Trigonometry Formula Sheet

Every formula you need for JEE Trigonometry on one page: ratios, identities, compound/multiple-angle formulas, transformation formulas, general solutions, inverse trig properties, and triangle rules. Use this as your last-minute revision sheet.

Basic Ratios & Standard Values

In a right triangle:

$$\sin\theta = \frac{\text{opposite}}{\text{hypotenuse}}, \quad \cos\theta = \frac{\text{adjacent}}{\text{hypotenuse}}, \quad \tan\theta = \frac{\text{opposite}}{\text{adjacent}}$$

Reciprocal ratios:

$$\csc\theta = \frac{1}{\sin\theta}, \quad \sec\theta = \frac{1}{\cos\theta}, \quad \cot\theta = \frac{1}{\tan\theta}$$

Quotient identities:

$$\tan\theta = \frac{\sin\theta}{\cos\theta}, \quad \cot\theta = \frac{\cos\theta}{\sin\theta}$$

Standard Angle Values

Angle	0°	30°	45°	60°	90°
$\sin$	0	1/2	$1/\sqrt{2}$	$\sqrt{3}/2$	1
$\cos$	1	$\sqrt{3}/2$	$1/\sqrt{2}$	1/2	0
$\tan$	0	$1/\sqrt{3}$	1	$\sqrt{3}$	$\infty$

Fundamental (Pythagorean) Identities

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

$$\boxed{1 + \tan^2\theta = \sec^2\theta}$$

$$\boxed{1 + \cot^2\theta = \csc^2\theta}$$

Identity	When to Use
$\sin^2\theta + \cos^2\theta = 1$	Find one ratio from another
$1 + \tan^2\theta = \sec^2\theta$	Tan–Sec conversions
$1 + \cot^2\theta = \csc^2\theta$	Cot–Cosec conversions

Compound (Sum & Difference) Formulas

$$\boxed{\sin(A \pm B) = \sin A \cos B \pm \cos A \sin B}$$

$$\boxed{\cos(A \pm B) = \cos A \cos B \mp \sin A \sin B}$$

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

Double Angle Formulas

$$\boxed{\sin 2\theta = 2\sin\theta\cos\theta}$$

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

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

Triple Angle Formulas

$$\boxed{\sin 3\theta = 3\sin\theta - 4\sin^3\theta}$$

$$\boxed{\cos 3\theta = 4\cos^3\theta - 3\cos\theta}$$

$$\boxed{\tan 3\theta = \frac{3\tan\theta - \tan^3\theta}{1 - 3\tan^2\theta}}$$

Half Angle Formulas

$$\sin^2\frac{\theta}{2} = \frac{1 - \cos\theta}{2}, \qquad \cos^2\frac{\theta}{2} = \frac{1 + \cos\theta}{2}$$$$\boxed{\sin\frac{\theta}{2} = \pm\sqrt{\frac{1-\cos\theta}{2}}, \quad \cos\frac{\theta}{2} = \pm\sqrt{\frac{1+\cos\theta}{2}}}$$$$\boxed{\tan\frac{\theta}{2} = \frac{1 - \cos\theta}{\sin\theta} = \frac{\sin\theta}{1 + \cos\theta}}$$

Transformation Formulas

Product → Sum

$$2\sin A \cos B = \sin(A+B) + \sin(A-B)$$

$$2\cos A \sin B = \sin(A+B) - \sin(A-B)$$

$$2\cos A \cos B = \cos(A+B) + \cos(A-B)$$

$$2\sin A \sin B = \cos(A-B) - \cos(A+B)$$

Sum → Product

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

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

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

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

Periodicity of Trig Functions

Function	Period
$\sin\theta,\ \cos\theta$	$2\pi$ (360°)
$\sec\theta,\ \csc\theta$	$2\pi$ (360°)
$\tan\theta,\ \cot\theta$	$\pi$ (180°)

General Solutions of Trig Equations

The Big 6 ($n \in \mathbb{Z}$, $\alpha$ = principal value):

$$\boxed{ \begin{align} \sin\theta = \sin\alpha &\implies \theta = n\pi + (-1)^n\alpha \\ \cos\theta = \cos\alpha &\implies \theta = 2n\pi \pm \alpha \\ \tan\theta = \tan\alpha &\implies \theta = n\pi + \alpha \\ \sin\theta = 0 &\implies \theta = n\pi \\ \cos\theta = 0 &\implies \theta = (2n+1)\tfrac{\pi}{2} \\ \tan\theta = 0 &\implies \theta = n\pi \end{align} }$$

Solutions for ±1

$$\boxed{ \begin{align} \sin\theta = 1 &\implies \theta = (4n+1)\tfrac{\pi}{2} \\ \sin\theta = -1 &\implies \theta = (4n-1)\tfrac{\pi}{2} \\ \cos\theta = 1 &\implies \theta = 2n\pi \\ \cos\theta = -1 &\implies \theta = (2n+1)\pi \\ \tan\theta = 1 &\implies \theta = n\pi + \tfrac{\pi}{4} \\ \tan\theta = -1 &\implies \theta = n\pi - \tfrac{\pi}{4} \end{align} }$$

Equation	General Solution
$\sin\theta = \sin\alpha$	$\theta = n\pi + (-1)^n\alpha$
$\cos\theta = \cos\alpha$	$\theta = 2n\pi \pm \alpha$
$\tan\theta = \tan\alpha$	$\theta = n\pi + \alpha$
$a\sin\theta + b\cos\theta = c$	Auxiliary angle: divide by $\sqrt{a^2+b^2}$
Quadratic in $\sin\theta$	Substitute $t = \sin\theta$, solve, then find $\theta$

Auxiliary Angle Method

For $a\sin\theta + b\cos\theta = c$:

$$\boxed{a\sin\theta + b\cos\theta = \sqrt{a^2 + b^2}\,\sin(\theta + \phi)}, \qquad \tan\phi = \frac{b}{a}$$

Inverse Trigonometric Functions

Domain & Range (Principal Values)

$$\boxed{ \begin{array}{|c|c|c|} \hline \textbf{Function} & \textbf{Domain} & \textbf{Range} \\ \hline \sin^{-1} x & [-1, 1] & [-\tfrac{\pi}{2}, \tfrac{\pi}{2}] \\ \cos^{-1} x & [-1, 1] & [0, \pi] \\ \tan^{-1} x & \mathbb{R} & (-\tfrac{\pi}{2}, \tfrac{\pi}{2}) \\ \cot^{-1} x & \mathbb{R} & (0, \pi) \\ \sec^{-1} x & (-\infty,-1]\cup[1,\infty) & [0,\pi]-\{\tfrac{\pi}{2}\} \\ \csc^{-1} x & (-\infty,-1]\cup[1,\infty) & [-\tfrac{\pi}{2},\tfrac{\pi}{2}]-\{0\} \\ \hline \end{array} }$$

Complementary Relations

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

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

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

Negative Argument

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

$$\cos^{-1}(-x) = \pi - \cos^{-1}x, \quad \sec^{-1}(-x) = \pi - \sec^{-1}x, \quad \cot^{-1}(-x) = \pi - \cot^{-1}x$$

Addition Formulas

$$\boxed{\tan^{-1}x + \tan^{-1}y = \begin{cases} \tan^{-1}\dfrac{x+y}{1-xy} & xy < 1 \\[4pt] \pi + \tan^{-1}\dfrac{x+y}{1-xy} & xy > 1,\ x,y > 0 \\[4pt] -\pi + \tan^{-1}\dfrac{x+y}{1-xy} & xy > 1,\ x,y < 0 \end{cases}}$$$$\tan^{-1}x - \tan^{-1}y = \tan^{-1}\frac{x-y}{1+xy}$$

Double-Argument Formulas

$$2\sin^{-1}x = \sin^{-1}\!\left(2x\sqrt{1-x^2}\right), \quad |x| \le \tfrac{1}{\sqrt{2}}$$

$$2\cos^{-1}x = \cos^{-1}\!\left(2x^2 - 1\right), \quad 0 \le x \le 1$$

$$2\tan^{-1}x = \begin{cases} \tan^{-1}\dfrac{2x}{1-x^2} & |x| < 1 \\[4pt] \pi + \tan^{-1}\dfrac{2x}{1-x^2} & x > 1 \\[4pt] -\pi + \tan^{-1}\dfrac{2x}{1-x^2} & x < -1 \end{cases}$$

Conversion Formulas

$$\sin^{-1}x = \cos^{-1}\sqrt{1-x^2} = \tan^{-1}\frac{x}{\sqrt{1-x^2}} = \cot^{-1}\frac{\sqrt{1-x^2}}{x}$$

$$\tan^{-1}x = \sin^{-1}\frac{x}{\sqrt{1+x^2}} = \cos^{-1}\frac{1}{\sqrt{1+x^2}} \ \ (x \ge 0)$$

Calculus Quick Reference

Derivatives:

$$\frac{d}{dx}\sin^{-1}x = \frac{1}{\sqrt{1-x^2}}, \quad \frac{d}{dx}\cos^{-1}x = \frac{-1}{\sqrt{1-x^2}}, \quad \frac{d}{dx}\tan^{-1}x = \frac{1}{1+x^2}$$

$$\frac{d}{dx}\cot^{-1}x = \frac{-1}{1+x^2}, \quad \frac{d}{dx}\sec^{-1}x = \frac{1}{|x|\sqrt{x^2-1}}, \quad \frac{d}{dx}\csc^{-1}x = \frac{-1}{|x|\sqrt{x^2-1}}$$

Standard integrals:

$$\int \frac{dx}{\sqrt{1-x^2}} = \sin^{-1}x + C, \quad \int \frac{dx}{1+x^2} = \tan^{-1}x + C, \quad \int \frac{dx}{x\sqrt{x^2-1}} = \sec^{-1}x + C$$

Properties of Triangles

Standard notation: side $a$ opposite $\angle A$, etc. Semi-perimeter $s = \dfrac{a+b+c}{2}$, and $A + B + C = \pi$.

Sine Rule

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

where $R$ is the circumradius.

Cosine Rule

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

$$b^2 = a^2 + c^2 - 2ac\cos B, \qquad c^2 = a^2 + b^2 - 2ab\cos C$$

Rearranged to find angles:

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

Projection Rule

$$a = b\cos C + c\cos B, \quad b = a\cos C + c\cos A, \quad c = a\cos B + b\cos A$$

Area Formulas

$$\boxed{\text{Area} = \sqrt{s(s-a)(s-b)(s-c)}} \quad \text{(Heron)}$$

$$\boxed{\text{Area} = \tfrac{1}{2}ab\sin C = \tfrac{1}{2}bc\sin A = \tfrac{1}{2}ac\sin B}$$

$$\text{Area} = \frac{abc}{4R} = \frac{a^2\sin B\sin C}{2\sin A} = rs$$

Circumradius & Inradius

$$\frac{a}{\sin A} = 2R, \qquad \text{Area} = \frac{abc}{4R}$$

$$\text{Area} = rs, \qquad r = (s-a)\tan\frac{A}{2} = (s-b)\tan\frac{B}{2} = (s-c)\tan\frac{C}{2}$$

$$r = 4R\sin\frac{A}{2}\sin\frac{B}{2}\sin\frac{C}{2}$$

m–n Theorem

Cevian from $A$ meets $BC$ at $D$ with $BD:DC = m:n$; $\theta$ = angle of $AD$ with $BC$, $\alpha = \angle BAD$, $\beta = \angle DAC$.

$$(m + n)\cot\theta = m\cot\alpha - n\cot\beta = n\cot B - m\cot C$$

(Not to be confused with Napier’s analogy $\dfrac{a+b}{a-b} = \dfrac{\tan\frac{A+B}{2}}{\tan\frac{A-B}{2}}$, which uses the side ratio.)

Angle-Sum Consequences

$$\sin(B+C) = \sin A, \quad \cos(B+C) = -\cos A, \quad \tan(B+C) = -\tan A$$

Special Triangles

Triangle	Key Results
Equilateral ($a=b=c$, all 60°)	$\text{Area} = \tfrac{\sqrt 3}{4}a^2$, $R = \tfrac{a}{\sqrt 3}$, $r = \tfrac{a}{2\sqrt 3} = \tfrac{R}{2}$
Isosceles ($a=b$, base $c$)	$A = B$, $\text{Area} = \tfrac{c}{4}\sqrt{4a^2 - c^2}$
Right ($C = 90°$)	$a^2+b^2=c^2$, $\text{Area} = \tfrac{1}{2}ab$, $R = \tfrac{c}{2}$, $r = \tfrac{a+b-c}{2}$

Choosing the Right Tool

Given	Use	To Find
SSS (3 sides)	Cosine rule	Angles
SSS	Heron’s formula	Area
SAS	Cosine rule	Third side
SAS	$\tfrac{1}{2}ab\sin C$	Area
ASA / AAS	Sine rule	Other sides
SSA	Sine rule (ambiguous: 0, 1, or 2 triangles)	Angle

Related Topics

• Trigonometric Identities
• Trigonometric Equations
• Inverse Trigonometric Functions
• Properties of Triangles