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Trigonometry Formulas For JEE 2027
Trigonometry is an important topic in JEE Mathematics for 2027 aspirants. It covers trigonometric ratios, identities, equations, inverse trigonometric functions, and their applications. Understanding these concepts also helps students solve problems in calculus and coordinate geometry. A JEE Mains Formula sheet can make it easier to organise and revise important results.
To prepare well, students should practise compound angle, double angle, half angle, and triple angle formulas, along with sum-to-product conversions and general solutions of trigonometric equations. Regular revision helps them recognise the right identity and apply it correctly. Taking a JEE Mains mock test helps students practise under timed conditions, identify mistakes, and improve their speed and accuracy.
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 2027: Conclusion
Trigonometry preparation for JEE 2027 requires a clear understanding of identities, equations, and inverse trigonometric functions. Regular revision with a JEE Maths Formula Sheet helps students recall important results and understand when to apply them. Focus on learning the conditions behind each identity along with practising its use.
Solving JEE Trigonometry PYQ problems helps students recognise common question types and strengthen their problem-solving skills. Review mistakes, reattempt difficult questions, and practise consistently to improve speed and accuracy. Combining concept revision with question practice can build confidence in Trigonometry for JEE 2027.
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