MathsCBSE Class 11 · Chapter 3

Trigonometric Functions Class 11: Notes, Identities and Solved Questions

Updated 14 October 2026 · Fact-checked

Trigonometric Functions is Chapter 3 of CBSE Class 11 Maths. It covers positive and negative angles, degree and radian measure, the six trigonometric functions defined with the unit circle, their signs, domains, ranges and graphs, the formulae for sin, cos and tan of sums and differences, sum-to-product identities, and identities for 2x and 3x.

What Trigonometric Functions is about

In Class 10 you met sine, cosine and tangent as ratios of sides of a right triangle, so they only made sense for acute angles. This chapter frees them from the triangle. An angle becomes an amount of rotation, positive anticlockwise and negative clockwise, of any size, measured in degrees or in radians ( $π$ radians = 180°).

Using a unit circle, the point reached after turning through x radians has coordinates $(cos x, sin x)$ . That defines sine and cosine for every real number x, and the other four functions follow. The identity $sin^{2} x + cos^{2} x = 1$ is then just Pythagoras on the unit circle. You learn the signs of the functions in each quadrant, their domains, ranges, periods and graphs, and how to reduce any angle to an acute one.

The second half is about identities: $cos (x \pm y)$ , $sin (x \pm y)$ , $tan (x \pm y)$ and $cot (x \pm y)$ ; the sum-to-product formulae; and formulae for $sin 2 x$ , $cos 2 x$ , $tan 2 x$ , $sin 3 x$ , $cos 3 x$ and $tan 3 x$ . For 2026-27, general solutions of trigonometric equations are assessed only formatively, and the sine and cosine rules for triangles are not in the syllabus.

Weightage: This chapter is part of Unit I, Sets and Functions, which the syllabus gives 23 of the 80 theory marks in the school annual exam (CBSE adds that there is no fixed chapter-wise weightage).Source: CBSE Mathematics (041) syllabus 2026-27, Class XI course structure, Unit I Sets and Functions

Key concepts

Angles: degree and radian measure

NCERT section 3.2

An angle is the amount of rotation of a ray from an initial side to a terminal side. Anticlockwise rotation gives a positive angle, clockwise a negative one. Angles can be larger than 360°.

Degree: one complete turn is 360°; $1° = 6 0^{'}$ (minutes), $1^{'} = 6 0^{''}$ (seconds).
Radian: the angle at the centre of a circle made by an arc whose length equals the radius. One complete turn is $2 π$ radians.

Conversion (from $π$ radians = 180°):

Radian measure = $\frac{π}{180} \times$ degree measure
Degree measure = $\frac{180}{π} \times$ radian measure; so 1 radian ≈ 57°16′.

Arc length: an arc of length l on a circle of radius r makes an angle of $θ$ radians at the centre, where $l = r θ$ . (θ must be in radians.)

When no unit is written, the angle is in radians: $sin 2$ means the sine of 2 radians.

Worked example

(a) Convert 75° and −37°30′ into radians. (b) Find the length of the arc that subtends 60° at the centre of a circle of radius 21 cm. (Use $π = \frac{22}{7}$ .) [3 marks]

Try it first, then show the solutionSolution

(a) $75 \times \frac{π}{180} = \frac{5 π}{12}$ ; $- 37 \frac{1}{2} ° = - \frac{75}{2} \times \frac{π}{180} = - \frac{5 π}{24}$
(b) $60° = \frac{π}{3}$ radians
$l = r θ = 21 \times \frac{22}{7} \times \frac{1}{3} = 22$ cm

Answer: $\frac{5 π}{12}$ , $- \frac{5 π}{24}$ ; arc 22 cm.

Trigonometric functions from the unit circle

NCERT section 3.3

Draw a circle of radius 1 centred at the origin. Start at (1, 0) and turn through x radians; call the point you reach P(a, b). Define
$cos x = a, sin x = b$

Then for every real x:

$tan x = \frac{s i n x}{c o s x}$ (when $cos x \neq = 0$ ), $cot x = \frac{c o s x}{s i n x}$ (when $sin x \neq = 0$ )
$sec x = \frac{1}{c o s x}$ , $csc x = \frac{1}{s i n x}$

Pythagorean identities: P is on the unit circle, so $a^{2} + b^{2} = 1$ :

$sin^{2} x + cos^{2} x = 1$
$1 + tan^{2} x = sec^{2} x$ (divide by $cos^{2} x$ )
$1 + cot^{2} x = csc^{2} x$ (divide by $sin^{2} x$ )

Zeros: $sin x = 0$ exactly when $x = nπ$ ; $cos x = 0$ exactly when $x = (2 n + 1) \frac{π}{2}$ , n any integer.

Periodicity: turning a further $2 π$ brings you back to P, so $sin (2 nπ + x) = sin x$ and $cos (2 nπ + x) = cos x$ . Also $sin (- x) = - sin x$ and $cos (- x) = cos x$ .

Worked example

Evaluate $sin^{2} \frac{π}{4} + cos^{2} \frac{π}{3} - tan^{2} \frac{π}{6}$ . [2 marks]

Try it first, then show the solutionSolution

$sin \frac{π}{4} = \frac{1}{2}$ , $cos \frac{π}{3} = \frac{1}{2}$ , $tan \frac{π}{6} = \frac{1}{3}$
$\frac{1}{2} + \frac{1}{4} - \frac{1}{3} = \frac{6 + 3 - 4}{12} = \frac{5}{12}$

Answer: $\frac{5}{12}$

Signs in the four quadrants

NCERT section 3.3.1

Since $cos x$ is the x-coordinate and $sin x$ the y-coordinate of P:

Quadrant	sin, cosec	cos, sec	tan, cot
I $(0, \frac{π}{2})$	+	+	+
II $(\frac{π}{2}, π)$	+	−	−
III $(π, \frac{3 π}{2})$	−	−	+
IV $(\frac{3 π}{2}, 2 π)$	−	+	−

Method for 'given one ratio, find the other five':
1. Use a Pythagorean identity to find the size of a second ratio.
2. Choose its sign from the quadrant.
3. Get the rest by reciprocals and $tan = \frac{s i n}{c o s}$ .

Worked example

If $sin x = \frac{8}{17}$ and x lies in the second quadrant, find the other five trigonometric functions. [3 marks]

Try it first, then show the solutionSolution

$cos^{2} x = 1 - \frac{64}{289} = \frac{225}{289}$ ; in quadrant II cosine is negative, so $cos x = - \frac{15}{17}$
$tan x = \frac{8/17}{- 15/17} = - \frac{8}{15}$ , $cot x = - \frac{15}{8}$
$csc x = \frac{17}{8}$ , $sec x = - \frac{17}{15}$

Answer: $cos x = - \frac{15}{17}$ , $tan x = - \frac{8}{15}$ , $cot x = - \frac{15}{8}$ , $csc x = \frac{17}{8}$ , $sec x = - \frac{17}{15}$

Domain, range, period and graphs

NCERT section 3.3.2

Function	Domain	Range	Period
sin x	R	[−1, 1]	$2 π$
cos x	R	[−1, 1]	$2 π$
tan x	$R - {(2 n + 1) \frac{π}{2}}$	R	$π$
cot x	$R - {nπ}$	R	$π$
sec x	$R - {(2 n + 1) \frac{π}{2}}$	$(- \infty, - 1] \cup [1, \infty)$	$2 π$
cosec x	$R - {nπ}$	$(- \infty, - 1] \cup [1, \infty)$	$2 π$

Graphs: $sin x$ is a wave through the origin, rising to 1 at $\frac{π}{2}$ ; $cos x$ is the same wave shifted to start at its top, (0, 1). $tan x$ climbs from $- \infty$ to $+ \infty$ between consecutive vertical asymptotes at odd multiples of $\frac{π}{2}$ .

How sin x changes over a turn: it increases from 0 to 1 in quadrant I, decreases from 1 to 0 in II, from 0 to −1 in III, and increases from −1 to 0 in IV. Cosine does the same, starting from 1.

Worked example

Find the values of $sin 870°$ , $cos 1140°$ and $tan (- 585°)$ . [3 marks]

Try it first, then show the solutionSolution

$870° = 2 \times 360° + 150°$ , so $sin 870° = sin 150° = sin (180° - 30°) = \frac{1}{2}$
$1140° = 3 \times 360° + 60°$ , so $cos 1140° = \frac{1}{2}$
$tan (- 585°) = - tan 585°$ ; tan has period 180°, and $585° = 3 \times 180° + 45°$ , so the value is $- 1$

Answer: $\frac{1}{2}$ , $\frac{1}{2}$ , $- 1$

Sum and difference formulae

NCERT section 3.4

The basic result, proved with the unit circle by comparing two equal chords, is
$cos (x + y) = cos x cos y - sin x sin y$

Everything else follows from it:

$cos (x - y) = cos x cos y + sin x sin y$ (replace y by −y)
$sin (x + y) = sin x cos y + cos x sin y$
$sin (x - y) = sin x cos y - cos x sin y$
$tan (x \pm y) = \frac{t a n x \pm t a n y}{1 \mp t a n x t a n y}$
$cot (x \pm y) = \frac{c o t x c o t y \mp 1}{c o t y \pm c o t x}$

Allied angles (special cases):

$cos (\frac{π}{2} - x) = sin x$ , $sin (\frac{π}{2} - x) = cos x$
$cos (\frac{π}{2} + x) = - sin x$ , $sin (\frac{π}{2} + x) = cos x$
$cos (π \pm x) = - cos x$ , $sin (π - x) = sin x$ , $sin (π + x) = - sin x$
$cos (2 π - x) = cos x$ , $sin (2 π - x) = - sin x$

Worked example

Find the exact values of $sin 105°$ and $tan 75°$ . [2 marks]

Try it first, then show the solutionSolution

$sin 105° = sin (60° + 45°) = \frac{3}{2} \cdot \frac{1}{2} + \frac{1}{2} \cdot \frac{1}{2} = \frac{3 + 1}{2 2} = \frac{6 + 2}{4}$
$tan 75° = tan (45° + 30°) = \frac{1 + \frac{1}{3}}{1 - \frac{1}{3}} = \frac{3 + 1}{3 - 1} = \frac{( 3 + 1 ) ^{2}}{2} = 2 + 3$

Answer: $sin 105° = \frac{6 + 2}{4}$ ; $tan 75° = 2 + 3$

Sum-to-product and product-to-sum identities

NCERT section 3.4

Adding and subtracting the formulae for $sin (x \pm y)$ and $cos (x \pm y)$ gives the product-to-sum forms:

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

Putting $x + y = α$ and $x - y = β$ gives the sum-to-product forms:

$sin α + sin β = 2 sin \frac{α + β}{2} cos \frac{α - β}{2}$
$sin α - sin β = 2 cos \frac{α + β}{2} sin \frac{α - β}{2}$
$cos α + cos β = 2 cos \frac{α + β}{2} cos \frac{α - β}{2}$
$cos α - cos β = - 2 sin \frac{α + β}{2} sin \frac{α - β}{2}$

These are the main tool for proving identities with sums like $sin 5 x + sin 3 x$ : pair terms so that the half-sums or half-differences match.

Worked example

Prove that $\frac{s i n 5 x + s i n 3 x}{c o s 5 x + c o s 3 x} = tan 4 x$ . [3 marks]

Try it first, then show the solutionSolution

Numerator: $sin 5 x + sin 3 x = 2 sin 4 x cos x$
Denominator: $cos 5 x + cos 3 x = 2 cos 4 x cos x$
Dividing, $\frac{2 s i n 4 x c o s x}{2 c o s 4 x c o s x} = \frac{s i n 4 x}{c o s 4 x} = tan 4 x$ = RHS

Answer: Proved.

Multiple-angle identities: 2x and 3x

NCERT section 3.4

Put y = x in the sum formulae:

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

Write 3x = 2x + x and simplify:

$sin 3 x = 3 sin x - 4 sin^{3} x$
$cos 3 x = 4 cos^{3} x - 3 cos x$
$tan 3 x = \frac{3 t a n x - t a n ^{3} x}{1 - 3 t a n ^{2} x}$

Half angles (from the cos 2x forms, with x replaced by $\frac{x}{2}$ ):
$sin^{2} \frac{x}{2} = \frac{1 - c o s x}{2}$ , $cos^{2} \frac{x}{2} = \frac{1 + c o s x}{2}$ . Pick the sign from the quadrant of $\frac{x}{2}$ , not of x.

Worked example

Prove that $\frac{1 - c o s 2 x}{s i n 2 x} = tan x$ . [2 marks]

Try it first, then show the solutionSolution

$1 - cos 2 x = 2 sin^{2} x$ and $sin 2 x = 2 sin x cos x$
So LHS $= \frac{2 s i n ^{2} x}{2 s i n x c o s x} = \frac{s i n x}{c o s x} = tan x$

Answer: Proved.

Maths in real life: Trigonometric Functions

Each concept, met in an everyday situation. Read the situation, try the problem, then check the solution. The one-line hook is what to remember.

Arc length

Suppose the minute hand of a clock tower is 1.4 m long.

How far does the tip of the minute hand move in 20 minutes? (Use $π = \frac{22}{7}$ .)

Try it first, then show the solutionSolution

In 60 minutes the hand turns $2 π$ , so in 20 minutes it turns $\frac{2 π}{3}$ radians
$l = r θ = 1.4 \times \frac{2}{3} \times \frac{22}{7} = \frac{8.8}{3} \approx 2.93$ m

Answer: About 2.93 m

Memory hook: Arc = radius × angle, with the angle in radians.

Angular speed in radians

Imagine a ceiling fan whose blades make 300 revolutions every minute.

Through how many radians does a blade turn in one second?

Try it first, then show the solutionSolution

300 revolutions per minute = 5 revolutions per second
Each revolution is $2 π$ radians, so $5 \times 2 π = 10 π$ radians

Answer: $10 π$ radians per second

Memory hook: One turn = 2π radians.

Cosine on a circle

Suppose a giant wheel has radius 10 m and its centre is 12 m above the ground. A seat starts at the lowest point and the wheel turns through an angle θ.

Show that the seat's height is $h = 12 - 10 cos θ$ metres and find h when θ = 60° and θ = 120°.

Try it first, then show the solutionSolution

Measured from the centre, the lowest point is 10 m down; after turning θ the seat is $10 cos θ$ below the centre (negative cos means above)
So $h = 12 - 10 cos θ$
θ = 60°: $h = 12 - 5 = 7$ m; θ = 120°: $h = 12 + 5 = 17$ m

Answer: 7 m and 17 m

Memory hook: A negative cosine past 90° lifts the seat above the centre.

Topics in this chapter

Trigonometric Functions formula sheet

Degree-radian conversion and arc length: $π rad = 180°, l = r θ (θ in radians)$
Pythagorean identities: $sin^{2} x + cos^{2} x = 1, 1 + tan^{2} x = sec^{2} x, 1 + cot^{2} x = csc^{2} x$
Sum and difference: $sin (x \pm y) = sin x cos y \pm cos x sin y, cos (x \pm y) = cos x cos y \mp sin x sin y$
Tangent and cotangent of sums: $tan (x \pm y) = \frac{t a n x \pm t a n y}{1 \mp t a n x t a n y}, cot (x \pm y) = \frac{c o t x c o t y \mp 1}{c o t y \pm c o t x}$
Sum to product: $sin α \pm sin β = 2 sin \frac{α \pm β}{2} cos \frac{α \mp β}{2}, cos α + cos β = 2 cos \frac{α + β}{2} cos \frac{α - β}{2}$; $cos α - cos β = - 2 sin \frac{α + β}{2} sin \frac{α - β}{2}$
Double angle: $sin 2 x = 2 sin x cos x, cos 2 x = 2 cos^{2} x - 1 = 1 - 2 sin^{2} x, tan 2 x = \frac{2 t a n x}{1 - t a n ^{2} x}$
Triple angle: $sin 3 x = 3 sin x - 4 sin^{3} x, cos 3 x = 4 cos^{3} x - 3 cos x, tan 3 x = \frac{3 t a n x - t a n ^{3} x}{1 - 3 t a n ^{2} x}$

How the school exam asks Trigonometric Functions

The question patterns to expect, and the method that works for each.

Question type	How it is asked	Method
Angle conversion and arc length	Convert degrees (with minutes) to radians and back; arcs, clock hands, wheels and pendulums with $l = r θ$ .	Convert to radians before using $l = r θ$ ; use $π = \frac{22}{7}$ only if told.
Given one ratio, find the others	One ratio and the quadrant are given; find the other five.	Pythagorean identity for size, quadrant for sign, then reciprocals.
Values of large or negative angles	Find values such as $sin 765°$ , $csc (- 1410°)$ , $tan \frac{19 π}{3}$ .	Remove whole turns ( $2 π$ ; π for tan and cot), use $sin (- x) = - sin x$ , then allied angles.
Exact values with compound angles	Find $sin 15°$ , $cos 105°$ , $tan \frac{13 π}{12}$ ; find $sin (A + B)$ from given ratios.	Split into standard angles (30°, 45°, 60°, 90°) and apply the sum or difference formula.
Proving identities	Prove identities using compound, sum-to-product and multiple-angle formulae; Miscellaneous Exercise style.	Start from the more complicated side; pair terms whose half-sum is common; change everything to sin and cos if stuck.
Half and multiple angles	Given a ratio of x and its quadrant, find $sin 2 x$ , $cos 2 x$ or $sin \frac{x}{2}$ , $cos \frac{x}{2}$ , $tan \frac{x}{2}$ .	Find $cos x$ first; use the half-angle forms and take the sign from the quadrant of $\frac{x}{2}$ .

Tricks to remember Trigonometric Functions

Short memory aids. Say each one aloud once, then use it on a question the same day.

All Silver Tea Cups
Quadrants I to IV: All positive; Sine (and cosec); Tan (and cot); Cos (and sec).
Use it when: Signs in quadrants.
Odd multiple of 90° swaps, even keeps
For $n \frac{π}{2} \pm x$ : if n is odd, sin ↔ cos (tan ↔ cot); if n is even, keep the function. Then fix the sign from the quadrant, treating x as acute.
Use it when: Allied and large angles.
cos is 'co-co minus si-si'
$cos (x + y) = cos cos - sin sin$ : cos keeps the order and flips the sign; sin mixes ('sin-cos + cos-sin') and keeps the sign.
Use it when: Sum formulae.
Sum to product: SC, CS, CC, minus SS
sin + sin → 2 S C; sin − sin → 2 C S; cos + cos → 2 C C; cos − cos → −2 S S, always half-sum first and half-difference second.
Use it when: Sum-to-product identities.
Triple angle: 3 then 4
$sin 3 x = 3 s - 4 s^{3}$ ; $cos 3 x = 4 c^{3} - 3 c$ (the order flips).
Use it when: Triple-angle formulae.

Trigonometric Functions: exam-level solved questions

Ordered from exam level to HOTS. Cover the solution, write your own, then compare step by step.

Example 1: Radian to degree · 1 mark · MCQExam level

The radian measure $\frac{5 π}{3}$ equals:
(a) 150°
(b) 300°
(c) 330°
(d) 270°

Try it first, then show the solutionSolution

$\frac{5 π}{3} \times \frac{180°}{π} = 300°$ .

Answer: (b) 300°

Example 2: Range of sine · 1 mark · Assertion-reasonExam level

Assertion (A): There is no real x with $sin x = \frac{3}{2}$ .
Reason (R): $sin x$ is the y-coordinate of a point on the unit circle.
(a) Both true, R explains A. (b) Both true, R does not explain A. (c) A true, R false. (d) A false, R true.

Try it first, then show the solutionSolution

Every point on the unit circle has $- 1 \leq y \leq 1$ , so $sin x$ can never be $\frac{3}{2}$ . R gives the reason.

Answer: (a)

Example 3: Compound angles from given ratios · 3 marks · Short answerHarder exam level

If $sin A = \frac{3}{5}$ and $cos B = \frac{5}{13}$ , where A and B are acute, find $sin (A + B)$ and $cos (A + B)$ . [3 marks]

Try it first, then show the solutionSolution

A and B acute: $cos A = \frac{4}{5}$ , $sin B = \frac{12}{13}$
$sin (A + B) = \frac{3}{5} \cdot \frac{5}{13} + \frac{4}{5} \cdot \frac{12}{13} = \frac{15 + 48}{65} = \frac{63}{65}$
$cos (A + B) = \frac{4}{5} \cdot \frac{5}{13} - \frac{3}{5} \cdot \frac{12}{13} = \frac{20 - 36}{65} = - \frac{16}{65}$

Answer: $\frac{63}{65}$ and $- \frac{16}{65}$ (so A + B is obtuse)

Example 4: Exact value via allied angles · 2 marks · Very short answerHarder exam level

Find the value of $tan \frac{19 π}{12}$ . [2 marks]

Try it first, then show the solutionSolution

$tan \frac{19 π}{12} = tan (π + \frac{7 π}{12}) = tan \frac{7 π}{12} = tan 105°$ (period π)
$tan (60° + 45°) = \frac{3 + 1}{1 - 3} = - \frac{( 3 + 1 ) ^{2}}{2} = - (2 + 3)$

Answer: $- (2 + 3)$

Example 5: Proof with sum to product · 3 marks · Short answerHarder exam level

Prove that $(cos 5 x + cos x) sin 2 x - (sin 5 x - sin x) cos 2 x = 0$ . [3 marks]

Try it first, then show the solutionSolution

$cos 5 x + cos x = 2 cos 3 x cos 2 x$ and $sin 5 x - sin x = 2 cos 3 x sin 2 x$
LHS $= 2 cos 3 x cos 2 x sin 2 x - 2 cos 3 x sin 2 x cos 2 x$
$= 0$ = RHS

Answer: Proved.

Example 6: Tangent of a sum · 2 marks · Very short answerHarder exam level

If $tan A = \frac{1}{2}$ and $tan B = \frac{1}{3}$ , with A, B acute, show that $A + B = \frac{π}{4}$ . [2 marks]

Try it first, then show the solutionSolution

$tan (A + B) = \frac{\frac{1}{2} + \frac{1}{3}}{1 - \frac{1}{6}} = \frac{5/6}{5/6} = 1$
A and B are acute and $tan (A + B) > 0$ with $A + B < π$ , so $A + B = \frac{π}{4}$

Answer: $A + B = \frac{π}{4}$

Example 7: Product of cosines · 3 marks · Short answerHOTS / competency

Prove that $cos 20° cos 40° cos 80° = \frac{1}{8}$ . [3 marks]

Try it first, then show the solutionSolution

Multiply and divide by $2 sin 20°$ : $2 sin 20° cos 20° = sin 40°$
Then $2 sin 40° cos 40° = sin 80°$ and $2 sin 80° cos 80° = sin 160°$ , so the product $= \frac{s i n 160°}{8 s i n 20°}$
$sin 160° = sin (180° - 20°) = sin 20°$ , so the product $= \frac{1}{8}$

Answer: Proved: $\frac{1}{8}$

Example 8: A sum of squares that is constant · 5 marks · Long answerHOTS / competency

Prove that $sin^{2} x + sin^{2} (x + \frac{π}{3}) + sin^{2} (x - \frac{π}{3}) = \frac{3}{2}$ . [5 marks]

Try it first, then show the solutionSolution

Use $sin^{2} θ = \frac{1 - c o s 2 θ}{2}$ on each term
LHS $= \frac{3}{2} - \frac{1}{2} [cos 2 x + cos (2 x + \frac{2 π}{3}) + cos (2 x - \frac{2 π}{3})]$
$cos (2 x + \frac{2 π}{3}) + cos (2 x - \frac{2 π}{3}) = 2 cos 2 x cos \frac{2 π}{3} = - cos 2 x$
So the bracket is $cos 2 x - cos 2 x = 0$
LHS $= \frac{3}{2}$ = RHS, for every x

Answer: Proved.

Case-study practice (Section E style)

Read the situation, then answer every part. In a school exam set on the board pattern this is a 4-mark question split into 1 + 1 + 1 + 1 marks.

A pendulum clock

Suppose the pendulum of a wall clock is 60 cm long and its tip swings along an arc. Treat the arc as part of a circle centred at the pivot.

(i) · 1 mark
Find the angle (in radians) of swing if the tip moves along an arc of 15 cm.
Answer
Answer: 0.25 rad
$θ = \frac{l}{r} = \frac{15}{60} = 0.25$
(ii) · 1 mark
Convert this angle into degrees (π ≈ 3.14).
Answer
Answer: About 14.3°
$0.25 \times \frac{180}{3.14} \approx 14.33°$
(iii) · 1 mark
If the angle of swing is $\frac{π}{15}$ , find the arc length.
Answer
Answer: 4π cm, about 12.6 cm
$l = 60 \times \frac{π}{15} = 4 π$
(iv) · 1 mark
Why must the angle be in radians in $l = r θ$ ?
Answer
Answer: The formula comes from the definition of a radian.
One radian is the angle for which l = r, so l is proportional to θ with constant r only in radians.

Sound waves and identities

Suppose two tuning forks produce sounds modelled by $y_{1} = sin 5 t$ and $y_{2} = sin 3 t$ , where t is time. When both sound together the combined disturbance is $y = sin 5 t + sin 3 t$ .

(i) · 1 mark
Write y as a product.
Answer
Answer: $y = 2 sin 4 t cos t$
Sum to product with half-sum 4t and half-difference t.
(ii) · 1 mark
Show that $∣ y ∣$ can never exceed 2.
Answer
Answer: $∣ y ∣ \leq 2$
$∣ y ∣ = 2∣ sin 4 t ∣∣ cos t ∣$ and both $∣ sin 4 t ∣$ and $∣ cos t ∣$ are at most 1.
(iii) · 1 mark
Find y when $t = \frac{π}{2}$ .
Answer
Answer: 0
$cos \frac{π}{2} = 0$
(iv) · 1 mark
Find y when $t = \frac{π}{8}$ .
Answer
Answer: $2 cos \frac{π}{8} \approx 1.85$
$sin \frac{π}{2} = 1$ , so $y = 2 cos \frac{π}{8}$

Where the marks are

How a typical answer is split into step marks. Show every step: a right final answer alone rarely earns full marks.

Find the radius of the circle in which a central angle of 45° cuts off an arc of length 11 cm. (Use $π = \frac{22}{7}$ .) [2 marks] (2 marks)

Step	Marks
$45° = \frac{π}{4} = \frac{22}{28} = \frac{11}{14}$ radians	1
$r = \frac{l}{θ} = 11 \times \frac{14}{11} = 14$ cm	1

If $cos x = - \frac{12}{13}$ and x lies in the third quadrant, find the other five trigonometric functions. [3 marks] (3 marks)

Step	Marks
$sin^{2} x = 1 - \frac{144}{169} = \frac{25}{169}$ ; sine is negative in quadrant III, so $sin x = - \frac{5}{13}$	1
$tan x = \frac{- 5/13}{- 12/13} = \frac{5}{12}$ , $cot x = \frac{12}{5}$	1
$sec x = - \frac{13}{12}$ , $csc x = - \frac{13}{5}$	1

If $sin x = \frac{3}{5}$ and $\frac{π}{2} < x < π$ , find (a) $sin 2 x$ , $cos 2 x$ and $tan 2 x$ (b) $sin \frac{x}{2}$ , $cos \frac{x}{2}$ and $tan \frac{x}{2}$ . [5 marks] (5 marks)

Step	Marks
x is in quadrant II, so $cos x = - 1 - \frac{9}{25} = - \frac{4}{5}$	1
$sin 2 x = 2 \cdot \frac{3}{5} \cdot (- \frac{4}{5}) = - \frac{24}{25}$ ; $cos 2 x = 1 - 2 \cdot \frac{9}{25} = \frac{7}{25}$ ; $tan 2 x = - \frac{24}{7}$	1
$\frac{π}{4} < \frac{x}{2} < \frac{π}{2}$ , so all ratios of $\frac{x}{2}$ are positive	1
$sin^{2} \frac{x}{2} = \frac{1 - c o s x}{2} = \frac{9}{10} \Rightarrow sin \frac{x}{2} = \frac{3}{10}$ ; $cos^{2} \frac{x}{2} = \frac{1 + c o s x}{2} = \frac{1}{10} \Rightarrow cos \frac{x}{2} = \frac{1}{10}$	1
$tan \frac{x}{2} = \frac{3/ 10}{1/ 10} = 3$	1

Where students lose marks in Trigonometric Functions

The slips examiners see most often, and what to write instead.

Mistake: Using degrees in $l = r θ$
Forgetting the formula comes from the radian definition.
Wrong working
$l = 21 \times 60 = 1260$ cm
Right working
$l = 21 \times \frac{π}{3} = 22$ cm
Do this: Convert to radians first.
Usually costs: 1-2 marks.
Mistake: Wrong sign from the quadrant
Taking the positive square root by habit.
Do this: After $cos^{2} x = \frac{16}{25}$ , decide the sign from the quadrant before writing $cos x$ .
Usually costs: 1 mark per wrong ratio.
Mistake: Writing $sin (x + y) = sin x + sin y$
Treating sin like multiplication by a number.
Do this: Check with x = y = 30°: $sin 60° \neq = 2 sin 30°$ .
Usually costs: Whole question.
Mistake: Half-angle sign taken from the quadrant of x
Not working out where $\frac{x}{2}$ lies.
Wrong working
x in quadrant III, so $sin \frac{x}{2} < 0$
Right working
$π < x < \frac{3 π}{2} \Rightarrow \frac{π}{2} < \frac{x}{2} < \frac{3 π}{4}$ : quadrant II, so $sin \frac{x}{2} > 0$
Do this: Halve the interval first.
Usually costs: 1-2 marks.
Mistake: Using period 2π for tan
Copying the rule for sin and cos.
Do this: tan and cot repeat every π (180°).
Usually costs: Usually no loss, but wastes steps; wrong if you also flip the sign.
Mistake: Proving an identity by working on both sides at once and cross-multiplying
Assuming what is to be proved.
Do this: Transform one side until it becomes the other, or reduce both sides separately to the same expression.
Usually costs: 1-2 marks.

NCERT exercise guide: Trigonometric Functions

What each exercise in the NCERT chapter tests, how to approach it, and a question of the same kind solved step by step. Questions here are written in our own words.

Exercise 3.1 (angles, radians and arcs)

What it tests: Converting degrees to radians and back, arcs and chords, wheels, clock hands, pendulums and ratios of radii.

How to approach it: Use π rad = 180° and $l = r θ$ with θ in radians.

Same kind as Exercise 3.1

Arcs of the same length in two circles subtend angles of 45° and 60° at their centres. Find the ratio of their radii.

Show the solutionSolution

$l = r_{1} \cdot \frac{π}{4} = r_{2} \cdot \frac{π}{3}$
$\frac{r _{1}}{r _{2}} = \frac{π /3}{π /4} = \frac{4}{3}$

Answer: 4 : 3

Exercise 3.2 (values and signs)

What it tests: Given one ratio and the quadrant, finding the other five; values of trigonometric functions of large and negative angles.

How to approach it: Identity for size, quadrant for sign; remove whole turns for large angles.

Same kind as Exercise 3.2

Find the value of $sec (- 1050°)$ .

Show the solutionSolution

sec is even and has period 360°: $sec (- 1050°) = sec 1050° = sec (3 \times 360° - 30°) = sec (- 30°)$
$= sec 30° = \frac{2}{3}$

Answer: $\frac{2}{3}$

Exercise 3.3 and Miscellaneous Exercise (identities)

What it tests: Exact values with compound angles, allied angles, proofs with sum-to-product and multiple-angle formulae, half angles.

How to approach it: Choose the formula that matches the pattern; simplify one side to the other.

Same kind as Exercise 3.3

Prove that $cos (\frac{π}{4} + x) + cos (\frac{π}{4} - x) = 2 cos x$ .

Show the solutionSolution

Sum to product: $2 cos \frac{π}{4} cos x$
$= 2 \cdot \frac{1}{2} cos x = 2 cos x$

Answer: Proved.

Practice questions on Trigonometric Functions

Write each answer on paper first. Open the hint only if you are stuck, and the answer only when you have finished.

Q1 · 1 mark · MCQExam level
The value of $sin (- \frac{13 π}{3})$ is:
(a) $- \frac{3}{2}$
(b) $\frac{3}{2}$
(c) $\frac{1}{2}$
(d) $- \frac{1}{2}$
Hint
sin is odd; remove whole turns.
Answer
(a): $sin (- \frac{13 π}{3}) = - sin (4 π + \frac{π}{3}) = - \frac{3}{2}$
Q2 · 1 mark · MCQExam level
In which quadrant does x lie if $sec x < 0$ and $tan x > 0$ ?
(a) I
(b) II
(c) III
(d) IV
Hint
cos negative, tan positive.
Answer
(c) III
Q3 · 1 mark · Assertion-reasonExam level
Assertion (A): $cos 2 x = 1 - 2 sin^{2} x$ for every real x.
Reason (R): $cos 2 x = cos^{2} x - sin^{2} x$ and $cos^{2} x = 1 - sin^{2} x$ .
(a) Both true, R explains A. (b) Both true, R does not explain A. (c) A true, R false. (d) A false, R true.
Hint
Substitute.
Answer
(a)
Q4 · 2 marks · Very short answerHarder exam level
Convert 2 radians into degrees, minutes and seconds (use $π = \frac{22}{7}$ ).
Hint
Multiply by $\frac{180 \times 7}{22}$ .
Answer
$\frac{2520}{22} ° = 114 \frac{6}{11} °$ ≈ 114°32′44″.
Q5 · 2 marks · Very short answerHarder exam level
Find $sin 75° - sin 15°$ without tables.
Hint
Sum to product.
Answer
$2 cos 45° sin 30° = 2 \cdot \frac{1}{2} \cdot \frac{1}{2} = \frac{1}{2}$ .
Q6 · 2 marks · Very short answerHarder exam level
Prove that $cos^{4} x - sin^{4} x = cos 2 x$ .
Hint
Difference of squares.
Answer
$(cos^{2} x - sin^{2} x) (cos^{2} x + sin^{2} x) = cos 2 x \times 1$ .
Q7 · 2 marks · Very short answerHarder exam level
Find the exact value of $cos 105°$ .
Hint
60° + 45°.
Answer
$cos 60° cos 45° - sin 60° sin 45° = \frac{1 - 3}{2 2} = \frac{2 - 6}{4}$ .
Q8 · 3 marks · Short answerHarder exam level
If $sin x = \frac{3}{5}$ with x in quadrant II, find $sin 3 x$ using the triple-angle formula.
Hint
Only sin x is needed.
Answer
$3 \cdot \frac{3}{5} - 4 \cdot \frac{27}{125} = \frac{225 - 108}{125} = \frac{117}{125}$ .
Q9 · 3 marks · Short answerHarder exam level
Prove that $sin 2 x + sin 4 x + sin 6 x + sin 8 x = 4 cos x cos 2 x sin 5 x$ .
Hint
Pair 8x with 2x and 6x with 4x.
Answer
$(sin 8 x + sin 2 x) + (sin 6 x + sin 4 x) = 2 sin 5 x cos 3 x + 2 sin 5 x cos x = 2 sin 5 x (cos 3 x + cos x) = 2 sin 5 x \cdot 2 cos 2 x cos x$ .
Q10 · 3 marks · Short answerHOTS / competency
Show that $3 cos x - sin x = 2 cos (x + \frac{π}{6})$ and hence find the greatest value of $3 cos x - sin x$ .
Hint
Expand the right side.
Answer
$2 cos (x + \frac{π}{6}) = 2 (\frac{3}{2} cos x - \frac{1}{2} sin x) = 3 cos x - sin x$ . Since cos is at most 1, the greatest value is 2 (at $x = - \frac{π}{6}$ ).
Q11 · 2 marks · Very short answerHOTS / competency
Without using tables, show that $sin 15° cos 15° = \frac{1}{4}$ and $cos^{2} 15° - sin^{2} 15° = \frac{3}{2}$ .
Hint
Double angles of 15°.
Answer
$sin 15° cos 15° = \frac{1}{2} sin 30° = \frac{1}{4}$ ; $cos^{2} 15° - sin^{2} 15° = cos 30° = \frac{3}{2}$ .

Test your recall

Answer in your head or on paper first, then open the card. Pulling an answer out of memory makes it stick far better than re-reading.

1.How many degrees is one radian, roughly?Show answer
About 57°16′.
2.Convert $\frac{7 π}{12}$ to degrees.Show answer
105°.
3.In which quadrant are both sin and cos negative?Show answer
The third.
4.What is the range of sec x?Show answer
$(- \infty, - 1] \cup [1, \infty)$ .
5.What is the period of tan x?Show answer
π.
6.State $cos (x - y)$ .Show answer
$cos x cos y + sin x sin y$ .
7.Write $cos 2 x$ in terms of $sin x$ only.Show answer
$1 - 2 sin^{2} x$ .
8.What is $cos 3 x$ in terms of $cos x$ ?Show answer
$4 cos^{3} x - 3 cos x$ .

Best order to study it

Learn positive and negative angles, degrees, radians and conversions.
Practise $l = r θ$ with clocks, wheels and pendulums.
Learn the unit-circle definitions and the three Pythagorean identities.
Learn the signs in each quadrant and practise 'given one, find five'.
Learn domains, ranges, periods and sketch the graphs of sin, cos and tan.
Learn reduction of large and negative angles with allied-angle rules.
Learn the sum and difference formulae, then sum-to-product.
Finish with 2x, 3x and half-angle identities and mixed proofs.

Quick revision checklist

The night before the exam: can you do each of these without looking?

Can you convert 40°20′ to radians and 6 radians to degrees?
Can you use $l = r θ$ for an arc, a clock hand or a pendulum?
Can you prove $1 + tan^{2} x = sec^{2} x$ from $sin^{2} x + cos^{2} x = 1$ ?
Can you give the sign of each function in each quadrant?
Can you state the domain and range of all six functions?
Can you evaluate trigonometric functions of angles like 1140° or $- \frac{11 π}{3}$ ?
Can you derive $sin (x + y)$ and $tan (x + y)$ from $cos (x + y)$ ?
Can you find sin 15°, cos 75° and tan 105° exactly?
Can you use sum-to-product formulae in proofs?
Can you write $sin 3 x$ , $cos 3 x$ and $tan 3 x$ and use half-angle forms with correct signs?

Trigonometric Functions: frequently asked questions

Are trigonometric equations in the 2026-27 Class 11 exam?

General solutions of equations such as \sin y = \sin a are taught but assessed only formatively, so they are not in the annual exam paper. The section on trigonometric equations was removed from the rationalised NCERT book.

Do I need the sine and cosine rules (sine formula, cosine formula)?

No. Their proofs and applications were removed in the NCERT rationalisation and are not in the 2026-27 syllabus.

Why do we use radians at all?

Radians make formulae simpler: arc length is r\theta and sector area is \frac{1}{2}r^2\theta only in radians. In calculus the derivative of sin x is cos x only when x is in radians.

Which identities must I memorise and which can I derive?

Memorise \cos(x + y) and \sin(x + y), the three forms of \cos 2x and the four sum-to-product formulae. The rest (tan of a sum, triple angles, half angles) can be derived in a line or two, though memorising them saves time.

How is the Class 11 Maths exam set?

It is a school annual exam of 80 marks (3 hours) plus 20 marks of internal assessment; Unit I, Sets and Functions, which includes this chapter, carries 23 marks. CBSE gives no Class 11 sample paper, so most schools use a mix of MCQs, assertion-reason, short and long answers and case-based questions.

Based on: CBSE Mathematics (041) syllabus 2026-27 (Classes XI-XII); NCERT Mathematics, Textbook for Class XI, Chapter 3 (Reprint 2026-27)