Mathematics · Chapter 3

Trigonometric Functions

20 min read 5 Topics Class 11 (JEE & NEET)

1. Chapter Overview

Every time your phone's GPS calculates the shortest route, or a Ferris wheel operator predicts exactly when a cabin will reach the top, trigonometric functions are running quietly in the background. Unlike the trigonometry you learned in Class 10 (limited to right triangles and acute angles), this chapter extends the idea to any angle — negative, greater than 360 degrees, measured in a completely different unit called radians. This generalization is what makes trigonometric functions periodic, and periodicity is the mathematical language behind waves, oscillations, and circular motion across Physics and Engineering.

In JEE Main, Trigonometric Functions and Trigonometric Equations together are one of the highest-yield topics, typically contributing 2-3 questions (roughly 6-8% weightage) every year, with compound angle identities and general solutions being favourite question types. JEE Advanced frequently combines trigonometric identities with Calculus and Complex Numbers in multi-step problems. For NEET, trigonometric ratios and identities are essential prerequisites for Physics chapters like Oscillations, Waves, and Alternating Current.

Before starting this chapter, you should be comfortable with basic Class 10 trigonometric ratios (sin, cos, tan for right triangles) and with the concept of the Cartesian coordinate plane, since every trigonometric function in this chapter is defined using a circle centered at the origin.

2. Angles, Radian Measure and Conversion

Concept Explanation: An angle is generated by the rotation of a ray about its starting point. A positive angle is measured in the anticlockwise direction; a negative angle is measured clockwise. Angles are measured in two systems: degrees (a full rotation = 360°) and radians (a full rotation = 2π radians), where one radian is the angle subtended at the center of a circle by an arc equal in length to the radius. Key Formula: $$ \theta \text{ (radians)} = \frac{\pi}{180} \times \theta \text{ (degrees)} $$, Arc length: $$ l = r\theta $$
SI Units: Radian (rad) is the SI unit of plane angle; degrees are a commonly used non-SI unit.
Common Exam Mistake: Forgetting to check whether a problem is set in degrees or radians before applying a formula. Correction: Always confirm the unit first — mixing degree and radian values in the same formula is one of the most common calculation errors in this chapter.
JEE/NEET Tip: Memorize the degree-radian pairs for standard angles: 30°=π/6, 45°=π/4, 60°=π/3, 90°=π/2, 180°=π — these appear in almost every question in this chapter.
Solved Example:
Convert 75° into radians, and find the arc length it subtends on a circle of radius 8 cm.
Step-by-Step Solution:
$$ \theta = \frac{\pi}{180} \times 75 = \frac{5\pi}{12} \text{ rad} $$. Arc length $$ l = r\theta = 8 \times \frac{5\pi}{12} = \frac{10\pi}{3} \text{ cm} $$.

3. Trigonometric Functions Using the Unit Circle

Concept Explanation: For any angle θ, consider a point P(x, y) on a circle of radius 1 (the unit circle) centered at the origin, obtained by rotating from the positive X-axis by angle θ. Then cos θ = x and sin θ = y by definition — this works for ANY angle, not just acute ones. The other four functions (tan, cot, sec, cosec) are derived from sine and cosine. The domain of sin and cos is all real numbers; tan and sec are undefined wherever cos θ = 0. The signs of the functions follow the "All Sin Tan Cos" rule across the four quadrants. Key Formula: $$ \sin^2\theta + \cos^2\theta = 1 $$, $$ \tan\theta = \frac{\sin\theta}{\cos\theta} $$, $$ 1 + \tan^2\theta = \sec^2\theta $$, $$ 1 + \cot^2\theta = \text{cosec}^2\theta $$
SI Units: Trigonometric function values are dimensionless ratios.
Common Exam Mistake: Assuming sine and cosine are only defined for angles between 0° and 90°. Correction: The unit-circle definition extends sine and cosine to ALL real angles, including negative angles and angles greater than 360°.
JEE/NEET Tip: "All Sin Tan Cos" (quadrants I, II, III, IV) tells you which function is positive in each quadrant — All positive in Q1, only Sin (and cosec) positive in Q2, only Tan (and cot) positive in Q3, only Cos (and sec) positive in Q4.
Solved Example:
If $$ \sin\theta = \frac{3}{5} $$ and θ lies in the second quadrant, find $$ \cos\theta $$.
Step-by-Step Solution:
Using $$ \sin^2\theta + \cos^2\theta = 1 $$: $$ \cos^2\theta = 1 - \frac{9}{25} = \frac{16}{25} \Rightarrow \cos\theta = \pm\frac{4}{5} $$. Since θ is in Quadrant II, cosine is negative, so $$ \cos\theta = -\frac{4}{5} $$.

4. Graphs of Trigonometric Functions

Concept Explanation: Trigonometric functions are periodic — they repeat their values after a fixed interval. Sine and cosine have period 2π and oscillate between -1 and 1. Tangent and cotangent have a smaller period of π, and are unbounded (they approach infinity near their undefined points). Understanding these graphs helps you instantly read off the domain, range, and periodicity of any trigonometric function without memorizing separately. Key Formula: Period of sin x, cos x: $$ 2\pi $$, Period of tan x, cot x: $$ \pi $$, Range of sin x, cos x: $$ [-1, 1] $$
SI Units: None (periods are measured in radians, a dimensionless ratio).
Common Exam Mistake: Assuming all six trigonometric functions have the same period of 2π. Correction: Tan and cot have a strictly smaller period of π, since $$ \tan(x+\pi) = \tan x $$.
JEE/NEET Tip: For a transformed function like $$ f(x) = a\sin(bx+c) $$, the new period is $$ \frac{2\pi}{|b|} $$ and the range is $$ [-|a|, |a|] $$ — this shortcut avoids re-deriving the graph from scratch every time.
Solved Example:
Find the period and range of $$ f(x) = 3\sin(2x) $$.
Step-by-Step Solution:
Here a=3, b=2. Period $$ = \frac{2\pi}{|2|} = \pi $$. Range $$ = [-|3|, |3|] = [-3, 3] $$.

5. Trigonometric Identities and Compound Angles

Concept Explanation: Compound angle formulas express trigonometric functions of a sum or difference of two angles in terms of the individual angles' functions. From these, we derive the double angle formulas (setting the two angles equal) and triple angle formulas. These identities are the single most tested skill in this chapter, since they let you simplify complex expressions and solve otherwise-impossible equations. Key Formula: $$ \sin(x \pm y) = \sin x \cos y \pm \cos x \sin y $$, $$ \cos(x \pm y) = \cos x \cos y \mp \sin x \sin y $$, $$ \sin 2x = 2\sin x \cos x $$, $$ \cos 2x = 1 - 2\sin^2 x $$
SI Units: None.
Common Exam Mistake: Writing $$ \cos(x+y) = \cos x \cos y + \sin x \sin y $$ (mixing up the sign with the sine formula). Correction: For cosine, the sign FLIPS: $$ \cos(x+y) = \cos x \cos y - \sin x \sin y $$.
JEE/NEET Tip: $$ \cos 2x $$ has THREE equivalent forms — $$ 1-2\sin^2x $$, $$ 2\cos^2x - 1 $$, and $$ \cos^2x - \sin^2x $$ — pick whichever form matches the rest of the expression you are simplifying.
Solved Example:
Find the value of $$ \sin 75^\circ $$ using the compound angle formula.
Step-by-Step Solution:
$$ \sin 75^\circ = \sin(45^\circ+30^\circ) = \sin45^\circ\cos30^\circ + \cos45^\circ\sin30^\circ = \frac{\sqrt{2}}{2}\cdot\frac{\sqrt{3}}{2} + \frac{\sqrt{2}}{2}\cdot\frac{1}{2} = \frac{\sqrt{6}+\sqrt{2}}{4} $$.

6. General Solutions of Trigonometric Equations

Concept Explanation: Since trigonometric functions are periodic, a trigonometric equation like $$ \sin\theta = k $$ has infinitely many solutions, not just one. The general solution captures all of them in a single formula using an integer n (n ∈ Z). Solving trigonometric equations always starts by finding the principal solution (the specific solution in a standard range) and then generalizing it using the appropriate periodicity formula. Key Formula: $$ \sin\theta = \sin\alpha \Rightarrow \theta = n\pi + (-1)^n\alpha $$, $$ \cos\theta = \cos\alpha \Rightarrow \theta = 2n\pi \pm \alpha $$, $$ \tan\theta = \tan\alpha \Rightarrow \theta = n\pi + \alpha $$
SI Units: None.
Common Exam Mistake: Using the same general solution formula format for sine, cosine, and tangent. Correction: Each function has its OWN distinct general solution pattern — sine uses $$ (-1)^n $$, cosine uses $$ \pm $$, and tangent simply adds $$ n\pi $$; mixing these up is the most common error in this topic.
JEE/NEET Tip: Always convert the equation to reference a STANDARD angle (like π/6, π/4, π/3) before applying the general solution formula — this avoids errors in identifying α.
Solved Example:
Find the general solution of $$ \cos\theta = \frac{1}{2} $$.
Step-by-Step Solution:
Since $$ \cos\frac{\pi}{3} = \frac{1}{2} $$, using $$ \cos\theta = \cos\alpha \Rightarrow \theta = 2n\pi \pm \alpha $$, the general solution is $$ \theta = 2n\pi \pm \frac{\pi}{3}, \ n \in \mathbb{Z} $$.

Quick Revision Cheat Sheet

**Top 10 Formulas:**
1. Degree to Radian: $$ \theta_{rad} = \frac{\pi}{180} \times \theta_{deg} $$
2. Arc length: $$ l = r\theta $$
3. Pythagorean identity: $$ \sin^2\theta + \cos^2\theta = 1 $$
4. Compound angle (sine): $$ \sin(x \pm y) = \sin x \cos y \pm \cos x \sin y $$
5. Compound angle (cosine): $$ \cos(x \pm y) = \cos x \cos y \mp \sin x \sin y $$
6. Double angle: $$ \sin 2x = 2\sin x \cos x $$
7. Double angle (cosine): $$ \cos 2x = 1 - 2\sin^2 x = 2\cos^2 x - 1 $$
8. General solution (sine): $$ \theta = n\pi + (-1)^n\alpha $$
9. General solution (cosine): $$ \theta = 2n\pi \pm \alpha $$
10. General solution (tangent): $$ \theta = n\pi + \alpha $$

**Memory Tricks:**
*   *Quadrant signs:* "All Sin Tan Cos" — going anticlockwise from Quadrant I, that is which function is positive in each quadrant.
*   *Cosine sign flip:* "Cos is the odd one out" — sine's compound formula keeps the same sign, cosine's flips it.

**Must-Memorize Facts:**
1. Standard angles: 30 deg = pi/6, 45 deg = pi/4, 60 deg = pi/3, 90 deg = pi/2, 180 deg = pi.
2. Sine and cosine range: always between -1 and 1, inclusive.
3. Tan and cot have period pi; sin, cos, sec, cosec have period 2 pi.
4. Each general solution formula (sin/cos/tan) has its own distinct pattern - never interchange them.

**Comparison Table: Sine vs Cosine General Solution**
| Feature | sin theta = sin alpha | cos theta = cos alpha |
| :--- | :--- | :--- |
| General solution | theta = n*pi + (-1)^n * alpha | theta = 2n*pi +/- alpha |
| Sign pattern | Alternates via (-1)^n | Uses plus/minus directly |
| n belongs to | Integers (Z) | Integers (Z) |

Chapter Practice Quiz (15 JEE/NEET Questions)

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