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Maths · Class 11 · Chapter 3

Trigonometric Functions

Class 10 trigonometry worked only for acute angles in a right triangle. Here the ratios become functions of any real number, defined through the unit circle, and you pick up the identities that the rest of JEE calculus and coordinate geometry lean on.

In this chapter: degree and radian measure, arc length, trigonometric functions through the unit circle, signs in the four quadrants, domain, range, periodicity and graphs, allied angles, sum and difference formulas, double and triple angle formulas, sum-to-product and product-to-sum formulas, general solutions.

Measuring angles

An angle is the amount of rotation of a ray about its initial point. Anticlockwise rotation gives a positive angle and clockwise rotation a negative one.

  • Degree: one complete revolution is 360°; 1° = 60′ (minutes) and 1′ = 60″ (seconds).
  • Radian: the angle subtended at the centre of a circle by an arc whose length equals the radius. One complete revolution is 2π radians.
π radian = 180°1 radian = 180°/π ≈ 57.3°; 1° = π/180 radian ≈ 0.0175 radianl = rθarc length, with θ in radians

To convert degrees to radians multiply by π/180; to go back multiply by 180/π. When no unit is written, the angle is in radians, so sin 1 means the sine of 1 radian, not 1°. For example, an arc of a circle of radius 21 cm subtending 60° at the centre has length 21 × π/3 = 7π = 22 cm (taking π = 22/7).

Trigonometric functions through the unit circle

Take the circle of radius 1 centred at the origin. Rotate the radius OA, starting from A(1, 0), through an angle x to reach a point P. The coordinates of P are defined to be (cos x, sin x). Since P is on the unit circle, for every real x:

cos²x + sin²x = 11 + tan²x = sec²x1 + cot²x = cosec²x

After one full turn P comes back to the same place, so sin(2nπ + x) = sin x and cos(2nπ + x) = cos x for every integer n. Also sin x = 0 exactly when x = nπ, and cos x = 0 exactly when x = (2n + 1)π/2. The other four are defined from sine and cosine: tan x = sin x/cos x, cot x = cos x/sin x, sec x = 1/cos x, cosec x = 1/sin x, each wherever its denominator is not zero.

Unit circle and the signs of trigonometric ratios in the four quadrantswww.iitmedicoguide.comxP(cos x, sin x)1cos xsin xA(1, 0)B(0, 1)C(−1, 0)D(0, −1)OQuadrant Iall positiveQuadrant IIsin, cosec positiveQuadrant IIItan, cot positiveQuadrant IVcos, sec positivewww.iitmedicoguide.com
The x-coordinate of P is cos x and the y-coordinate is sin x. Reading the signs of these coordinates quadrant by quadrant gives the familiar rule: all positive in I, sine in II, tangent in III, cosine in IV.

Values you should know without thinking

x0π/6π/4π/3π/2π3π/22π
sin x01/21/√2√3/210−10
cos x1√3/21/√21/20−101
tan x01/√31√3not defined0not defined0

Domain, range and period

FunctionDomainRangePeriod
sin xR[−1, 1]2π
cos xR[−1, 1]2π
tan xR − {(2n + 1)π/2 : n ∈ Z}Rπ
cot xR − {nπ : n ∈ Z}Rπ
sec xR − {(2n + 1)π/2 : n ∈ Z}(−∞, −1] ∪ [1, ∞)2π
cosec xR − {nπ : n ∈ Z}(−∞, −1] ∪ [1, ∞)2π

Sine is an odd function and cosine is even: sin(−x) = −sin x and cos(−x) = cos x. Tangent, cotangent and cosecant are odd; secant is even.

Graphs of sin x and cos x over one periodwww.iitmedicoguide.comxyπ/2π3π/22π1−1Oy = sin xy = cos xwww.iitmedicoguide.com
Both graphs repeat every 2π and stay between −1 and 1. The cosine curve is the sine curve shifted left by π/2, because cos x = sin(x + π/2).

Allied angles

For angles of the form nπ/2 ± x: if n is odd, sin changes to cos (and tan to cot); if n is even, the function stays the same. The sign in front is the sign of the original function in the quadrant where the angle falls, treating x as acute.

Anglesincos
π/2 − xcos xsin x
π/2 + xcos x−sin x
π − xsin x−cos x
π + x−sin x−cos x
2π − x−sin xcos x

Sum and difference formulas

cos(x + y) = cos x cos y − sin x sin ycos(x − y) = cos x cos y + sin x sin ysin(x + y) = sin x cos y + cos x sin ysin(x − y) = sin x cos y − cos x sin ytan(x + y) = (tan x + tan y)/(1 − tan x tan y)tan(x − y) = (tan x − tan y)/(1 + tan x tan y)cot(x + y) = (cot x cot y − 1)/(cot y + cot x)cot(x − y) = (cot x cot y + 1)/(cot y − cot x)

The tangent formulas need none of x, y, x ± y to be an odd multiple of π/2; the cotangent formulas need none of them to be a multiple of π.

Worked example: Find sin 15° and tan 15°.
Solution: sin 15° = sin(45° − 30°) = (1/√2)(√3/2) − (1/√2)(1/2) = (√3 − 1)/(2√2). For the tangent, tan 15° = (tan 45° − tan 30°)/(1 + tan 45° tan 30°) = (1 − 1/√3)/(1 + 1/√3) = (√3 − 1)/(√3 + 1). Multiplying top and bottom by √3 − 1 gives (4 − 2√3)/2 = 2 − √3.

Double and triple angle formulas

cos 2x = cos²x − sin²x = 2cos²x − 1 = 1 − 2sin²x = (1 − tan²x)/(1 + tan²x)sin 2x = 2 sin x cos x = 2 tan x/(1 + tan²x)tan 2x = 2 tan x/(1 − tan²x)sin 3x = 3 sin x − 4 sin³xcos 3x = 4 cos³x − 3 cos xtan 3x = (3 tan x − tan³x)/(1 − 3 tan²x)

Rearranging cos 2x gives the power-reducing forms 2cos²x = 1 + cos 2x and 2sin²x = 1 − cos 2x, which you will use again and again in integration.

Worked example: sin x = 3/5 and x lies in the second quadrant. Find cos x, tan x and sin 2x.
Solution: cos²x = 1 − 9/25 = 16/25, so cos x = ±4/5. Cosine is negative in the second quadrant, so cos x = −4/5. Then tan x = (3/5)/(−4/5) = −3/4 and sin 2x = 2(3/5)(−4/5) = −24/25.

Sum-to-product and product-to-sum

cos x + cos y = 2 cos((x + y)/2) cos((x − y)/2)cos x − cos y = −2 sin((x + y)/2) sin((x − y)/2)sin x + sin y = 2 sin((x + y)/2) cos((x − y)/2)sin x − sin y = 2 cos((x + y)/2) sin((x − y)/2)
2 cos x cos y = cos(x + y) + cos(x − y)−2 sin x sin y = cos(x + y) − cos(x − y)2 sin x cos y = sin(x + y) + sin(x − y)2 cos x sin y = sin(x + y) − sin(x − y)

A typical use: (sin 5x + sin 3x)/(cos 5x + cos 3x) = (2 sin 4x cos x)/(2 cos 4x cos x) = tan 4x. Whenever you see a sum of two sines or cosines in a fraction, try converting to products and look for a common factor.

One more result used constantly in JEE: a sin x + b cos x always lies between −√(a² + b²) and √(a² + b²), and both extremes are attained. So 3 sin x + 4 cos x has maximum 5 and minimum −5.

General solutions

The rationalised NCERT textbook has dropped the section on trigonometric equations, but general solutions are part of the MHT‑CET syllabus (through the Maharashtra board Class 12 chapter on trigonometric functions), are listed in the JEE Advanced syllabus, and keep turning up inside JEE problems. For n ∈ Z:

sin x = 0 ⇒ x = nπ; cos x = 0 ⇒ x = (2n + 1)π/2; tan x = 0 ⇒ x = nπsin x = sin y ⇒ x = nπ + (−1)nycos x = cos y ⇒ x = 2nπ ± ytan x = tan y ⇒ x = nπ + y

For example, sin x = √3/2 = sin(π/3) gives x = nπ + (−1)nπ/3; the solutions in [0, 2π) are π/3 and 2π/3.

Common mistakes: (1) Taking the positive square root out of habit: after finding cos²x, fix the sign from the quadrant. (2) Treating sin 2 as sin 2°; without a degree sign the angle is in radians. (3) Writing tan(x + y) with a plus sign in the denominator; it is 1 − tan x tan y. (4) Changing sin to cos for π ± x; the swap happens only for odd multiples of π/2. (5) Forgetting the minus sign in cos x − cos y = −2 sin((x + y)/2) sin((x − y)/2). (6) Giving the range of sec x as [−1, 1], which is exactly the set it avoids (apart from ±1).

Exam focus

  • Radian and degree conversion, and arc length l = rθ.
  • Signs of the ratios in each quadrant and values at allied angles, including large angles such as 765° or −1710°.
  • Sum, difference, double and triple angle formulas, used to evaluate expressions like tan 75° or cos²15° − sin²15°.
  • Sum-to-product conversions to simplify and prove identities.
  • Maximum and minimum of a sin x + b cos x; domain, range and period of the six functions.
  • General solutions of simple equations such as sin x = k or cos 2x = cos x.

Practice questions

40° 20′ expressed in radians is:

  1. 121π/180
  2. 121π/540
  3. 2π/9
  4. 11π/27
Show answer
B. 40° 20′ = (121/3)°, and (121/3) × π/180 = 121π/540.

The value of sin 765° is:

  1. √3/2
  2. −1/√2
  3. 1/√2
  4. 1/2
Show answer
C. 765° = 2 × 360° + 45°, so sin 765° = sin 45° = 1/√2.

The value of cos(−1710°) is:

  1. 0
  2. 1
  3. −1
  4. 1/2
Show answer
A. cos is even, and 1710° = 4 × 360° + 270°, so the value is cos 270° = 0.

cos²15° − sin²15° equals:

  1. 1/2
  2. √3/2
  3. 1
  4. 0
Show answer
B. It is cos 30° = √3/2.

The maximum value of 3 sin x + 4 cos x is:

  1. 7
  2. 1
  3. 12
  4. 5
Show answer
D. √(3² + 4²) = 5.

tan 75° equals:

  1. 2 − √3
  2. √3 + 1
  3. 2 + √3
  4. (√3 + 1)/2
Show answer
C. tan(45° + 30°) = (1 + 1/√3)/(1 − 1/√3) = (√3 + 1)/(√3 − 1) = 2 + √3.

If tan x = −4/3 and x lies in the second quadrant, then sin x is:

  1. −4/5
  2. 4/5
  3. −3/5
  4. 3/5
Show answer
B. sec²x = 1 + 16/9 = 25/9, so cos x = −3/5 in quadrant II and sin x = tan x cos x = 4/5.

cos 2x is equal to:

  1. 2 tan x/(1 + tan²x)
  2. (1 + tan²x)/(1 − tan²x)
  3. 2 sin²x − 1
  4. (1 − tan²x)/(1 + tan²x)
Show answer
D. Divide cos²x − sin²x by cos²x + sin²x = 1 and then divide top and bottom by cos²x. Option A is sin 2x, and C is −cos 2x.
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