Trigonometric Functions ISC Class 11: Notes and Graphs

What Trigonometric Functions is about

Weightage: Trigonometric Functions belongs to the unit Sets and Functions, which CISCE weights at 18 of the 80 marks in the ISC Class XI Mathematics Paper I (Theory). Class XI is examined by the school, so your school sets the paper within this weightage.Source: CISCE ISC Mathematics syllabus, examination year 2028, Class XI unit weightage

Key concepts

Angles: sign, degree and radian measure

An angle is formed by rotating a ray from an initial side to a terminal side. Anticlockwise rotation gives a positive angle and clockwise a negative angle; rotations can exceed 360°. Degree measure: 1 revolution = 360°, 1° = 60′, 1′ = 60″. Radian (circular) measure: 1 radian is the angle subtended at the centre of a circle by an arc whose length equals the radius. One revolution is 2π radians, so π radians = 180°, 1 rad = 180°/π ≈ 57°16′ and 1° = π/180 rad ≈ 0.01746 rad. To convert degrees to radians multiply by π/180; radians to degrees multiply by 180/π.

Worked example

Convert 40°20′ into radians, and 6 radians into degrees (to the nearest minute).

Try it first, then show the solutionSolution

1. 40°20′ = 40⅓° = (121/3)° = (121/3) × π/180 = 121π/540 rad ≈ 0.704 rad
2. 6 rad = 6 × 180/π = 343.77° = 343° + 0.77 × 60′ ≈ 343°46′

Answer: 121π/540 rad; about 343°46′

Arc length and area of a sector

In a circle of radius r, an arc of length s subtends an angle θ (in radians) at the centre, where s = rθ. The area of the sector bounded by two radii and the arc is $A=\frac{1}{2}{r}^2\theta =\frac{1}{2}rs$ (a fraction θ/2π of the whole area πr²). These formulae need θ in radians; for degrees use $s=\frac{\pi r\theta °}{180}$ and $A=\frac{\pi r^2\theta °}{360}$. The syllabus explicitly includes questions on the area of a sector.

Worked example

In a circle of radius 21 cm, an arc subtends 60° at the centre. Find the arc length and the area of the sector. (π = 22/7)

Try it first, then show the solutionSolution

1. θ = 60° = π/3 rad; s = 21 × π/3 = 7π = 22 cm
2. A = ½ × 21² × π/3 = 441π/6 = 231 cm²

Answer: 22 cm; 231 cm²

Trigonometric functions on the unit circle; the basic identities

Place the angle x in standard position (vertex at the origin, initial side along the positive x-axis) and let its terminal side meet the unit circle x² + y² = 1 at P(a, b). Define cos x = a and sin x = b; then tan x = b/a (a ≠ 0), cot x = a/b (b ≠ 0), sec x = 1/a, cosec x = 1/b. Because a² + b² = 1, sin²x + cos²x = 1 for all x. Dividing by cos²x gives 1 + tan²x = sec²x; dividing by sin²x gives 1 + cot²x = cosec²x. Angles x and x + 2nπ have the same terminal side, so they have the same values. sin x = 0 when x = nπ; cos x = 0 when x = (2n + 1)π/2.

Worked example

If sin x = 3/5 and x lies in the second quadrant, find the other five trigonometric functions.

Try it first, then show the solutionSolution

1. cos²x = 1 − 9/25 = 16/25; in the second quadrant cos x < 0, so cos x = −4/5
2. tan x = (3/5)/(−4/5) = −3/4; cot x = −4/3
3. sec x = −5/4; cosec x = 5/3

Answer: cos x = −4/5, tan x = −3/4, cot x = −4/3, sec x = −5/4, cosec x = 5/3

Signs of trigonometric functions in the four quadrants

The signs follow from the coordinates of P: Quadrant I (0 to π/2): all positive. Quadrant II (π/2 to π): sin and cosec positive, the rest negative. Quadrant III (π to 3π/2): tan and cot positive. Quadrant IV (3π/2 to 2π): cos and sec positive. Remember 'All Students Take Coffee' (All, Sin, Tan, Cos). Negative angles: sin(−x) = −sin x, cos(−x) = cos x, tan(−x) = −tan x; so sin and tan are odd functions and cos is even.

Worked example

State the signs of sin 300°, cos 200°, tan 200° and sec(−50°).

Try it first, then show the solutionSolution

1. 300° is in Q IV: sin negative. 200° is in Q III: cos negative, tan positive.
2. −50° lies in Q IV: cos and sec positive.

Answer: −, −, +, +

Trigonometric functions of all angles (allied angles)

Values for any angle reduce to values for acute angles. Periodicity: sin(2nπ + x) = sin x, cos(2nπ + x) = cos x, tan(nπ + x) = tan x. Allied angles (think of x as acute when fixing signs): for π/2 − x, the co-function appears with a plus sign (cos x for sin, sin x for cos); for π/2 + x, sin becomes cos x while cos becomes −sin x; for π − x, sin stays sin x while cos becomes −cos x; for π + x, both sin and cos change sign; for 2π − x, cos stays cos x while sin becomes −sin x. Rule: with π or 2π the function keeps its name; with π/2 or 3π/2 it changes to the co-function; the sign is that of the original function in the quadrant where the angle lies.

Worked example

Find the values of sin 765°, cos(−1710°) and tan 945°.

Try it first, then show the solutionSolution

1. 765° = 2 × 360° + 45°: sin 765° = sin 45° = 1/√2
2. cos(−1710°) = cos 1710° = cos(4 × 360° + 270°) = cos 270° = 0
3. 945° = 5 × 180° + 45°: tan 945° = tan 45° = 1

Answer: 1/√2; 0; 1

Domain, range and periods

A function f is periodic with period T if f(x + T) = f(x) for all x, T being the smallest such positive number. sin x, cos x: domain R, range [−1, 1], period 2π. tan x: domain R − {(2n + 1)π/2}, range R, period π. cot x: domain R − {nπ}, range R, period π. sec x: domain R − {(2n + 1)π/2}, range (−∞, −1] ∪ [1, ∞), period 2π. cosec x: domain R − {nπ}, range (−∞, −1] ∪ [1, ∞), period 2π. If f has period T, then f(ax + b) has period T/|a|: sin 3x has period 2π/3 and tan(x/2) has period 2π.

Worked example

Find the period of sin 3x and of tan 2x, and the range of 3 sin x + 4.

Try it first, then show the solutionSolution

1. sin 3x: 2π/3. tan 2x: π/2.
2. −1 ≤ sin x ≤ 1, so 1 ≤ 3 sin x + 4 ≤ 7: range [1, 7]

Answer: 2π/3; π/2; [1, 7]

Graphs of the six trigonometric functions

y = sin x: a smooth wave through (0, 0), maximum 1 at π/2, back to 0 at π, minimum −1 at 3π/2, repeating every 2π. y = cos x: the same wave shifted left by π/2, starting at (0, 1). y = tan x: increasing branches through (nπ, 0), with vertical asymptotes at x = (2n + 1)π/2; period π. y = cot x: decreasing branches with asymptotes at x = nπ. y = sec x: U-shaped branches above y = 1 and below y = −1, touching where cos x = ±1, with asymptotes where cos x = 0. y = cosec x: similar branches touching where sin x = ±1, asymptotes where sin x = 0. Only sketches are needed, showing key points, asymptotes and the period.

Worked example

Using the graphs, state where tan x has vertical asymptotes, and how many times y = sin x meets the line y = ½ for 0 ≤ x ≤ 2π.

Try it first, then show the solutionSolution

1. tan x = sin x/cos x is undefined where cos x = 0: x = ±π/2, ±3π/2, …
2. y = ½ cuts the first arch of sin x (between 0 and π) twice and misses the negative arch: at x = π/6 and 5π/6.

Answer: x = (2n + 1)π/2; twice (π/6 and 5π/6)

Proving simple identities

To prove an identity, start from the more complicated side and simplify it to the other side (or reduce both sides to the same expression); do not move terms across the equals sign as if solving an equation. Useful moves: write everything in terms of sin and cos; use sin²x + cos²x = 1, 1 + tan²x = sec²x, 1 + cot²x = cosec²x; take a common denominator; factorise (difference of squares); multiply numerator and denominator by a conjugate.

Worked example

Prove that tan x + cot x = sec x cosec x.

Try it first, then show the solutionSolution

1. LHS = sin x/cos x + cos x/sin x = (sin²x + cos²x)/(sin x cos x)
2. = 1/(sin x cos x) = (1/cos x)(1/sin x) = sec x cosec x = RHS

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.

Topics in this chapter

• Radian Measure, Arc Length and Sector Area
• Trigonometric Ratios, Allied Angles, Graphs and Identities

Trigonometric Functions formula sheet

Degree-radian conversion

$\pi \text{rad}=180^{\circ };1\text{rad}\approx 57^{\circ }16^{\prime }$

Degrees × π/180 = radians.

Arc and sector

$s=r\theta ,A=\frac{1}{2}{r}^2\theta =\frac{1}{2}rs$

θ in radians.

Pythagorean identities

${\sin }^{2}x+{\cos }^{2}x=1;1+{\tan }^{2}x={\sec }^{2}x;1+{\cot }^{2}x={\csc }^{2}x$

True for all x where defined.

Negative angles

$\sin (-x)=-\sin x,\cos (-x)=\cos x,\tan (-x)=-\tan x$

cos is even; sin, tan odd.

Allied angles

$\sin (\pi -x)=\sin x;\cos (\pi -x)=-\cos x;\sin (\frac{\pi }{2}+x)=\cos x;\cos (\frac{\pi }{2}+x)=-\sin x$

π: same name; π/2: co-function.

Periods

$\sin ,\cos ,\sec ,\csc :2\pi ;\tan ,\cot :\pi ;f(ax):T/|a|$

sin 3x has period 2π/3.

Ranges

$\sin ,\cos :[-1,1];\tan ,\cot :R;\sec ,\csc :(-\infty ,-1]\cup [1,\infty )$

From the unit circle.

Trigonometric Functions: diagrams to practise

Practise each one on paper: draw it from the checklist, then tick off every label.

How the school exam asks Trigonometric Functions

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

Tricks to remember Trigonometric Functions

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

• All Students Take Coffee

  Quadrants I to IV: All, Sin, Tan, Cos are positive.

  Use it when: Signs of trigonometric functions.

• π keeps, π/2 swaps

  With π or 2π the function name stays; with π/2 or 3π/2 it swaps to the co-function. Then fix the sign by the quadrant.

  Use it when: Allied angles.

• Radian means radius

  One radian: the arc is one radius long, so s = rθ.

  Use it when: Arc and sector problems.

• Tan and cot are quick

  tan and cot repeat every π; the others every 2π.

  Use it when: Periods.

• Sec hugs cos

  sec x branches touch cos x at ±1 and fly off where cos x = 0.

  Use it when: Sketching sec and cosec graphs.

Trigonometric Functions: exam-level solved questions

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

Case-study practice (Section D style)

Read the situation, then answer every part. In a school exam set on the board pattern this is a 5-mark question split into 1 + 1 + 1 + 2 marks.

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.

Where students lose marks in Trigonometric Functions

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

• ✗Mistake: Using degrees in s = rθ

  The formula needs radians; degrees give an answer about 57 times too large.

  Wrong working

  r = 21 cm, θ = 60°: s = 21 × 60 = 1260 cm.

  Right working

  θ = π/3: s = 21 × π/3 = 22 cm.

  ✓Do this: Convert to radians before using s = rθ or A = ½r²θ.

  Usually costs: 1-2 marks.

• ✗Mistake: Ignoring the quadrant when taking square roots

  Students take cos x = +4/5 from sin x = 3/5 whatever the quadrant.

  Wrong working

  x in Q II, sin x = 3/5: cos x = 4/5.

  Right working

  In Q II cos x < 0: cos x = −4/5.

  ✓Do this: Write the quadrant's signs before computing.

  Usually costs: 1 mark per wrong value.

• ✗Mistake: Wrong sign or name with allied angles

  Students write cos(90° + x) = sin x.

  Wrong working

  cos(90° + x) = sin x.

  Right working

  90° + x lies in Q II, where cos is negative: cos(90° + x) = −sin x.

  ✓Do this: Name change for 90° and 270°; sign from the quadrant of the original angle.

  Usually costs: 1 mark.

• ✗Mistake: Proving identities by cross-multiplying

  Students treat the identity as an equation and work on both sides at once.

  Wrong working

  (1 − sin x)(1 + sin x) = cos²x, so the identity is true.

  Right working

  Start from one side and transform it step by step into the other side.

  ✓Do this: Work on one side; end with '= RHS'.

  Usually costs: 1-2 marks.

• ✗Mistake: Wrong range for sec and cosec

  Students write [−1, 1] or R for sec x.

  Wrong working

  Range of sec x = [−1, 1].

  Right working

  Range of sec x = (−∞, −1] ∪ [1, ∞): since |cos x| ≤ 1, |sec x| ≥ 1.

  ✓Do this: Reciprocals of values in [−1, 1] lie outside (−1, 1).

  Usually costs: 1 mark.

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.

1. Q1 · 1 mark · MCQExam level

   The degree measure of 5π/6 radians is:
   (a) 120°
   (b) 135°
   (c) 150°
   (d) 165°

   Hint

   Multiply by 180/π.

   Answer

   (c)

2. Q2 · 1 mark · MCQExam level

   In which quadrant are sin x and cos x both negative?
   (a) I
   (b) II
   (c) III
   (d) IV

   Hint

   Only tan is positive.

   Answer

   (c)

3. Q3 · 1 mark · Very short answerExam level

   Write the period of cos 4x.

   Hint

   2π/4.

   Answer

   π/2

4. Q4 · 1 mark · Very short answerExam level

   Find the value of sin(−π/6).

   Hint

   sin is odd.

   Answer

   −½

5. Q5 · 2 marks · Short answerExam level

   Find the length of an arc of a circle of radius 10 cm that subtends 1.2 rad at the centre, and the area of the sector.

   Hint

   s = rθ.

   Answer

   s = 12 cm; A = ½ × 100 × 1.2 = 60 cm².

6. Q6 · 2 marks · Short answerHarder exam level

   Find the values of cos 315° and tan(−585°).

   Hint

   Reduce each angle.

   Answer

   cos 315° = cos(360° − 45°) = 1/√2; tan(−585°) = −tan 585° = −tan(3 × 180° + 45°) = −1.

7. Q7 · 2 marks · Short answerHarder exam level

   Find the range of 5 − 2 cos x.

   Hint

   cos x ∈ [−1, 1].

   Answer

   −2 cos x ∈ [−2, 2], so the range is [3, 7].

8. Q8 · 2 marks · Short answerHarder exam level

   Prove that (1 + tan²x) cos²x = 1.

   Hint

   1 + tan²x = sec²x.

   Answer

   LHS = sec²x cos²x = (1/cos²x) cos²x = 1 = RHS.

9. Q9 · 2 marks · Short answerHarder exam level

   The angles of a triangle are in the ratio 2 : 3 : 4. Find them in radians.

   Hint

   Sum is π.

   Answer

   2π/9, π/3 and 4π/9.

10. Q10 · 3 marks · Short answerHarder exam level

    If sin x = −24/25 and x lies in the third quadrant, find cos x, tan x and cosec x + cot x.

    Hint

    In Q III cos is negative.

    Answer

    cos x = −7/25; tan x = 24/7; cosec x = −25/24, cot x = 7/24, so cosec x + cot x = −18/24 = −3/4.

11. Q11 · 3 marks · Short answerHarder exam level

    Prove that (sin x + cos x)² + (sin x − cos x)² = 2, and that sin⁴x − cos⁴x = sin²x − cos²x.

    Hint

    Expand; difference of squares.

    Answer

    First: 2 sin²x + 2 cos²x = 2. Second: (sin²x − cos²x)(sin²x + cos²x) = sin²x − cos²x.

12. Q12 · 5 marks · Long answerHarder exam level

    (a) A railway line is to be laid on a curve of radius 1500 m. Through what angle must it turn so that a train covers 1.5 km on the curve? Give the answer in degrees. [2]
    (b) Sketch y = cos x and y = sec x for 0 ≤ x ≤ 2π on the same axes. [2]
    (c) State the domain of sec x. [1]

    Hint

    θ = s/r.

    Answer

    (a) θ = 1500/1500 = 1 rad ≈ 57°16′. (b) cos x: wave from 1 down to −1 at π and back; sec x: branches touching cos x at 0, π and 2π, with asymptotes at π/2 and 3π/2. (c) R − {(2n + 1)π/2 : n ∈ Z}.

13. Q13 · 3 marks · Short answerHOTS / competency

    Find all values of x in [0, 2π] for which sin x = cos x, and explain your answer using the graphs.

    Hint

    tan x = 1.

    Answer

    Dividing by cos x (non-zero there): tan x = 1, so x = π/4 or 5π/4. The sin and cos waves cross twice per period, once in Q I (both positive) and once in Q III (both negative).

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. 1.π radians in degrees?Show answer

   180°.

2. 2.Arc length formula?Show answer

   s = rθ (θ in radians).

3. 3.Area of a sector?Show answer

   ½r²θ.

4. 4.Which functions are positive in Q III?Show answer

   tan and cot.

5. 5.cos(π − x)?Show answer

   −cos x.

6. 6.Period of tan x?Show answer

   π.

7. 7.Range of cosec x?Show answer

   (−∞, −1] ∪ [1, ∞).

8. 8.Is cos x odd or even?Show answer

   Even: cos(−x) = cos x.

Quick revision checklist

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

• ☐Can you convert between degrees and radians?
• ☐Can you find arc length and sector area?
• ☐Can you find all six ratios from one, given the quadrant?
• ☐Can you state signs in each quadrant?
• ☐Can you evaluate sin, cos, tan of angles like 765° and −1125°?
• ☐Can you give the period and range of each function?
• ☐Can you sketch tan, sec and cosec with asymptotes?
• ☐Can you prove simple identities working on one side?

Trigonometric Functions: frequently asked questions