Trigonometry

Trigonometry

🟢 Lite — Quick Review (1h–1d)

Rapid summary for last-minute revision before your exam.

Six Trigonometric Ratios: For angle $\theta$ in right triangle: $\sin\theta = \frac{\text{opposite}}{\text{hypotenuse}}$, $\cos\theta = \frac{\text{adjacent}}{\text{hypotenuse}}$, $\tan\theta = \frac{\sin\theta}{\cos\theta}$. Reciprocals: $\csc\theta = \frac{1}{\sin\theta}$, $\sec\theta = \frac{1}{\cos\theta}$, $\cot\theta = \frac{1}{\tan\theta} = \frac{\cos\theta}{\sin\theta}$.

Fundamental Identities:

• $\sin^2\theta + \cos^2\theta = 1$
• $1 + \tan^2\theta = \sec^2\theta$
• $1 + \cot^2\theta = \csc^2\theta$

Signs in Quadrants:

Quadrant	sin	cos	tan
I	+	+	+
II	+	-	-
III	-	-	+
IV	-	+	-

Key Values to Memorise:

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

⚡ JEE Tip: For JEE, the identity $\tan\theta = \frac{\sin\theta}{\cos\theta}$ and knowing exact values is crucial. Also remember $\sin(2\theta) = 2\sin\theta\cos\theta$.

⚡ Common Mistake: $\sin(30°) = 1/2$, NOT $1/3$. $\cos(60°) = 1/2$, NOT $1/3$. Always double-check angle conventions (degrees vs radians).

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🟡 Standard — Regular Study (2d–2mo)

For students who want genuine understanding.

Compound Angle Formulas:

• $\sin(A \pm B) = \sin A \cos B \pm \cos A \sin B$
• $\cos(A \pm B) = \cos A \cos B \mp \sin A \sin B$
• $\tan(A \pm B) = \frac{\tan A \pm \tan B}{1 \mp \tan A \tan B}$

Double Angle:

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

Multiple Angle Formulas:

• $\sin 3A = 3\sin A - 4\sin^3 A$
• $\cos 3A = 4\cos^3 A - 3\cos A$
• $\tan 3A = \frac{3\tan A - \tan^3 A}{1 - 3\tan^2 A}$

Transformation Formulas:

• $\sin A + \sin B = 2\sin\left(\frac{A+B}{2}\right)\cos\left(\frac{A-B}{2}\right)$
• $\sin A - \sin B = 2\cos\left(\frac{A+B}{2}\right)\sin\left(\frac{A-B}{2}\right)$
• $\cos A + \cos B = 2\cos\left(\frac{A+B}{2}\right)\cos\left(\frac{A-B}{2}\right)$
• $\cos A - \cos B = -2\sin\left(\frac{A+B}{2}\right)\sin\left(\frac{A-B}{2}\right)$

Worked Examples:

Example 1 (JEE 2022): If $\sin A + \sin B = 1$ and $\cos A + \cos B = \sqrt{2}$, find $\cos(A-B)$.

Square and add: $(\sin A + \sin B)^2 + (\cos A + \cos B)^2 = 1^2 + (\sqrt{2})^2 = 1 + 2 = 3$. Left side: $\sin^2 A + \sin^2 B + 2\sin A \sin B + \cos^2 A + \cos^2 B + 2\cos A \cos B = (\sin^2 A + \cos^2 A) + (\sin^2 B + \cos^2 B) + 2(\sin A \sin B + \cos A \cos B) = 1 + 1 + 2\cos(A-B) = 2 + 2\cos(A-B)$. So $2 + 2\cos(A-B) = 3$ → $2\cos(A-B) = 1$ → $\cos(A-B) = 1/2$.

Example 2: Prove that if $A + B + C = \pi$, then $\tan A + \tan B + \tan C = \tan A \tan B \tan C$. (This is the standard “angles of a triangle” identity that JEE problems rely on.)

Since $A + B + C = \pi$, we have $A + B = \pi - C$. Taking the tangent of both sides and using $\tan(\pi - C) = -\tan C$: $$\tan(A + B) = -\tan C.$$ Expand the left side with the compound-angle formula: $$\frac{\tan A + \tan B}{1 - \tan A \tan B} = -\tan C.$$ Cross-multiplying, $$\tan A + \tan B = -\tan C,(1 - \tan A \tan B) = -\tan C + \tan A \tan B \tan C.$$ Adding $\tan C$ to both sides gives the result: $$\tan A + \tan B + \tan C = \tan A \tan B \tan C. \quad \blacksquare$$

⚡ Caution: this identity requires $A + B + C = \pi$. It does not hold for arbitrary angles — for example $20°, 40°, 60°$ sum to $120°$ (not $180°$), and there the sum $\approx 2.94$ while the product $\approx 0.53$, so they are not equal.

Useful conditional identity: When $A + B + C = \pi$, $$\tan A + \tan B + \tan C = \tan A \tan B \tan C.$$ This follows from $A + B = \pi - C$, so $\tan(A+B) = -\tan C$, giving $\tan A + \tan B = \tan C(\tan A \tan B - 1)$ and therefore the stated product relation.

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🔴 Extended — Deep Study (3mo+)

Comprehensive theory for serious JEE Advanced preparation.

Euler’s Formula: $e^{i\theta} = \cos\theta + i\sin\theta$. From this: $\cos\theta = \frac{e^{i\theta} + e^{-i\theta}}{2}$, $\sin\theta = \frac{e^{i\theta} - e^{-i\theta}}{2i}$.

De Moivre’s Theorem: $(\cos\theta + i\sin\theta)^n = \cos(n\theta) + i\sin(n\theta)$.

Summation of Trigonometric Series: Series such as $\sum \cos(k\theta)$ and $\sum \sin(k\theta)$ are evaluated by treating them as the real and imaginary parts of a geometric series in $e^{i\theta}$, or equivalently by telescoping with product-to-sum transformation formulas.

Inverse Trigonometric Functions:

• $\sin(\arcsin x) = x$, $\arcsin(\sin x) = x$ (principal value)
• Domain and range: $\arcsin: [-1,1] \to [-\pi/2, \pi/2]$
• $\arctan: \mathbb{R} \to (-\pi/2, \pi/2)$

Principal Values:

• $\arcsin x \in [-\pi/2, \pi/2]$
• $\arccos x \in [0, \pi]$
• $\arctan x \in (-\pi/2, \pi/2)$

Advanced Problems:

Problem (JEE Advanced 2020): Solve $\sin^{-1} x + \sin^{-1} 2x = \pi/3$.

First fix the domain. Both $\sin^{-1} x$ and $\sin^{-1} 2x$ require $|x| \le 1$ and $|2x| \le 1$, so the feasible set is $|x| \le 1/2$.

Apply the addition formula for inverse sine. When the sum lies in $[-\pi/2, \pi/2]$, $$\sin^{-1} x + \sin^{-1} 2x = \sin^{-1}!\big(x\sqrt{1-4x^2} + 2x\sqrt{1-x^2}\big).$$ Setting this equal to $\pi/3$ and taking the sine of both sides gives $$x\sqrt{1-4x^2} + 2x\sqrt{1-x^2} = \sin(\pi/3) = \frac{\sqrt{3}}{2}.$$ Factor out $x$: $$x\big[\sqrt{1-4x^2} + 2\sqrt{1-x^2}\big] = \frac{\sqrt{3}}{2}.$$ Since the bracket is positive on the domain, $x > 0$. Square both sides: $$x^2\big[,1-4x^2 + 4(1-x^2) + 4\sqrt{(1-4x^2)(1-x^2)},\big] = \frac{3}{4}.$$ Simplify the bracket to $5 - 8x^2 + 4\sqrt{1 - 5x^2 + 4x^4}$. Substituting $u = x^2$ with $u \in [0, 1/4]$: $$5u - 8u^2 + 4u\sqrt{1 - 5u + 4u^2} = \frac{3}{4}.$$ Isolate the radical: $$4u\sqrt{1 - 5u + 4u^2} = 8u^2 - 5u + \frac{3}{4}.$$ Squaring once more, $$16u^2(1 - 5u + 4u^2) = \Big(8u^2 - 5u + \tfrac{3}{4}\Big)^2,$$ which reduces to a polynomial equation in $u$. Solving it and discarding roots that violate $u \in [0, 1/4]$ (and that fail to keep the original inverse-sine sum within its principal range) yields the admissible value of $x$ in the interval $0 < x \le 1/2$. Always back-substitute the candidate root into the original equation, since squaring can introduce extraneous solutions.

JEE Advanced Patterns (2018–2024):

• Inverse trigonometric equations and identities are very common
• Sum-to-product and product-to-sum transformations are frequent
• Trigonometric equations with multiple angles tested in 2021, 2023
• Maximum/minimum of trigonometric expressions using AM-GM or Cauchy
• Conditional trigonometric identities (like the $A+B+C=\pi$ type) are common

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