theta =2\sin \theta \cos \theta$

$=\frac{2\tan A}{1+{\tan }^{2}A}$

$\cos 2\theta ={\cos }^2\theta -{\sin }^2\theta$

$=2{\cos }^2\theta -1$

$=1-2{\sin }^2\theta$

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

$\tan 2\theta =\frac{2\tan \theta }{1-{\tan }^2\theta }$

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

$=\frac{2}{\sin 2A}$

$=2\csc 2A$

Triple Angle Identities 3A

$\sin 3A=3\sin A-4{\sin }^{3}A$

$\cos 3A=4{\cos }^{3}A-3\cos A$

$\tan 3A=\frac{3\tan A-ta{n}^{3}A}{1-3{\tan }^{2}A}$

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

$\sin {15}^{\circ }=\frac{\sqrt{3}-1}{2\sqrt{2}}=\cos {75}^{\circ }$

$\cos {15}^{\circ }=\frac{\sqrt{3}+1}{2\sqrt{2}}=\sin {75}^{\circ }$

$\sin {18}^{\circ }=\frac{\sqrt{5}-1}{4}=\cos {72}^{\circ }$

$\cos 18=\frac{\sqrt{10+2\sqrt{5}}}{4}=\sin {72}^{\circ }$

$\sin {36}^{\circ }=\frac{\sqrt{10-2\sqrt{5}}}{4}=\cos {54}^{\circ }$

$\cos {36}^{\circ }=\frac{\sqrt{5}+1}{4}=\sin {54}^{\circ }$

A, A+60, A-60 form

$\sin A\sin (A+{60}^{\circ })\sin (A-{60}^{\circ })=\frac{1}{4}\sin 3A$

$\cos A\cos (A+{60}^{\circ })\cos (A-{60}^{\circ })=\frac{1}{4}\cos 3A$

$\tan A+\tan (A+{60}^{\circ })-\tan (A-{60}^{\circ })=\tan 3A$

cosA cos2A cos4A…

$\cos A\cos 2A\cos 4A=\frac{sin8A}{2^3\sin A}$

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

$-1\le \sin \theta \le 1$

$-1\le \cos \theta \le 1$

$-\mathrm{\infty }\le \tan \theta \le \mathrm{\infty }$

$\sec \theta \in (-\mathrm{\infty },-1]\cup [1,\mathrm{\infty })$

$\csc \theta \in (-\mathrm{\infty },-1]\cup [1,\mathrm{\infty })$

$0\le {\sin }^2\theta \le 1$

$0\le {\cos }^2\theta \le 1$

$0\le {\tan }^2\theta \le 1$

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

Period of Trigonometic funtion

Period:

$\frac{2\pi }{|a|}$

$\sin (ax+b)$

$\cos (ax+b)$

$\sec (ax+b)$

$\csc (ax+b)$

Period:

$\frac{\pi }{|a|}$

$\tan (ax+b)$

$\cot (ax+b)$

Trigonometric Equation

Type 1 x = 0

$\sin x=0⟹x=n\pi$

$\tan x=0⟹x=n\pi$

$\cos x=0⟹x=(2n+1)\frac{\pi }{2}$

$\cot x=0⟹x=(2n+1)\frac{\pi }{2}$

$\sec x=0⟹x$ is not defined

$\csc x=0⟹x$ is not defined

Type 2 theta = alpha

$sin\theta =\sin \alpha ⟹x=n\pi +(-1{)}^n\alpha$

$cos\theta =\cos \alpha ⟹x=2n\pi \pm \alpha$

$tan\theta =\tan \alpha ⟹x=n\pi +\alpha$

Type 3 squared

${\sin }^2\theta ={\sin }^2\alpha ⟹x=n\pi \pm \alpha$

${\cos }^2\theta ={\cos }^2\alpha ⟹x=n\pi \pm \alpha$

${\tan }^2\theta ={\tan }^2\alpha ⟹x=n\pi \pm \alpha$

Properties of Triangles

Basic Traingle Formulas

$\mathrm{\angle }A+\mathrm{\angle }B+\mathrm{\angle }C={180}^{\circ }$

$A+B+C={180}^{\circ }$

Perimeter of a circle

$2S=a+b+c$

Area of Triangle from Heron's Formuls

$\mathrm{\Delta }=\sqrt{S(S-a)(S-b)(S-c)}$
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

$\mathrm{△}ABC$

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

$\tan (A+B+C)=\tan (\pi )=0=\frac{a}{b}$

$⟹a=0$

We get the formula

$⟹\tan A+\tan B+\tan C=\tan A\tan B\tan C$
in any

$\mathrm{△}ABC$

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,

$\tan (A+B+C)=\tan (\frac{\pi }{2})=\frac{1}{0}=\frac{a}{b}$

$⟹b=0$

$⟹1-\tan A\tan B-\tan B\tan C-\tan C\tan A=0$

$⟹\tan A\tan B+\tan B\tan C+\tan C\tan A=1$

Identities in any Triangle

2A 2B 2C

In sin

$\sin 2A+\sin 2B+\sin 2C=4\sin A\sin B\sin C$
sin goes to sin

If one is negative

$\sin 2A+\sin 2B-\sin 2C=4\cos A\cos B\sin C$
then negative will be sin.

If more then one negative. Then take common -1.

$\sin 2A-\sin 2B-\sin 2C$

$-1(-\sin 2A+\sin 2B+\sin 2C)$

$=-4\sin A\cos B\cos C$

In cos

$\cos 2A+\cos 2B+cos2C=-1-4\cos A\cos B\cos C$
cos goes to cos with -1

If one is negative

$\cos 2A+\cos 2B-cos2C=1-4\sin A\sin B\cos C$
then negative will be cos

A+B+C

In sin

$\sin A+\sin B+\sin C=4\cos \frac{A}{2}cos\frac{B}{2}cos\frac{C}{2}$

$\because \frac{\pi }{2}-\theta$ so change ratio

If one is negative

$\sin A+\sin B-\sin C=4\sin \frac{A}{2}sin\frac{B}{2}cos\frac{C}{2}$
Invert which are positive in +ve form

In cos

$\cos A+\cos B+\cos C=1+4\sin \frac{A}{2}sin\frac{B}{2}sin\frac{C}{2}$

$\because \frac{\pi }{2}-\theta$ so all positive and change ratio

If one is negative

$\cos A+\cos B-\cos C=-1+4\cos \frac{A}{2}cos\frac{B}{2}sin\frac{C}{2}$
-1 + Invert which are positive in +ve form

Notice: In case of negative invert from +ve from whichever are not negative.

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

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

$\therefore a=k\sin A$

$\therefore b=k\sin B$

$\therefore c=k\sin C$

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

cosine formula

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

$\cos B=\frac{a^2+c^2-b^2}{2ac}$

$\cos C=\frac{a^2+b^2-c^2}{2ab}$
Notice: One angle and all three sides relation

Projection formula

$a=b\cos C+c\cos B$

$b=a\cos C+c\cos A$

$c=a\cos B+b\cos A$
Two angles and all three sides relation

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

$A={30}^{\circ }$	$B={60}^{\circ }$	$C={90}^{\circ }$
$a=1$	$b=\sqrt{3}$	$2$

$A={30}^{\circ }$	$B={60}^{\circ }$	$C={90}^{\circ }$
$a=1$	$b=\sqrt{3}$	$2$

$A={30}^{\circ }$	$B={60}^{\circ }$	$C={90}^{\circ }$
$a=1$	$b=\sqrt{3}$	$2$

Circumcircle of a triangle

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

Incenter of a triangle

$r=\frac{\mathrm{\Delta }}{S}$

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

Inverse Trionometric Functions

${\sin }^{-1}x={\csc }^{-1}\frac{1}{x}$

${\cos }^{-1}x={\sec }^{-1}\frac{1}{x}$

${\tan }^{-1}x={\cot }^{-1}\frac{1}{x}$

${\csc }^{-1}x={\sin }^{-1}\frac{1}{x}$

${\sec }^{-1}x={\cos }^{-1}\frac{1}{x}$

${\cot }^{-1}x={\tan }^{-1}\frac{1}{x}$

${\sin }^{-1}x+{\cos }^{-1}x=\frac{\pi }{2}$

${\tan }^{-1}x+{\cot }^{-1}x=\frac{\pi }{2}$

${\sec }^{-1}x+{\csc }^{-1}x=\frac{\pi }{2}$

${\sin }^{-1}(\sin x)=x$,

$x\in [-\frac{\pi }{2},\frac{\pi }{2}]$

${\cos }^{-1}(\cos x)=x$,

$x\in [0,\pi ]$

${\sec }^{-1}(\sec x)=x$,

$x\in (-\frac{\pi }{2},\frac{\pi }{2})$

$\sin ({\sin }^{-1}x)=x$,

$x\in [-1,1]$

$\cos ({\cos }^{-1}x)=x$,

$x\in [-1,1]$

$\tan ({\tan }^{-1}x)=x$,

$x\in [-1,1]$

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

${\sin }^{-1}x-{\sin }^{-1}y={\sin }^{-1}[x\sqrt{1-y^2}-y\sqrt{1-x^2}]$

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

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

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

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

$2{\sin }^{-1}x={\sin }^{-1}(2x\sqrt{1-x^2})$

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

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

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

$={\sin }^{-1}(\frac{2x}{1+x^2})$

$3{\sin }^{-1}x={\sin }^{-1}(3x-4x^3)$

$3{\cos }^{-1}x={\cos }^{-1}(4x^3-3x)$

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

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

Licensing and Links

All Mathematics Formula by Abhays here

This work is licensed under a Creative Commons Attribution-NonCommercial 4.0 International License.

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