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Trigonometry Formulas For JEE 2027
Trigonometry is an important and high-weightage chapter in JEE Mathematics for 2027 aspirants. Around 3–4 questions are often asked from this topic in JEE Main and JEE Advanced, making it a scoring area for students. This chapter includes key concepts such as trigonometric ratios, identities, equations, inverse trigonometric functions, and their applications. A clear understanding of these basics helps students solve a wide range of problems and also makes revision easier with a well-organized JEE Mains Maths Formula 2027 resource.
To perform well in Trigonometry, students should be comfortable with important formulas such as compound angles, double angle, half angle, triple angle formulas, sum-to-product conversions, and general solutions of trigonometric equations. Since many questions are based on the direct application of formulas, regular practice and revision are very important. A good JEE Formula 2027 PDF helps aspirants quickly revise key formulas, improve speed, and build confidence during practice and last-minute preparation.
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.
