Trigonometry Formulas For JEE 2026
Trigonometry is an important chapter in JEE Mathematics and is known for its high weightage, with around 3–4 questions asked every year. This topic includes key concepts such as trigonometric ratios, identities, equations, and inverse trigonometric functions. When students understand these basics clearly, it becomes easier to solve different types of problems and also helps in other related topics.
To perform well, students should be comfortable with formulas like compound angles, double, half, and triple angle formulas, sum-to-product conversions, and the general solutions of trigonometric equations. Since many questions are based on direct application of formulas, regular practice is very important. For quick revision, students can also use a well-organized JEE Mains Mathematics Formula PDF to go through important formulas and concepts easily.
Angles and Measurement
Degree–Radian Conversion
$$180° = \pi \text{ radians}$$- Degrees to radians: multiply by $\dfrac{\pi}{180}$
- Radians to degrees: multiply by $\dfrac{180}{\pi}$
- Arc length: $\ell = r\theta$ (where $\theta$ is in radians)
Trigonometric Ratios
Six Trigonometric Ratios
- $\sin\theta = \dfrac{\text{Opposite}}{\text{Hypotenuse}}$, $\csc\theta = \dfrac{1}{\sin\theta}$
- $\cos\theta = \dfrac{\text{Adjacent}}{\text{Hypotenuse}}$, $\sec\theta = \dfrac{1}{\cos\theta}$
- $\tan\theta = \dfrac{\text{Opposite}}{\text{Adjacent}} = \dfrac{\sin\theta}{\cos\theta}$, $\cot\theta = \dfrac{1}{\tan\theta}$
Standard Angle Values
| Ratio | $0°$ | $30°$ | $45°$ | $60°$ | $90°$ |
|---|---|---|---|---|---|
| $\sin\theta$ | $0$ | $\dfrac{1}{2}$ | $\dfrac{1}{\sqrt{2}}$ | $\dfrac{\sqrt{3}}{2}$ | $1$ |
| $\cos\theta$ | $1$ | $\dfrac{\sqrt{3}}{2}$ | $\dfrac{1}{\sqrt{2}}$ | $\dfrac{1}{2}$ | $0$ |
| $\tan\theta$ | $0$ | $\dfrac{1}{\sqrt{3}}$ | $1$ | $\sqrt{3}$ | undefined |
Memory trick: For $\sin$, write $\sqrt{0}, \sqrt{1}, \sqrt{2}, \sqrt{3}, \sqrt{4}$ and divide each by 2. For $\cos$, reverse the order.
Signs in Quadrants
Signs in Quadrants — "All Students Take Coffee"
| Ratio | Q I ($0°$–$90°$) | Q II ($90°$–$180°$) | Q III ($180°$–$270°$) | Q IV ($270°$–$360°$) |
|---|---|---|---|---|
| $\sin, \csc$ | $+$ | $+$ | $-$ | $-$ |
| $\cos, \sec$ | $+$ | $-$ | $-$ | $+$ |
| $\tan, \cot$ | $+$ | $-$ | $+$ | $-$ |
ASTC: All positive in Q I, Sin positive in Q II, Tan positive in Q III, Cos positive in Q IV.
Allied Angles
Allied Angle Rules
Rule 1: For $(90° \pm \theta)$ or $(270° \pm \theta)$: function changes ($\sin \leftrightarrow \cos$, $\tan \leftrightarrow \cot$, $\sec \leftrightarrow \csc$).
Rule 2: For $(180° \pm \theta)$ or $(360° \pm \theta)$: function stays the same.
Rule 3: Sign is determined by the quadrant of the original angle.
Key results:
- $\sin(90° - \theta) = \cos\theta$, $\cos(90° - \theta) = \sin\theta$
- $\sin(180° - \theta) = \sin\theta$, $\cos(180° - \theta) = -\cos\theta$
- $\sin(180° + \theta) = -\sin\theta$, $\cos(180° + \theta) = -\cos\theta$
- $\sin(-\theta) = -\sin\theta$, $\cos(-\theta) = \cos\theta$, $\tan(-\theta) = -\tan\theta$
Fundamental Identities
Pythagorean Identities
- $\sin^2\theta + \cos^2\theta = 1$
- $1 + \tan^2\theta = \sec^2\theta$
- $1 + \cot^2\theta = \csc^2\theta$
Compound Angle
Sum and Difference Formulas
- $\sin(A + B) = \sin A \cos B + \cos A \sin B$
- $\sin(A - B) = \sin A \cos B - \cos A \sin B$
- $\cos(A + B) = \cos A \cos B - \sin A \sin B$
- $\cos(A - B) = \cos A \cos B + \sin A \sin B$
- $\tan(A + B) = \dfrac{\tan A + \tan B}{1 - \tan A \tan B}$
- $\tan(A - B) = \dfrac{\tan A - \tan B}{1 + \tan A \tan B}$
Tip: In the $\sin$ formulas, the sign between terms matches the sign in the argument. In the $\cos$ formulas, the sign is opposite.
Worked Example
Find $\cos 75°$.
$\cos 75° = \cos(45° + 30°) = \cos 45° \cos 30° - \sin 45° \sin 30° = \dfrac{\sqrt{3} - 1}{2\sqrt{2}} = \boldsymbol{\dfrac{\sqrt{6} - \sqrt{2}}{4}}$
Double Angle Formula
- $\sin 2\theta = 2\sin\theta\cos\theta$
- $\cos 2\theta = \cos^2\theta - \sin^2\theta = 2\cos^2\theta - 1 = 1 - 2\sin^2\theta$
- $\tan 2\theta = \dfrac{2\tan\theta}{1 - \tan^2\theta}$
Useful rearrangements:
- $\cos^2\theta = \dfrac{1 + \cos 2\theta}{2}$, $\sin^2\theta = \dfrac{1 - \cos 2\theta}{2}$
- $1 + \sin 2\theta = (\sin\theta + \cos\theta)^2$
- $1 - \sin 2\theta = (\sin\theta - \cos\theta)^2$
Half Angle Formula
- $\sin\dfrac{\theta}{2} = \pm\sqrt{\dfrac{1 - \cos\theta}{2}}$
- $\cos\dfrac{\theta}{2} = \pm\sqrt{\dfrac{1 + \cos\theta}{2}}$
- $\tan\dfrac{\theta}{2} = \pm\sqrt{\dfrac{1 - \cos\theta}{1 + \cos\theta}} = \dfrac{\sin\theta}{1 + \cos\theta} = \dfrac{1 - \cos\theta}{\sin\theta}$
(The $\pm$ sign depends on the quadrant in which $\theta/2$ lies.)
Triple Angle Formula
- $\sin 3\theta = 3\sin\theta - 4\sin^3\theta$
- $\cos 3\theta = 4\cos^3\theta - 3\cos\theta$
- $\tan 3\theta = \dfrac{3\tan\theta - \tan^3\theta}{1 - 3\tan^2\theta}$
Sum-to-Product and Product-to-Sum Formula
Product-to-Sum Formulas
- $2\sin A\cos B = \sin(A+B) + \sin(A-B)$
- $2\cos A\sin B = \sin(A+B) - \sin(A-B)$
- $2\cos A\cos B = \cos(A-B) + \cos(A+B)$
- $2\sin A\sin B = \cos(A-B) - \cos(A+B)$
Sum-to-Product Formulas
- $\sin C + \sin D = 2\sin\dfrac{C+D}{2}\cos\dfrac{C-D}{2}$
- $\sin C - \sin D = 2\cos\dfrac{C+D}{2}\sin\dfrac{C-D}{2}$
- $\cos C + \cos D = 2\cos\dfrac{C+D}{2}\cos\dfrac{C-D}{2}$
- $\cos C - \cos D = -2\sin\dfrac{C+D}{2}\sin\dfrac{C-D}{2}$
Worked Example
Express $\sin 75° + \sin 15°$ as a product.
$= 2\sin 45° \cos 30° = 2 \cdot \dfrac{1}{\sqrt{2}} \cdot \dfrac{\sqrt{3}}{2} = \boldsymbol{\dfrac{\sqrt{6}}{2}}$
Trigonometric Equations
General Solutions
| Equation | General Solution |
|---|---|
| $\sin\theta = 0$ | $\theta = n\pi$ |
| $\cos\theta = 0$ | $\theta = (2n+1)\dfrac{\pi}{2}$ |
| $\tan\theta = 0$ | $\theta = n\pi$ |
| $\sin\theta = \sin\alpha$ | $\theta = n\pi + (-1)^n \alpha$ |
| $\cos\theta = \cos\alpha$ | $\theta = 2n\pi \pm \alpha$ |
| $\tan\theta = \tan\alpha$ | $\theta = n\pi + \alpha$ |
where $n \in \mathbb{Z}$.
Worked Example
Solve $2\cos^2\theta - 3\cos\theta + 1 = 0$.
Factor: $(2\cos\theta - 1)(\cos\theta - 1) = 0$
Case 1: $\cos\theta = \dfrac{1}{2} \Rightarrow \theta = 2n\pi \pm \dfrac{\pi}{3}$
Case 2: $\cos\theta = 1 \Rightarrow \theta = 2n\pi$
Tip: For equations like $a\sin\theta + b\cos\theta = c$, write the LHS as $R\sin(\theta + \phi)$ where $R = \sqrt{a^2 + b^2}$ and $\tan\phi = b/a$. A solution exists only when $|c| \leq R$.
Inverse Trigonometric Functions
Domain and Range of Inverse Trig Functions
| Function | Domain | Range (Principal Value) |
|---|---|---|
| $\sin^{-1}x$ | $[-1, 1]$ | $\left[-\dfrac{\pi}{2}, \dfrac{\pi}{2}\right]$ |
| $\cos^{-1}x$ | $[-1, 1]$ | $[0, \pi]$ |
| $\tan^{-1}x$ | $\mathbb{R}$ | $\left(-\dfrac{\pi}{2}, \dfrac{\pi}{2}\right)$ |
| $\cot^{-1}x$ | $\mathbb{R}$ | $(0, \pi)$ |
| $\sec^{-1}x$ | $(-\infty,-1] \cup [1,\infty)$ | $[0,\pi] \setminus \left\{\dfrac{\pi}{2}\right\}$ |
| $\csc^{-1}x$ | $(-\infty,-1] \cup [1,\infty)$ | $\left[-\dfrac{\pi}{2}, \dfrac{\pi}{2}\right] \setminus \{0\}$ |
Important: $\sin^{-1}x$ does NOT mean $\dfrac{1}{\sin x}$. It means "the angle whose sine is $x$." Also written as $\arcsin x$.
Negative Argument Properties
- $\sin^{-1}(-x) = -\sin^{-1}(x)$ (odd function)
- $\cos^{-1}(-x) = \pi - \cos^{-1}(x)$
- $\tan^{-1}(-x) = -\tan^{-1}(x)$ (odd function)
- $\cot^{-1}(-x) = \pi - \cot^{-1}(x)$
Complementary Relationships
- $\sin^{-1}x + \cos^{-1}x = \dfrac{\pi}{2}$, for $x \in [-1, 1]$
- $\tan^{-1}x + \cot^{-1}x = \dfrac{\pi}{2}$, for all $x \in \mathbb{R}$
- $\sec^{-1}x + \csc^{-1}x = \dfrac{\pi}{2}$, for $|x| \geq 1$
Sum and Difference of Inverse Tangents
$\tan^{-1}x + \tan^{-1}y = \begin{cases} \tan^{-1}\!\left(\dfrac{x+y}{1-xy}\right) & \text{if } xy < 1 \\ \pi + \tan^{-1}\!\left(\dfrac{x+y}{1-xy}\right) & \text{if } xy > 1,\; x > 0 \\ -\pi + \tan^{-1}\!\left(\dfrac{x+y}{1-xy}\right) & \text{if } xy > 1,\; x < 0 \end{cases}$
$\tan^{-1}x - \tan^{-1}y = \tan^{-1}\!\left(\dfrac{x-y}{1+xy}\right)$, if $xy > -1$
Worked Example
Evaluate $\tan^{-1}1 + \tan^{-1}2 + \tan^{-1}3$.
$\tan^{-1}2 + \tan^{-1}3$: since $xy = 6 > 1$ and $x > 0$:
$= \pi + \tan^{-1}\!\left(\dfrac{5}{-5}\right) = \pi + \tan^{-1}(-1) = \pi - \dfrac{\pi}{4} = \dfrac{3\pi}{4}$
$\tan^{-1}1 + \dfrac{3\pi}{4} = \dfrac{\pi}{4} + \dfrac{3\pi}{4} = \boldsymbol{\pi}$
Double Angle Identities for Inverse Trig
- $2\tan^{-1}x = \sin^{-1}\!\left(\dfrac{2x}{1+x^2}\right)$ if $|x| \leq 1$
- $2\tan^{-1}x = \cos^{-1}\!\left(\dfrac{1-x^2}{1+x^2}\right)$ if $x \geq 0$
- $2\tan^{-1}x = \tan^{-1}\!\left(\dfrac{2x}{1-x^2}\right)$ if $|x| < 1$
Conversion Between Inverse Trig Functions
- $\sin^{-1}x = \cos^{-1}\!\sqrt{1-x^2} = \tan^{-1}\!\dfrac{x}{\sqrt{1-x^2}}$ (for $x \geq 0$)
- $\tan^{-1}x = \sin^{-1}\!\dfrac{x}{\sqrt{1+x^2}} = \cos^{-1}\!\dfrac{1}{\sqrt{1+x^2}}$
Worked Example
Simplify $\sin(\tan^{-1}x)$.
Let $\theta = \tan^{-1}x$, so $\tan\theta = x$. Right triangle: opposite $= x$, adjacent $= 1$, hypotenuse $= \sqrt{1+x^2}$.
$\sin(\tan^{-1}x) = \boldsymbol{\dfrac{x}{\sqrt{1+x^2}}}$
Tip: Many JEE problems on inverse trig reduce to: (1) applying the sum formula for $\tan^{-1}$, (2) using $\sin^{-1}x + \cos^{-1}x = \pi/2$, or (3) converting between inverse trig functions via right triangles. Master these three techniques.
Trigonometry Formulas For JEE 2026: Conclusion
Trigonometry is a core topic in JEE Mathematics that requires a clear understanding of formulas and their applications. Concepts like identities, compound angles, and trigonometric equations form the base for solving a variety of problems efficiently.
Consistent practice and regular revision play a key role in mastering this chapter. By focusing on standard formulas and improving problem-solving skills, students can boost their accuracy and confidence in the exam.
