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State Board · Class 12 · Mathematics Part 1

Lesson 3: Trigonometric Functions

4 concepts · 32 practice questions with answers · Free to read

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Trigonometric equations — principal and general solutions

From daily life

A clock's minute hand returns to the same position again and again, so sin θ = ½ has not one answer but a repeating family of angles. Building on class 11 : trigonometric ratios of angles were studied in class 11.

Simple definition

A trigonometric equation contains functions of θ. A principal solution lies in 0 ≤ θ < 2π. The general solution is a formula giving all solutions, with n ∈ Z.

Rule 1: Sine — sin θ = sin α ⇒ θ = nπ + (−1)ⁿ α.

Rule 2: Cosine — cos θ = cos α ⇒ θ = 2nπ ± α.

Rule 3: Tangent — tan θ = tan α ⇒ θ = nπ + α.

Rule 4: Squares — sin²θ = sin²α ⇒ θ = nπ ± α; cos²θ = cos²α ⇒ θ = nπ ± α; tan²θ = tan²α ⇒ θ = nπ ± α.

How is the formula obtained? : sin θ = √3/2 = sin(π/3) ⇒ θ = nπ + (−1)ⁿ π/3; n = 0 gives π/3, n = 1 gives 2π/3, n = 2 gives 7π/3 ... all have sin = √3/2.

  • Easy : tan θ = 1 ⇒ θ = nπ + π/4.
  • Medium : cos θ = ½ ⇒ θ = 2nπ ± π/3; principal solutions π/3 and 5π/3.
  • Exam-type : sin²θ = ¼ ⇒ θ = nπ ± π/6.
Remember

The general solution has n ∈ Z; principal solutions lie in [0, 2π).

Common mistake

Giving only π/6 for sin θ = ½; do not forget 5π/6.

Shortcut

sin and tan formulas start with nπ, cos with 2nπ. Memory clue : "Sin-tan: nπ; cos: 2nπ".

CET trap

In cos θ = cos α the ± gives two families; (−1)ⁿ belongs only to sin.

CET-style example : The principal solutions of sin θ = −½ are: (A) π/6, 5π/6 (B) 7π/6, 11π/6 (C) π/3, 2π/3 (D) 4π/3, 5π/3

Answer : (B). A negative sine means quadrants III and IV. Time-saving method : Fix the quadrants and eliminate options.

Spot the difference

सामान्य उत्तरGeneral solution
nπ + (−1)ⁿ αnπ + (−1)ⁿ α
2nπ ± α2nπ ± α
nπ + αnπ + α

Practice questions

Easy
  1. 1General solution of sin θ = 0?
  2. 2General solution of cos θ = 0?
  3. 3General solution of tan θ = 1?
Medium
  1. 4General solution of sin θ = √3/2?
  2. 5General solution of cos θ = ½?
  3. 6General solution of sin²θ = ¼?
Challenging
  1. 7General solution of cos θ = −½?
  2. 8General solution of tan θ = −√3?
Show answers (try them yourself first)
  1. 1θ = nπ.
  2. 2θ = (2n + 1)π/2.
  3. 3θ = nπ + π/4.
  4. 4θ = nπ + (−1)ⁿ π/3.
  5. 5θ = 2nπ ± π/3.
  6. 6θ = nπ ± π/6.
  7. 7θ = 2nπ ± 2π/3.
  8. 8tan(−π/3) = −√3 ⇒ θ = nπ − π/3.

Methods for solving equations

From daily life

To open a lock you try keys until one fits; in a trigonometric equation the right rewriting is the key.

Simple definition

Bring the equation to one function (quadratic, factorise) or write a sin θ + b cos θ = c as R sin(θ + φ) with R = √(a² + b²). Then use the basic formulas.

Rule 1: Quadratic / factors — Use sin²θ = 1 − cos²θ to make a quadratic in one function and factorise.

Rule 2: a sin θ + b cos θ = c — Divide by R: sin(θ + φ) = c/R; solve. If |c| > R there is no solution.

How is the formula obtained? : sin θ + cos θ = 1: R = √2, so sin(θ + π/4) = 1/√2 = sin(π/4) ⇒ θ + π/4 = nπ + (−1)ⁿ π/4 ⇒ θ = 2mπ or 2mπ + π/2.

  • Easy : 2 sin²θ − 3 sin θ + 1 = 0 ⇒ (2 sin θ − 1)(sin θ − 1) = 0 ⇒ sin θ = ½ or 1.
  • Medium : 2 cos²θ − cos θ − 1 = 0 ⇒ (2 cos θ + 1)(cos θ − 1) = 0 ⇒ θ = 2nπ ± 2π/3 or 2nπ.
  • Exam-type : sin θ + cos θ = 1 ⇒ θ = 2mπ or 2mπ + π/2 (method above).
Remember

After factorising, set each factor to 0 and write separate solutions.

Common mistake

Accepting an impossible answer such as sin θ = 2 (|sin| ≤ 1).

Shortcut

Reject any root of the quadratic that lies outside −1 to 1. Memory clue : "One function, one equation, then factors".

CET trap

Do not divide both sides by sin θ; solutions are lost (factorise instead).

CET-style example : The principal solution of sin θ + cos θ = √2 is: (A) π/4 (B) π/2 (C) π/3 (D) π/6

Answer : (A). R = √2, so sin(θ + π/4) = 1 ⇒ θ = π/4. Time-saving method : When R = c there is just one solution.

Spot the difference

निष्कर्षConclusion
दोन उत्तर कुटुंबेtwo families of solutions
एक कुटुंबone family
उत्तर नाहीno solution

Practice questions

Easy
  1. 1Solve 2 sin θ − 1 = 0 (principal).
  2. 2Does sin θ = 2 have a solution?
  3. 3General solution of 2 cos θ = 1?
Medium
  1. 4Principal solutions of 2 sin²θ − 3 sin θ + 1 = 0?
  2. 5General solution of 2 cos²θ − cos θ − 1 = 0?
  3. 6Find R for sin θ + cos θ = 1.
Challenging
  1. 7General solution of sin θ + cos θ = 1?
  2. 8Does √3 sin θ + cos θ = 3 have a solution?
Show answers (try them yourself first)
  1. 1θ = π/6, 5π/6.
  2. 2No, since |sin θ| ≤ 1.
  3. 3θ = 2nπ ± π/3.
  4. 4sin θ = ½: π/6, 5π/6; sin θ = 1: π/2.
  5. 5θ = 2nπ or 2nπ ± 2π/3.
  6. 6R = √(1 + 1) = √2.
  7. 7θ = 2mπ or 2mπ + π/2.
  8. 8No, since R = 2 < 3.

Solution of a triangle — sine, cosine and projection rules

From daily life

A surveyor finds the width of a lake without crossing it by measuring one distance on the bank and two angles — that is the sine rule.

Simple definition

In △ABC the sides a, b, c are opposite to angles A, B, C. Solving the triangle means finding the remaining sides and angles. 2s = a + b + c.

Rule 1: Sine rule — a/sin A = b/sin B = c/sin C = 2R (R = circumradius).

Rule 2: Cosine rule — a² = b² + c² − 2bc cos A; cos A = (b² + c² − a²)/(2bc).

Rule 3: Projection rule — a = b cos C + c cos B (and similarly for b, c).

Rule 4: Area — ½ ab sin C; Heron: √(s(s − a)(s − b)(s − c)).

ABCcbaA

In the figure : Side a is opposite angle A.

How is the formula obtained? : a = 7, b = 8, c = 9: cos A = (64 + 81 − 49)/144 = 2/3; s = 12; area = √(12·5·4·3) = 12√5 ≈ 26.83.

  • Easy : A = 30°, B = 45°, a = 10: b = a sin B / sin A = 10·(√2/2)/(½) = 10√2 ≈ 14.14.
  • Medium : a = 5, b = 7, C = 60°: c² = 25 + 49 − 35 = 39, so c = √39 ≈ 6.24.
  • Exam-type : Area of the same triangle = ½·5·7·sin 60° = 35√3/4 ≈ 15.16.
Remember

First spot the pair of an angle and its opposite side.

Common mistake

Leaving the calculator in radian mode while the angle is in degrees.

Shortcut

Two sides + included angle ⇒ cosine; two angles + one side ⇒ sine. Memory clue : "Included angle = cosine; a matching pair = sine".

CET trap

The sine rule can give two triangles (ambiguous case) when the given angle is opposite the shorter given side.

CET-style example : If a = 2, b = 3, C = 90°, then c² = ? (A) 5 (B) 13 (C) 1 (D) 25

Answer : (B). cos 90° = 0, so c² = 4 + 9. Time-saving method : At 90° it is Pythagoras.

Spot the difference

नियमRule
cosinecosine
cosinecosine
sinesine

Practice questions

Easy
  1. 1If A = 90°, what is a² by the cosine rule?
  2. 2For a = 6, A = 30° find 2R.
  3. 3Sides 3, 4, 5: find s.
Medium
  1. 4Area of the triangle with sides 3, 4, 5?
  2. 5For a = 5, b = 7, C = 60° find c.
  3. 6For A = 30°, B = 45°, a = 10 find b.
Challenging
  1. 7Sides 7, 8, 9: find cos A.
  2. 8Sides 7, 8, 9: value of b cos C + c cos B by the projection rule?
Show answers (try them yourself first)
  1. 1b² + c².
  2. 26/(½) = 12.
  3. 36.
  4. 4√(6·3·2·1) = 6.
  5. 5√39 ≈ 6.24.
  6. 610√2 ≈ 14.14.
  7. 72/3.
  8. 8= a = 7.

Inverse trigonometric functions

From daily life

The sin⁻¹ key on a calculator answers "which angle has this sine?". Many angles share one sine, so only the angle in a fixed range (the principal value) is returned.

Simple definition

sin⁻¹: [−1, 1] → [−π/2, π/2]; cos⁻¹: [−1, 1] → [0, π]; tan⁻¹: R → (−π/2, π/2); cot⁻¹: R → (0, π). These ranges give the principal values.

Rule 1: Negative arguments — sin⁻¹(−x) = −sin⁻¹x; tan⁻¹(−x) = −tan⁻¹x; cos⁻¹(−x) = π − cos⁻¹x.

Rule 2: Sums — sin⁻¹x + cos⁻¹x = π/2; tan⁻¹x + tan⁻¹y = tan⁻¹[(x + y)/(1 − xy)] (xy < 1).

Rule 3: Double — 2 tan⁻¹x = tan⁻¹[2x/(1 − x²)] (|x| < 1).

yx1−1π/2−π/2

In the figure : The graph of sin⁻¹ stays between −π/2 and π/2.

How is the formula obtained? : tan⁻¹(½) + tan⁻¹(⅓): x = ½, y = ⅓, xy = 1/6 < 1, so (5/6)/(1 − 1/6) = 1 and tan⁻¹1 = π/4.

  • Easy : sin⁻¹(½) = π/6; cos⁻¹(½) = π/3.
  • Medium : cos⁻¹(−½) = π − π/3 = 2π/3; tan⁻¹(−1) = −π/4.
  • Exam-type : sin⁻¹(sin 5π/6) = π/6 (5π/6 is outside the range; sin 5π/6 = ½).
Remember

sin⁻¹(sin x) = x only for x ∈ [−π/2, π/2].

Common mistake

Taking sin⁻¹x as 1/sin x; that is cosec x, not sin⁻¹.

Shortcut

Use π − x or −x to bring the value into the range. Memory clue : "sin and tan: −π/2 to π/2; cos and cot: 0 to π".

CET trap

In tan⁻¹x + tan⁻¹y with xy > 1, π must be added to the answer.

CET-style example : cos⁻¹(−½) = ? (A) π/3 (B) 2π/3 (C) −π/3 (D) 5π/6

Answer : (B). Time-saving method : cos⁻¹ answers lie between 0 and π.

Spot the difference

परिसरRange
[−π/2, π/2][−π/2, π/2]
[0, π][0, π]
(−π/2, π/2)(−π/2, π/2)

Practice questions

Easy
  1. 1Find sin⁻¹(1).
  2. 2Find cos⁻¹(0).
  3. 3Find tan⁻¹(1).
Medium
  1. 4Find cos⁻¹(−½).
  2. 5Find tan⁻¹(−1).
  3. 6Find sin⁻¹(3/5) + cos⁻¹(3/5).
Challenging
  1. 7Find tan⁻¹(½) + tan⁻¹(⅓).
  2. 8Show that 2 tan⁻¹(½) = tan⁻¹(4/3).
Show answers (try them yourself first)
  1. 1π/2.
  2. 2π/2.
  3. 3π/4.
  4. 42π/3.
  5. 5−π/4.
  6. 6π/2.
  7. 7π/4.
  8. 82·½/(1 − ¼) = 1/(¾) = 4/3, so tan⁻¹(4/3).