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State Board · Class 11 · Mathematics Part 1

Lesson 2: Trigonometry - I

5 concepts · 40 practice questions with answers · Free to read

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Trigonometric functions of any angle and their signs

From daily life

As a Ferris-wheel seat goes round, its sideways and upward distances from the centre keep changing. For every turning angle, cos and sin record exactly these two distances. Building on classes 9–10 : There sin, cos, tan were ratios of the sides of a right triangle, so only for 0° to 90°; now the unit circle gives them for every angle.

Simple definition

The unit circle has its centre at the origin and radius 1. The terminal ray of an angle θ in standard position meets it at P(x, y). Then cos θ = x and sin θ = y. The other four functions: tan θ = y/x, cot θ = x/y, sec θ = 1/x, cosec θ = 1/y.

Rule 1: Definition from any point — If P(x, y) is any point on the terminal ray and r = √(x² + y²) is its distance from the origin, then sin θ = y/r, cos θ = x/r, tan θ = y/x. Taking r = 1 gives x = cos θ and y = sin θ.

XYθPOxyr = 1x = cos θy = sin θtan θ = y/x

In the figure : P(x, y) is a point on the unit circle; x = cos θ (orange), y = sin θ (green), r = 1.

  • Easy : P(0.6, 0.8) on the unit circle: sin θ = 0.8, cos θ = 0.6, tan θ = 0.8/0.6 = 4/3.
  • Medium : P(−5, 12): r = √(25 + 144) = 13; sin θ = 12/13, cos θ = −5/13, tan θ = −12/5.
  • Exam-type : The terminal ray passes through (−√3, −1): r = √(3 + 1) = 2; sin θ = −1/2, cos θ = −√3/2, tan θ = 1/√3.

How is the formula obtained? : Whichever point we pick on the terminal ray, similar triangles keep y/r, x/r and y/x the same. So we take the point with r = 1 (on the unit circle) and get x = cos θ, y = sin θ directly. Since x and y may be negative, every angle gets a value with its sign.

Rule 2: Signs in the quadrants — r is always positive, so the sign comes from x and y. sin and cosec take the sign of y, cos and sec the sign of x, tan and cot the sign of y/x.

XYIAll +all six +IIsin +(cosec +)IIItan +(cot +)IVcos +(sec +)O

In the figure : All functions are positive in quadrant I; sin (and cosec) in II; tan (and cot) in III; cos (and sec) in IV.

  • Easy : sin 100°: 100° is in quadrant II, so sin 100° is positive and cos 100° is negative.
  • Medium : tan 250°: 250° is in quadrant III, so it is positive; cos(−40°): −40° is in quadrant IV, so it is positive.
  • Exam-type : 3 radians ≈ 171.9° (π/2 ≈ 1.57 < 3 < π ≈ 3.14) is in quadrant II: sin 3 > 0, cos 3 < 0. 4 radians ≈ 229.2° is in quadrant III: sin 4 < 0, tan 4 > 0.
Remember

r is always positive; the sign comes only from x and y. All are positive in quadrant I; sin (and cosec) in II; tan (and cot) in III; cos (and sec) in IV.

Common mistake

Taking the sign of cosec from cos. cosec is the reciprocal of sin, so it has the sign of sin; likewise sec has the sign of cos and cot the sign of tan.

Shortcut

Go anticlockwise from quadrant I saying A-S-T-C: All, Sin, Tan, Cos. The named function (and its reciprocal) is positive there, the rest are negative. Memory clue : "All Students Take Coffee" — A, S, T, C.

CET trap

If the angle is in radians (sin 3, cos 4), recall π ≈ 3.14, π/2 ≈ 1.57, 3π/2 ≈ 4.71 to find the quadrant; do not treat 3 as "quadrant I". A negative angle (−40°) is measured clockwise, hence quadrant IV.

CET-style example : If sin θ < 0 and cot θ > 0, in which quadrant does θ lie? (A) I (B) II (C) III (D) IV

Answer : (C). sin θ < 0 → quadrant III or IV; cot θ > 0 → quadrant I or III; the common one is III. Time-saving method : Write the quadrants for each condition and pick the common one: sin negative = {III, IV}, cot positive = {I, III} → III.

Spot the difference

QuadrantAngleSigns of x, yPositive functions
I0°–90°+, +all six
II90°–180°−, +sin, cosec
III180°–270°−, −tan, cot
IV270°–360°+, −cos, sec

Practice questions

Easy
  1. 1For the point (0.8, 0.6) on the unit circle write sin θ and cos θ.
  2. 2Which two functions are positive in quadrant III?
  3. 3What is the sign of cos 200°?
Medium
  1. 4The terminal ray passes through (−12, −5); find sin θ, cos θ, tan θ.
  2. 5If sin θ > 0 and cos θ < 0, in which quadrant is θ?
  3. 6Is sec 300° positive or negative?
Challenging
  1. 7If tan θ = −5/12 and sin θ < 0, find cos θ and cosec θ.
  2. 8State the signs of sin 5 and cos 5 (5 radians).
Show answers (try them yourself first)
  1. 1sin θ = 0.6, cos θ = 0.8.
  2. 2tan, cot.
  3. 3Negative (200° is in quadrant III).
  4. 4r = 13; sin θ = −5/13, cos θ = −12/13, tan θ = 5/12.
  5. 5Quadrant II.
  6. 6Positive (300° is in quadrant IV, where cos is positive).
  7. 7θ is in quadrant IV: cos θ = 12/13, cosec θ = −13/5.
  8. 85 radians ≈ 286.5° (quadrant IV): sin 5 < 0, cos 5 > 0.

Values at standard angles; negative and large angles

From daily life

On a clock we know 12, 3, 6 and 9 by heart and read every other time from them. In the same way, once the values at 0°, 30°, 45°, 60°, 90° are known, every other angle reduces to them. Building on classes 9–10 : In class 10 we learnt the values from 0° to 90°; now we get 150°, 225°, −30°, 1000° and every other angle from them.

Simple definition

Learn the table below. tan 90° is not defined because x = cos 90° = 0 and y/x would need division by zero.

θ0°30°45°60°90°
sin θ01/21/√2√3/21
cos θ1√3/21/√21/20
tan θ01/√31√3not defined
XY30° : (√3/2, 1/2)45° : (1/√2, 1/√2)60° : (1/2, √3/2)0°90°O

In the figure : The unit circle in quadrant I; each point has co-ordinates (cos θ, sin θ).

Rule 1: Reference angle — For an angle in quadrant II, III or IV take the acute angle α it makes with the X-axis (the reference angle). The value is that of α and the sign is that of the quadrant. The angles 180° − α, 180° + α, 360° − α lie in quadrants II, III, IV respectively.

  • Easy : sin 150° = sin(180° − 30°) = + sin 30° = 1/2 (quadrant II, sin positive).
  • Medium : cos 225° = cos(180° + 45°) = − cos 45° = −1/√2 (quadrant III, cos negative).
  • Exam-type : tan 330° = tan(360° − 30°) = − tan 30° = −1/√3 (quadrant IV, tan negative).

Rule 2: Negative angles — sin(−θ) = −sin θ, cos(−θ) = cos θ, tan(−θ) = −tan θ. So cos (and sec) are even functions; sin, tan, cot, cosec are odd functions.

  • Easy : sin(−30°) = −sin 30° = −1/2.
  • Medium : cos(−120°) = cos 120° = −cos 60° = −1/2.
  • Exam-type : tan(−225°) = −tan 225° = −tan(180° + 45°) = −1.

How is the formula obtained? : The reflection of P(x, y) in the X-axis is P′(x, −y). The terminal ray of −θ passes through P′, so cos(−θ) = x = cos θ and sin(−θ) = −y = −sin θ. Dividing, tan(−θ) = −y/x = −tan θ.

Rule 3: Large angles (periodicity) — After 360° (2π) the terminal ray returns to the same place, so for sin, cos, cosec, sec subtract multiples of 360°; for tan and cot multiples of 180° (π) can be subtracted.

  • Easy : cos 420° = cos(420° − 360°) = cos 60° = 1/2.
  • Medium : sin 1125° = sin(1125° − 1080°) = sin 45° = 1/√2.
  • Exam-type : tan 1230° = tan(1230° − 1080°) = tan 150° = −1/√3; sin(25π/6) = sin(25π/6 − 4π) = sin(π/6) = 1/2.
Remember

Only cos and sec are even; all others are odd (the sign flips for a negative angle). Divide a large angle by 360° (180° for tan, cot) and use the remainder.

Common mistake

Writing tan(−θ) = tan θ. The sign of sin, tan, cot, cosec flips; only cos and sec stay the same. Also sin(180° − θ) is sin θ, not −sin θ.

Shortcut

Large angle: divide by 360° and use the remainder (add 360° if it is negative). For tan, dividing by 180° also works. Memory clue : "Only cos stays the same" — cos keeps its sign for −θ, all the others flip.

CET trap

sin 150° = +1/2 but sin 210° = −1/2; the size is the same and the options list both signs. Always decide the sign from the quadrant.

CET-style example : What is the value of sin 765° + cos(−330°)? (A) (√2 + √3)/2 (B) (√2 − √3)/2 (C) (√3 − √2)/2 (D) (2 + √3)/2

Answer : (A). 765° − 720° = 45°, sin 45° = 1/√2 = √2/2; cos(−330°) = cos 330° = cos 30° = √3/2; the sum is (√2 + √3)/2. Time-saving method : Reduce each term to a small angle; both come out positive, so the sum is positive and only (A) and (D) are possible; (D) is too large (1.93), so (A).

Spot the difference

Anglesincostan
180° − θ+ sin θ− cos θ− tan θ
180° + θ− sin θ− cos θ+ tan θ
360° − θ− sin θ+ cos θ− tan θ
−θ− sin θ+ cos θ− tan θ

Practice questions

Easy
  1. 1Find the value of sin 120°.
  2. 2What is the value of cos 180°?
  3. 3Find the value of tan 135°.
Medium
  1. 4Find the value of sin(−150°).
  2. 5Find the value of cos 840°.
  3. 6Find the value of tan(5π/4).
Challenging
  1. 7Find the value of sin 2010°.
  2. 8Find the value of sin²135° + cos²(−120°).
Show answers (try them yourself first)
  1. 1√3/2 (120° = 180° − 60°, sin positive).
  2. 2−1.
  3. 3−1 (135° = 180° − 45°, tan negative).
  4. 4−1/2 (−sin 150° = −1/2).
  5. 5−1/2 (840° − 720° = 120°).
  6. 61 (180° + 45°, tan positive).
  7. 7−1/2 (2010° − 1800° = 210°).
  8. 83/4 (1/2 + 1/4).

Fundamental trigonometric identities

From daily life

A ladder leaning on a wall makes a right triangle of ladder length, height on the wall and distance on the ground, and Pythagoras ties the three together every time. Take the ladder length as 1 and the relation is in terms of sin and cos. Building on classes 9–10 : In class 10 we learnt sin²θ + cos²θ = 1 for 0° to 90°; now the same holds for every angle.

Simple definition

An identity is an equality that is true for all (defined) values of the variable. The most basic one is sin²θ + cos²θ = 1.

Rule 1: Reciprocal and quotient relations — cosec θ = 1/sin θ, sec θ = 1/cos θ, cot θ = 1/tan θ, tan θ = sin θ/cos θ, cot θ = cos θ/sin θ.

Rule 2: Pythagorean identities — sin²θ + cos²θ = 1, 1 + tan²θ = sec²θ, 1 + cot²θ = cosec²θ. Knowing one function gives all the others; decide the sign after the square root from the quadrant.

cos θsin θ1θsin²θ + cos²θ = 1(Pythagoras)

In the figure : The right triangle in the unit circle: sides cos θ, sin θ and hypotenuse 1, so sin²θ + cos²θ = 1.

How is the formula obtained? : For P(x, y) on the unit circle x² + y² = 1 (r = 1). Putting x = cos θ, y = sin θ gives sin²θ + cos²θ = 1. Divide both sides by cos²θ: tan²θ + 1 = sec²θ. Divide by sin²θ: 1 + cot²θ = cosec²θ.

  • Easy : sin θ = 3/5 and θ acute: cos²θ = 1 − 9/25 = 16/25, so cos θ = 4/5 and tan θ = 3/4.
  • Medium : tan θ = 5/12 and θ in quadrant III: sec²θ = 1 + 25/144 = 169/144; cos is negative in quadrant III, so sec θ = −13/12, cos θ = −12/13 and sin θ = tan θ × cos θ = −5/13.
  • Exam-type : If sin θ + cos θ = 7/5, find sin θ cos θ: (sin θ + cos θ)² = 1 + 2 sin θ cos θ; 49/25 = 1 + 2 sin θ cos θ; so sin θ cos θ = 12/25.
Remember

All three Pythagorean identities hold for any angle. Pairs: tan ↔ sec and cot ↔ cosec. A ± appears after the square root; fix the sign from the quadrant.

Common mistake

Writing "sin θ + cos θ = 1" and forgetting the squares. Also sin²θ means (sin θ)², not sin(θ²).

Shortcut

Right-triangle triples: (3, 4, 5), (5, 12, 13), (8, 15, 17), (7, 24, 25). If sin θ = 5/13 then cos θ = 12/13 at once, with no square root. Memory clue : "1 + tan² = sec²" — the tan–sec pair; "1 + cot² = cosec²" — the cot–cosec pair. The "co" function pairs with the "co" function.

CET trap

CET often gives "sec θ + tan θ = k" and asks for sec θ − tan θ; since sec²θ − tan²θ = 1 the answer is 1/k. Do not waste time finding θ.

CET-style example : If sec θ + tan θ = 2, then sec θ − tan θ equals? (A) 1/2 (B) 2 (C) 1 (D) 3/4

Answer : (A). (sec θ + tan θ)(sec θ − tan θ) = sec²θ − tan²θ = 1, so the other factor is 1/2. Time-saving method : Spot that the product is 1; answer = 1/given value. Solved in 10 seconds.

Spot the difference

IdentityObtained by dividing byWhen to use
sin²θ + cos²θ = 1basic identityrelation between sin and cos
1 + tan²θ = sec²θcos²θwhen tan or sec is given
1 + cot²θ = cosec²θsin²θwhen cot or cosec is given

Practice questions

Easy
  1. 1If cos θ = 12/13 (θ acute), find sin θ.
  2. 2Write 1 + tan²θ as a single function.
  3. 3If sin θ = 0.6, what is the value of sin²θ + cos²θ?
Medium
  1. 4cot θ = 4/3 and θ in quadrant III; find sin θ and cos θ.
  2. 5If sin θ − cos θ = 1/5, find sin θ cos θ.
  3. 6Simplify (1 − cos²θ) cosec²θ.
Challenging
  1. 7If cosec θ − cot θ = 1/3, find cosec θ + cot θ.
  2. 8tan θ = −3/4 and θ in quadrant IV; find sec θ and sin θ.
Show answers (try them yourself first)
  1. 1sin θ = 5/13.
  2. 2sec²θ.
  3. 31.
  4. 4Quadrant III: sin θ = −3/5, cos θ = −4/5.
  5. 5sin θ cos θ = 12/25.
  6. 61 (since 1 − cos²θ = sin²θ and sin²θ × cosec²θ = 1).
  7. 73.
  8. 8sec θ = 5/4, sin θ = −3/5.

Graphs, domain, range and period

From daily life

Sea waves, heartbeats, swing movements — all repeat after a fixed time. The graphs of sin and cos are the mathematical picture of such repetition. Building on classes 9–10 : In classes 9–10 we drew y = x and y = x²; now for y = sin x the X-axis carries the angle (in radians).

Simple definition

The domain of a function is the set of all allowed x; the range is the set of all values y takes; the period is the smallest positive T for which f(x + T) = f(x).

Rule 1: Properties of sin, cos and tan — sin x and cos x: domain all real numbers, range [−1, 1], period 2π. tan x: domain R − {(2n + 1)π/2}, range all real numbers, period π.

π/2π3π/22π1−1sin xcos x

In the figure : Graphs of sin x and cos x from 0 to 2π; the wave stays within [−1, 1] and repeats after 2π.

−π−π/2π/2πtan x

In the figure : Graph of tan x: vertical lines (asymptotes) at x = ±π/2 which the curve never touches; the period is π.

Rule 2: Finding period and range — The period of sin(kx) and cos(kx) is 2π/|k|; of tan(kx) it is π/|k|. The range of a sin x + b is [b − |a|, b + |a|]. The period of a sum of two functions is the LCM of their periods.

How is the formula obtained? : For sin(kx), sin(k(x + T)) = sin(kx + kT). This equals sin(kx) when kT = 2π, so T = 2π/k. Range: multiply −1 ≤ sin x ≤ 1 by a and add b to get b − |a| to b + |a|.

  • Easy : The period of sin 4x = 2π/4 = π/2.
  • Medium : Range of 5 cos x − 2: cos x ∈ [−1, 1], so 5 cos x ∈ [−5, 5] and 5 cos x − 2 ∈ [−7, 3].
  • Exam-type : The period of sin(x/2) + cos(x/3): the component periods are 4π and 6π; their LCM is 12π.
Remember

sin, cos are bounded in [−1, 1]; tan, cot take any value. The tan graph breaks at its asymptotes. Period is 2π for sin, cos and π for tan, cot.

Common mistake

Writing the period of sin 3x as 6π (multiplying). The period comes by dividing: 2π/3. A bigger k packs the wave tighter and shortens the period.

Shortcut

Look at the multiplier k of x; period = 2π ÷ k (π ÷ k for tan). For a range, put sin x = +1 and −1 and read the two extreme values. Memory clue : "Bigger k, tighter wave — shorter period."

CET trap

The period of |sin x| and of sin²x is π, not 2π, because the negative half is flipped upwards. The options deliberately include 2π.

CET-style example : What is the range of 3 − 4 sin x? (A) [−1, 7] (B) [−4, 4] (C) [3, 7] (D) [−1, 3]

Answer : (A). At sin x = 1: 3 − 4 = −1; at sin x = −1: 3 + 4 = 7. Time-saving method : Substitute the two extreme values; they give the maximum and minimum. Answer in 5 seconds.

Spot the difference

Propertysin xcos xtan x
Domainall Rall RR − {(2n + 1)π/2}
Range[−1, 1][−1, 1]all R
Period2π2ππ
Cuts the X-axis atx = nπx = (2n + 1)π/2x = nπ
Even / oddoddevenodd

Practice questions

Easy
  1. 1Write the range of cos x.
  2. 2What is the period of tan x?
  3. 3Find the period of cos 2x.
Medium
  1. 4Find the maximum and minimum of 2 sin x + 1.
  2. 5Find the period of sin(x/3).
  3. 6Find the period of tan 3x.
Challenging
  1. 7Find the period of sin 2x + cos 4x.
  2. 8Find the range of 7 − 3 cos x.
Show answers (try them yourself first)
  1. 1[−1, 1].
  2. 2π.
  3. 3π.
  4. 4Maximum 3, minimum −1.
  5. 56π.
  6. 6π/3.
  7. 7π (component periods are π and π/2).
  8. 8[4, 10].

Polar co-ordinates

From daily life

A ship's radar does not say "3 km east, 4 km north"; it says "5 km away, in the 53° direction". Giving a position by distance and direction is exactly the polar method. Building on classes 9–10 : In classes 9–10 a point was (x, y) and distance came from Pythagoras; now the same point is also written (r, θ).

Simple definition

Take a fixed point O (the pole) and a fixed ray from it (the polar axis, the positive X-axis). The polar co-ordinates of P are (r, θ): r = the distance OP, θ = the angle measured anticlockwise from the polar axis.

XYθP(r, θ)Orx = r cos θy = r sin θ

In the figure : Polar co-ordinates (r, θ) and Cartesian co-ordinates x = r cos θ, y = r sin θ of the point P.

Rule 1: Polar to Cartesian — x = r cos θ, y = r sin θ. Given (r, θ), x and y follow directly.

Rule 2: Cartesian to polar — r = √(x² + y²) and tan θ = y/x. Fix the quadrant of θ from the signs of (x, y); do not rely on tan⁻¹ alone.

How is the formula obtained? : In the figure drop a perpendicular PM from P to the X-axis. In the right triangle OMP the hypotenuse is OP = r, OM = x, PM = y. So cos θ = x/r and sin θ = y/r, i.e. x = r cos θ, y = r sin θ. Pythagoras gives r² = x² + y².

  • Easy : (8, 60°) → x = 8 cos 60° = 4, y = 8 sin 60° = 4√3; the Cartesian point is (4, 4√3).
  • Medium : (−3, 3) → r = √(9 + 9) = 3√2; tan θ = −1 and the point is in quadrant II, so θ = 135°; polar (3√2, 135°).
  • Exam-type : (−2√3, −2) → r = √(12 + 4) = 4; tan θ = 1/√3, reference angle 30° and quadrant III, so θ = 180° + 30° = 210°; polar (4, 210°).
Remember

r is always positive (a distance); θ is measured anticlockwise from the polar axis. For the origin (0, 0), r = 0 and θ is undefined. x² + y² = r².

Common mistake

Taking θ = tan⁻¹(y/x) as the final answer. For (−1, 1), tan⁻¹(−1) = −45°, but the point lies in quadrant II, so the true θ is 135°.

Shortcut

First find the quadrant of the point; then the reference angle α = tan⁻¹|y/x|; θ = α, 180° − α, 180° + α or 360° − α (quadrants I, II, III, IV). Memory clue : "Quadrant first, angle next."

CET trap

The options often show (2, 60°), (2, 120°), (2, 240°) with the same r. Check the quadrant from the signs and two options drop out at once.

CET-style example : What are the polar co-ordinates of the point (−1, √3)? (A) (2, 120°) (B) (2, 60°) (C) (2, 240°) (D) (√2, 120°)

Answer : (A). r = √(1 + 3) = 2; tan θ = −√3 and quadrant II, so θ = 120°. Time-saving method : x negative, y positive → quadrant II: 60° and 240° are out; r = 2 rules out (D). Answer (A).

Spot the difference

PointCartesian (x, y)Polar (r, θ)
Gives position bydistances from the axesdistance and direction from the pole
Origin is calledoriginpole
Entries for one pointa single pairθ + 360° gives the same point
Conversionr = √(x² + y²)x = r cos θ, y = r sin θ

Practice questions

Easy
  1. 1Write the Cartesian co-ordinates of (10, 0°).
  2. 2Find r for (6, 8).
  3. 3What are the polar co-ordinates of (0, −4) (0° ≤ θ < 360°)?
Medium
  1. 4Find the Cartesian co-ordinates of (12, 30°).
  2. 5Find the polar co-ordinates of (−5, 5).
  3. 6Write x² + y² = 36 in polar form.
Challenging
  1. 7Find the Cartesian co-ordinates of (2, 4π/3).
  2. 8Find the polar co-ordinates of (1, −1) (0° ≤ θ < 360°).
Show answers (try them yourself first)
  1. 1(10, 0).
  2. 2r = 10.
  3. 3(4, 270°).
  4. 4(6√3, 6).
  5. 5(5√2, 135°).
  6. 6r = 6.
  7. 7(−1, −√3).
  8. 8(√2, 315°).

Lesson 2: Trigonometry - I — quick revision

TopicFormula / value
Unit circlecos θ = x, sin θ = y, tan θ = y/x
From any pointsin θ = y/r, cos θ = x/r, r = √(x² + y²)
Negative anglesin(−θ) = −sin θ, cos(−θ) = cos θ, tan(−θ) = −tan θ
Basic identitiessin²θ + cos²θ = 1, 1 + tan²θ = sec²θ, 1 + cot²θ = cosec²θ
Periodsin kx, cos kx: 2π/k; tan kx: π/k (k > 0)
Polarx = r cos θ, y = r sin θ

🎯 Most important points for the exam

  • All positive in quadrant I; sin in II; tan in III; cos in IV (A-S-T-C).
  • Learn the values at 30°, 45°, 60°: sin = 1/2, 1/√2, √3/2.
  • For a negative angle only cos keeps its sign; divide a large angle by 360° and use the remainder.
  • The products (sec θ + tan θ)(sec θ − tan θ) and (cosec θ + cot θ)(cosec θ − cot θ) are 1.
  • The period comes by dividing (2π ÷ k); the period of |sin x| is π.
  • For the polar angle find the quadrant first, then use tan⁻¹.
Study mantra

Learn the identities, picture the wave, and quadrant first — then the angle.