ilovepdf_merged (6).pdf). Content covers sections 24.1–24.x.Inverse trig functions undo sine, cosine, tangent on restricted domains so each is one–one and onto onto its principal range.
Formulas for sin⁻¹ x ± sin⁻¹ y, tan⁻¹ x ± tan⁻¹ y need domain checks (sign of products, etc.).
Most exam-important points from this chapter:
Start every answer with the key definition or standard form from Inverse Trigonometric Functions.
Copy the formula, then substitute values; never jump to the number alone.
Check domain, quadrant, non-zero denominators, and applicability of the theorem.
Memorise and apply: 1, sin 2
Memorise and apply: 1 , sin 4
English questions from NIOS Mathematics (311) public / sample papers (Sample 2024, Apr 2024, Oct 2024, Apr 2025; board-style where scanned papers had no extractable text). Mapped exclusively to this chapter — no cross-chapter duplicates. Use Model Answer for the marking key and Explanation for working.
4 question(s) · Sources: Apr 2024, Board-style, Board-style identity
PYQ1. The principal value of cos⁻¹(−1/2) is:
Model Answer
1) 2π/3
Explanation
1) cos(2π/3)=−1/2 and 2π/3 ∈ [0,π], the principal range of cos⁻¹.
PYQ2. Find the principal value of sin⁻¹(1/2).
Model Answer
1) π/6
Explanation
1) sin(π/6)=1/2 and π/6 ∈ [−π/2, π/2].
PYQ3. Find tan⁻¹(1) (principal value).
Model Answer
1) π/4
Explanation
1) tan(π/4)=1 and π/4 ∈ (−π/2, π/2).
PYQ4. If sin⁻¹ x = π/6, find cos⁻¹ x.
Model Answer
1) π/3
Explanation
1) sin⁻¹x + cos⁻¹x = π/2 for x∈[−1,1]
2) cos⁻¹x = π/2 − π/6 = π/3.
10 English exam-style problems for this chapter (NIOS Mathematics 311). Each question states the formula, gives full step-by-step working with the final answer, plus formal/easy explanations. English only.
Find the principal value of sin⁻¹(1/2).
1) sin(π/6)=1/2 and π/6 ∈ [−π/2,π/2].
2) Hence sin⁻¹(1/2)=π/6.
Answer: π/6
Principal value is the unique angle in [−π/2,π/2] with that sine.
30° is π/6 radians in the principal range.
Not 5π/6 (outside principal range for sin⁻¹).
Always use radians unless asked otherwise.
Find cos⁻¹(−1/2) (principal value).
1) cos(2π/3)=−1/2 and 2π/3 ∈ [0,π].
2) Answer 2π/3.
Answer: 2π/3
Principal cos⁻¹ lands in [0,π].
120° = 2π/3 is the principal value.
Not −π/3 for cos⁻¹.
Check range first.
Evaluate tan⁻¹(1).
1) tan(π/4)=1 and π/4 is in (−π/2,π/2).
2) tan⁻¹1=π/4.
Answer: π/4
Principal arctangent of 1 is π/4.
45° as radians π/4.
tan⁻¹ domain is all reals.
Do not give 5π/4.
If sin⁻¹x = π/6, find cos⁻¹x.
1) cos⁻¹x = π/2 − sin⁻¹x = π/2 − π/6 = π/3.
Answer: π/3
Using the cofunction identity for inverse sine and cosine.
They add to a right angle (π/2).
Valid for all x in [−1,1].
Identity is not sin⁻¹x+cos⁻¹x=π.
Find tan⁻¹(1/2)+tan⁻¹(1/3) given both acute and product <1.
1) tan(A+B)=(tanA+tanB)/(1−tanA tanB)=(1/2+1/3)/(1−1/6)=(5/6)/(5/6)=1.
2) A+B=π/4 since A,B>0 and A+B<π/2.
3) Sum = π/4.
Answer: π/4
Tangent-addition formula with principal values yields π/4.
Formula for tan(A+B) gives 1, so angle π/4.
If xy>1 signs/branches need care.
Check 1−xy≠0.
Why is sin⁻¹(√3/2) equal to π/3 and not 2π/3?
1) sin(2π/3)=√3/2 also, but 2π/3 ∉ [−π/2,π/2].
2) Principal value must lie in [−π/2,π/2], hence π/3.
Answer: π/3 (2π/3 not in principal range)
Definition forces the unique principal value in the closed interval [−π/2,π/2].
Many angles share a sine; inverse picks the special one.
Always state the range of the inverse function.
Listing any preimage is incorrect for sin⁻¹.
Find the principal value of sin⁻¹(√3/2).
1) sin(π/3)=√3/2 and π/3 ∈ [−π/2,π/2].
2) sin⁻¹(√3/2)=π/3.
Answer: π/3
Principal arcsine uses the standard range.
60 degrees in radians.
Not 2π/3 (outside range).
Check each algebraic step carefully.
Find cos⁻¹(−1/2).
1) cos(2π/3)=−1/2 and 2π/3 ∈ [0,π].
2) cos⁻¹(−1/2)=2π/3.
Answer: 2π/3
Principal arccosine range is [0,π].
120 degrees.
Not −π/3.
Check each algebraic step carefully.
Evaluate sin⁻¹(3/5) + cos⁻¹(3/5).
1) By identity on [−1,1], the sum is π/2.
Answer: π/2
Standard complementary identity for principal values.
No calculator needed.
x=3/5 is in the domain.
Check each algebraic step carefully.
Find tan⁻¹(1).
1) tan(π/4)=1 and π/4 is in the principal range.
2) tan⁻¹1=π/4.
Answer: π/4
Principal arctangent of 1.
45 degrees.
Not 5π/4.
Check each algebraic step carefully.