NCERT Chapter 3

Trigonometric Functions

The biggest and highest-weightage chapter: master radians, the unit circle, ASTC quadrant rules, and all 25+ essential identities.

~55 min read 8–10 marks in exam kemh103.pdf

What you will learn in this chapter

  1. Angles & Measurements — Degrees, Radians & $l = r\theta$
  2. Unit Circle & The ASTC Quadrant Rule
  3. Domain, Range & Values at Quadrantal Angles
  4. Sum & Difference Identities (Compound Angles)
  5. Multiple & Submultiple Angles ($2x, 3x, x/2$)
  6. Product-to-Sum & C-D Transformation Formulas
  7. Worked Examples — Easy to Hard
  8. NCERT Key Exercise Questions Solved
  9. Common Exam Traps & Pitfalls
  10. Quick Self-Check Quiz

1. Angles & Measurement: Degrees vs Radians

Definition of Radian Measure

One radian ($1\text{ rad}$) is the angle subtended at the center of a circle of radius $r$ by an arc of length equal to $r$.

$$l = r \cdot \theta \quad (\text{where } \theta \text{ MUST be in radians!})$$

The Bridge Between Degrees and Radians

A full circle of $360^\circ$ corresponds to a circumference of $2\pi r$. Hence:

$$2\pi \text{ radians} = 360^\circ \iff \pi \text{ radians} = 180^\circ$$

Degree $0^\circ$ $30^\circ$ $45^\circ$ $60^\circ$ $90^\circ$ $180^\circ$ $270^\circ$ $360^\circ$
Radian $0$ $\frac{\pi}{6}$ $\frac{\pi}{4}$ $\frac{\pi}{3}$ $\frac{\pi}{2}$ $\pi$ $\frac{3\pi}{2}$ $2\pi$

2. Unit Circle & The ASTC Quadrant Rule

Trig on the Unit Circle

Consider a circle of radius $r = 1$ centered at origin $(0,0)$. For any angle $x$, the terminal ray hits the circle at point $P(a, b)$:

$$\cos x = a \quad (\text{the x-coordinate}), \qquad \sin x = b \quad (\text{the y-coordinate})$$

Since $x^2 + y^2 = 1$ on the circle, we immediately get the fundamental identity: $\cos^2 x + \sin^2 x = 1$.

Quadrant I ($0$ to $\pi/2$) — "ALL"

Both $x > 0$ and $y > 0$.
ALL trigonometric functions are POSITIVE ($+$, $+$, $+$, $+$, $+$, $+$).

Quadrant II ($\pi/2$ to $\pi$) — "SILVER / STUDENTS"

$x < 0$, $y > 0$.
Only SINE and COSECANT are POSITIVE. All others are negative.

Quadrant III ($\pi$ to $3\pi/2$) — "TEA / TAKE"

$x < 0$, $y < 0$.
Only TANGENT and COTANGENT are POSITIVE. All others are negative.

Quadrant IV ($3\pi/2$ to $2\pi$) — "CUPS / CALCULUS"

$x > 0$, $y < 0$.
Only COSINE and SECANT are POSITIVE. All others are negative.

Mnemonic: "All Silver Tea Cups" or "Add Sugar To Coffee"

$$Q_1: \mathbf{A}\text{ll} \quad \longrightarrow \quad Q_2: \mathbf{S}\text{in} \quad \longrightarrow \quad Q_3: \mathbf{T}\text{an} \quad \longrightarrow \quad Q_4: \mathbf{C}\text{os}$$

Negative Angles Rule:

$$\cos(-x) = \cos x, \quad \sec(-x) = \sec x \quad (\text{Even functions — eat the minus!})$$

$$\sin(-x) = -\sin x, \quad \tan(-x) = -\tan x, \quad \cot(-x) = -\cot x, \quad \csc(-x) = -\csc x \quad (\text{Odd functions})$$

3. Domain, Range & Allied Angles

Function Domain Range Period
$\sin x$ $\mathbb{R}$ $[-1, 1]$ $2\pi$
$\cos x$ $\mathbb{R}$ $[-1, 1]$ $2\pi$
$\tan x$ $\mathbb{R} \setminus \{(2n+1)\frac{\pi}{2} : n \in \mathbb{Z}\}$ $\mathbb{R}$ $\pi$
$\cot x$ $\mathbb{R} \setminus \{n\pi : n \in \mathbb{Z}\}$ $\mathbb{R}$ $\pi$
$\sec x$ $\mathbb{R} \setminus \{(2n+1)\frac{\pi}{2} : n \in \mathbb{Z}\}$ $(-\infty, -1] \cup [1, \infty)$ $2\pi$
$\csc x$ $\mathbb{R} \setminus \{n\pi : n \in \mathbb{Z}\}$ $(-\infty, -1] \cup [1, \infty)$ $2\pi$

Golden Rule for Allied Angles: $\frac{n\pi}{2} \pm x$

4. Sum & Difference of Two Angles

The Master Formulas (Must Memorize!)

1. Cosine of Sum & Difference:

$$\cos(x + y) = \cos x \cos y - \sin x \sin y$$

$$\cos(x - y) = \cos x \cos y + \sin x \sin y$$

2. Sine of Sum & Difference:

$$\sin(x + y) = \sin x \cos y + \cos x \sin y$$

$$\sin(x - y) = \sin x \cos y - \cos x \sin y$$

3. Tangent of Sum & Difference:

$$\tan(x + y) = \frac{\tan x + \tan y}{1 - \tan x \tan y}$$

$$\tan(x - y) = \frac{\tan x - \tan y}{1 + \tan x \tan y}$$

4. Cotangent:

$$\cot(x + y) = \frac{\cot x \cot y - 1}{\cot y + \cot x}, \qquad \cot(x - y) = \frac{\cot x \cot y + 1}{\cot y - \cot x}$$

5. Multiple & Submultiple Angles ($2x, 3x, x/2$)

Double Angle Formulas ($2x$)

1. Cosine has 4 distinct forms:

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

Power-Reduction (Lifesavers in proofs and calculus!):

$$\sin^2 x = \frac{1 - \cos 2x}{2}, \qquad \cos^2 x = \frac{1 + \cos 2x}{2}$$

2. Sine:

$$\sin 2x = 2\sin x \cos x = \frac{2\tan x}{1 + \tan^2 x}$$

3. Tangent:

$$\tan 2x = \frac{2\tan x}{1 - \tan^2 x}$$

Triple Angle Formulas ($3x$)

$$\sin 3x = 3\sin x - 4\sin^3 x$$

$$\cos 3x = 4\cos^3 x - 3\cos x$$

$$\tan 3x = \frac{3\tan x - \tan^3 x}{1 - 3\tan^2 x}$$

6. Product-to-Sum & C-D Formulas

Product to Sum / Difference

$$2\sin x \cos y = \sin(x + y) + \sin(x - y)$$

$$2\cos x \sin y = \sin(x + y) - \sin(x - y)$$

$$2\cos x \cos y = \cos(x + y) + \cos(x - y)$$

$$-2\sin x \sin y = \cos(x + y) - \cos(x - y)$$

Sum to Product (The C-D Formulas)

$$\sin C + \sin D = 2\sin\frac{C+D}{2}\cos\frac{C-D}{2}$$

$$\sin C - \sin D = 2\cos\frac{C+D}{2}\sin\frac{C-D}{2}$$

$$\cos C + \cos D = 2\cos\frac{C+D}{2}\cos\frac{C-D}{2}$$

$$\cos C - \cos D = -2\sin\frac{C+D}{2}\sin\frac{C-D}{2} = 2\sin\frac{C+D}{2}\sin\frac{D-C}{2}$$

7. Worked Examples — Easy to Hard

Example 1: Degree to Radian Conversion with Minutes Easy
Convert $40^\circ 20'$ into radian measure.
Step 1: Convert minutes into fractions of a degree: $$20' = \frac{20}{60}^\circ = \frac{1}{3}^\circ$$ $$40^\circ 20' = 40 + \frac{1}{3} = \frac{121^\circ}{3}$$
Step 2: Multiply by $\frac{\pi}{180^\circ}$: $$\text{Radian measure} = \frac{121}{3} \times \frac{\pi}{180} = \frac{121\pi}{540} \text{ rad}$$
Example 2: Arc Length Formula $l = r\theta$ Easy
A wheel makes 360 revolutions in one minute. Through how many radians does it turn in one second?
Step 1: Number of revolutions in 1 second: $$\text{Revolutions per sec} = \frac{360}{60} = 6$$
Step 2: In one complete revolution, angle turned is $2\pi$ radians. $$\text{In 6 revolutions, angle turned} = 6 \times 2\pi = 12\pi \text{ radians}$$
Example 3: Finding All 5 Ratios from One Value & Quadrant Medium
If $\cos x = -\frac{3}{5}$ and $x$ lies in the 3rd quadrant, find the values of the other five trigonometric functions.
Step 1: Secant $$\sec x = \frac{1}{\cos x} = -\frac{5}{3}$$
Step 2: Sine $$\sin^2 x = 1 - \cos^2 x = 1 - \left(-\frac{3}{5}\right)^2 = 1 - \frac{9}{25} = \frac{16}{25}$$ Since $x$ is in Quadrant III, $\sin x$ is negative: $$\sin x = -\sqrt{\frac{16}{25}} = -\frac{4}{5}$$
Step 3: Cosecant, Tangent, Cotangent $$\csc x = \frac{1}{\sin x} = -\frac{5}{4}$$ $$\tan x = \frac{\sin x}{\cos x} = \frac{-4/5}{-3/5} = +\frac{4}{3} \quad (\text{positive in Q3!})$$ $$\cot x = \frac{1}{\tan x} = +\frac{3}{4}$$
Example 4: Evaluating $\sin 15^\circ$ and $\tan 15^\circ$ Medium
Find the exact value of $\sin 15^\circ$.
Step 1: Write $15^\circ$ as a difference of standard angles: $45^\circ - 30^\circ$. $$\sin 15^\circ = \sin(45^\circ - 30^\circ) = \sin 45^\circ \cos 30^\circ - \cos 45^\circ \sin 30^\circ$$
Step 2: Substitute standard values: $$\sin 15^\circ = \left(\frac{1}{\sqrt{2}}\right)\left(\frac{\sqrt{3}}{2}\right) - \left(\frac{1}{\sqrt{2}}\right)\left(\frac{1}{2}\right) = \frac{\sqrt{3} - 1}{2\sqrt{2}}$$
Example 5: Proving Identity Using C-D Formulas Hard
Prove that: $$\frac{\sin 5x - 2\sin 3x + \sin x}{\cos 5x - \cos x} = \tan x$$
Step 1: Regroup numerator terms: $$\text{LHS} = \frac{(\sin 5x + \sin x) - 2\sin 3x}{\cos 5x - \cos x}$$
Step 2: Apply C-D formulas: For numerator: $\sin 5x + \sin x = 2\sin\left(\frac{5x+x}{2}\right)\cos\left(\frac{5x-x}{2}\right) = 2\sin 3x \cos 2x$. For denominator: $\cos 5x - \cos x = -2\sin\left(\frac{5x+x}{2}\right)\sin\left(\frac{5x-x}{2}\right) = -2\sin 3x \sin 2x$.
Step 3: Factor out $2\sin 3x$: $$\text{LHS} = \frac{2\sin 3x (\cos 2x - 1)}{-2\sin 3x \sin 2x} = \frac{\cos 2x - 1}{-\sin 2x} = \frac{1 - \cos 2x}{\sin 2x}$$
Step 4: Use double angle identities: $$1 - \cos 2x = 2\sin^2 x, \qquad \sin 2x = 2\sin x \cos x$$ $$\text{LHS} = \frac{2\sin^2 x}{2\sin x \cos x} = \frac{\sin x}{\cos x} = \tan x = \text{RHS}$$ Hence proved!
Example 6: Finding $\tan \frac{\pi}{8}$ (Submultiple Half Angle) Hard
Find the exact value of $\tan \frac{\pi}{8}$ ($22.5^\circ$).
Step 1: Let $x = \frac{\pi}{8}$, then $2x = \frac{\pi}{4}$. We know: $\tan 2x = \frac{2\tan x}{1 - \tan^2 x}$. $$\tan \frac{\pi}{4} = 1 \implies 1 = \frac{2\tan x}{1 - \tan^2 x}$$
Step 2: Form quadratic in $t = \tan x$: $$1 - t^2 = 2t \implies t^2 + 2t - 1 = 0$$
Step 3: Solve using quadratic formula: $$t = \frac{-2 \pm \sqrt{4 - 4(1)(-1)}}{2} = \frac{-2 \pm \sqrt{8}}{2} = -1 \pm \sqrt{2}$$ Since $\frac{\pi}{8}$ is in Quadrant I ($22.5^\circ$), $\tan \frac{\pi}{8} > 0$. $$\mathbf{\tan \frac{\pi}{8} = \sqrt{2} - 1}$$

8. NCERT Key Exercise Questions Solved

NCERT Exercise 3.1 — Question 6

Q: If in two circles, arcs of the same length subtend angles $60^\circ$ and $75^\circ$ at the centre, find the ratio of their radii.
View Full Solution

Convert angles to radians: $\theta_1 = 60^\circ \times \frac{\pi}{180} = \frac{\pi}{3}$, $\theta_2 = 75^\circ \times \frac{\pi}{180} = \frac{5\pi}{12}$.

Given arc lengths are equal: $l_1 = l_2 \implies r_1 \theta_1 = r_2 \theta_2$.

$$\frac{r_1}{r_2} = \frac{\theta_2}{\theta_1} = \frac{5\pi/12}{\pi/3} = \frac{5}{12} \times \frac{3}{1} = \frac{5}{4}$$

Ratio of radii $r_1 : r_2 = 5 : 4$.

NCERT Exercise 3.2 — Question 7

Q: Find the value of $\csc(-1410^\circ)$.
View Full Solution

Using $\csc(-x) = -\csc(x)$: $\csc(-1410^\circ) = -\csc(1410^\circ)$.

Notice that $4 \times 360^\circ = 1440^\circ$.

So $1410^\circ = 4 \times 360^\circ - 30^\circ$.

Since $\csc(2n\pi - \theta) = -\csc \theta$, we have:

$$\csc(1410^\circ) = \csc(1440^\circ - 30^\circ) = -\csc 30^\circ = -2$$

$$\csc(-1410^\circ) = -(-2) = 2$$

NCERT Exercise 3.3 — Question 21

Q: Prove that $\frac{\cos 4x + \cos 3x + \cos 2x}{\sin 4x + \sin 3x + \sin 2x} = \cot 3x$.
View Full Solution

Pair terms with symmetric angles $(4x, 2x)$ so that their average is $3x$:

Numerator: $(\cos 4x + \cos 2x) + \cos 3x = 2\cos 3x \cos x + \cos 3x = \cos 3x(2\cos x + 1)$.

Denominator: $(\sin 4x + \sin 2x) + \sin 3x = 2\sin 3x \cos x + \sin 3x = \sin 3x(2\cos x + 1)$.

Dividing numerator by denominator: $\frac{\cos 3x(2\cos x + 1)}{\sin 3x(2\cos x + 1)} = \frac{\cos 3x}{\sin 3x} = \cot 3x$.

9. Common Exam Traps & Pitfalls

Trap 1: Forgetting that $\theta$ MUST be in Radians in $l = r\theta$

If a question says radius $r = 10\text{ cm}$ and angle $\theta = 30^\circ$, NEVER calculate $l = 10 \times 30 = 300\text{ cm}$! You must convert $30^\circ$ to radians first: $\frac{\pi}{6}$, giving $l = \frac{10\pi}{6} = \frac{5\pi}{3}\text{ cm}$.

Trap 2: $\cos C - \cos D$ Negative Sign

The identity has a minus sign: $\cos C - \cos D = -2\sin\frac{C+D}{2}\sin\frac{C-D}{2}$. If you write positive $2$, you lose marks! Alternatively, write $2\sin\frac{C+D}{2}\sin\frac{D-C}{2}$.

Trap 3: Parity of Cosine vs Sine

$\cos(-x) = \cos x$ (absorbs minus), whereas $\sin(-x) = -\sin x$. Many students erroneously write $\cos(-\pi/3) = -\cos(\pi/3)$, which leads to sign errors.

10. Quick Self-Check Quiz

Q1: What is the degree measure corresponding to $-\frac{2\pi}{3}$ radians?

Answer
$-\frac{2\pi}{3} \times \frac{180^\circ}{\pi} = -2 \times 60^\circ = -120^\circ$.

Q2: If $\tan x = \frac{3}{4}$ and $x$ is in the 3rd quadrant, what is the value of $\sin x$?

Answer
In Q3, sine is negative. Hypotenuse = $\sqrt{3^2 + 4^2} = 5$. Thus, $\sin x = -\frac{3}{5}$.

Q3: What is the maximum value of $f(x) = 3\sin x + 4\cos x$?

Answer
For $a\sin x + b\cos x$, maximum is $\sqrt{a^2 + b^2} = \sqrt{3^2 + 4^2} = \sqrt{25} = 5$. Minimum is $-5$.

Q4: What is $\cos 2x$ in terms of $\tan x$?

Answer
$\cos 2x = \frac{1 - \tan^2 x}{1 + \tan^2 x}$.

Q5: What is the value of $\sin 75^\circ \cos 15^\circ + \cos 75^\circ \sin 15^\circ$?

Answer
Using $\sin(A + B)$: $\sin(75^\circ + 15^\circ) = \sin 90^\circ = 1$.
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