Inverse Trigonometric Functions

Inverse trigonometric functions (arcsin, arccos, arctan and their partners) appear in the CUET Mathematics paper as value-evaluation and identity questions. The key is respecting the restricted principal-value ranges.

Principal value ranges (memorise)

• sin-inverse x: range [-pi/2, pi/2], domain [-1, 1].
• cos-inverse x: range [0, pi], domain [-1, 1].
• tan-inverse x: range (-pi/2, pi/2), domain all reals.
  The answer of an inverse-trig expression must lie in these ranges.

Core identities

• sin-inverse x + cos-inverse x = pi/2; tan-inverse x + cot-inverse x = pi/2; sec-inverse x + cosec-inverse x = pi/2.
• tan-inverse x + tan-inverse y = tan-inverse((x + y)/(1 - xy)), valid when xy < 1.
• sin-inverse(-x) = -sin-inverse x (odd); cos-inverse(-x) = pi - cos-inverse x.

Method

For "evaluate cos-inverse(cos(7pi/6))", do not just cancel - bring the angle into the principal range first. Here cos(7pi/6) = cos(5pi/6) whose value lies in [0, pi], giving 5pi/6.

Exam Tricks & Tips

• 🎯 Always check the principal-value range before finalising an inverse-trig value.
• 🎯 Use sin-inverse x + cos-inverse x = pi/2 to convert between the two instantly.
• 🎯 For the tan-inverse sum formula, confirm the condition xy < 1 before applying it.
• 🎯 When simplifying f-inverse(f(x)), reduce the inner angle to the principal range first.
• 🎯 Exploit odd/even symmetry: sin-inverse and tan-inverse are odd; cos-inverse(-x) = pi - cos-inverse x.
• ❌ Common mistake: writing cos-inverse(cos theta) = theta for every theta - it equals theta only when theta lies in [0, pi]; otherwise adjust into that range.

Expected exam pattern

Evaluate a principal value, simplify a composite inverse-trig expression, and apply the sum/complementary identities. Moderate.

Quick recap

Learn the three principal ranges, the pi/2 complementary pairs, and the tan-inverse sum formula (xy < 1). For f-inverse(f(x)), always reduce the angle into the principal range before cancelling.

Cover the answer, recall, then check. 11 cards on inverse trigonometric functions for CUET Mathematics.

Q1. State the principal-value range of sin-inverse x.
A1. [-pi/2, pi/2], with domain [-1, 1].

Q2. State the principal-value range of cos-inverse x.
A2. [0, pi], with domain [-1, 1].

Q3. State the range of tan-inverse x.
A3. (-pi/2, pi/2), domain all real numbers.

Q4. What does sin-inverse x + cos-inverse x equal?
A4. pi/2.

Q5. Give the tan-inverse addition formula.
A5. tan-inverse x + tan-inverse y = tan-inverse((x + y)/(1 - xy)), when xy < 1.

Q6. Evaluate cos-inverse(cos(7pi/6)).
A6. 5pi/6 (the value in [0, pi] with the same cosine).

Q7. Simplify sin-inverse(-x).
A7. -sin-inverse x (it is an odd function).

Q8. Simplify cos-inverse(-x).
A8. pi - cos-inverse x.

Q9. What does tan-inverse x + cot-inverse x equal?
A9. pi/2.

Q10. Is cos-inverse(cos theta) always equal to theta?
A10. No - only when theta lies in [0, pi]; otherwise reduce into that range.

Q11. What is the value of sin-inverse(1)?
A11. pi/2.

Inverse trig functions answer the question "which angle has this sine (or cosine, or tangent)?" The whole topic hinges on one subtlety — principal value ranges — which is exactly what CUET loves to test. Nail the ranges and the identities follow.

What this tests / core idea: evaluating inverse trigonometric functions within their principal-value branches, and applying standard identities to simplify expressions.

Deep explanation

Beginner — principal value ranges

Because sine, cosine and tangent repeat, we restrict each inverse to a single branch:

Function	Principal range
sin⁻¹x	[−π/2, π/2]
cos⁻¹x	[0, π]
tan⁻¹x	(−π/2, π/2)
So sin⁻¹(1/2) = π/6, and cos⁻¹(−1/2) = 2π/3 (not −π/3, which lies outside [0, π]).

Intermediate — key identities

• sin⁻¹x + cos⁻¹x = π/2; tan⁻¹x + cot⁻¹x = π/2.
• sin⁻¹(−x) = −sin⁻¹x (odd); cos⁻¹(−x) = π − cos⁻¹x.
• tan⁻¹x + tan⁻¹y = tan⁻¹((x + y)/(1 − xy)), valid when xy < 1.

Advanced — simplifying composite expressions

For expressions like sin(cos⁻¹x), set θ = cos⁻¹x so cos θ = x; then sin θ = √(1 − x²), giving sin(cos⁻¹x) = √(1 − x²). Substitution (let the inverse function equal an angle θ) turns messy composites into right-triangle relationships.

Worked example

Q. Evaluate cos⁻¹(−1/2).
We need the angle in [0, π] whose cosine is −1/2. Cosine is negative in the second quadrant; cos(2π/3) = −1/2, and 2π/3 lies in [0, π]. So cos⁻¹(−1/2) = 2π/3. (Not −π/3 — that is outside the principal range.) Answer: 2π/3.

CUET relevance

A guaranteed Mathematics topic: evaluate a principal value, or simplify using the sin⁻¹ + cos⁻¹ = π/2 identity. Short, formula-based questions (2–4 marks).

Speed tricks & shortcuts

• Memorise the three principal ranges — most errors are range errors.
• Complementary identity: sin⁻¹x + cos⁻¹x = π/2 solves many one-liners instantly.
• For composites, substitute the inverse = θ and draw the right triangle.

Key formulas

• Principal ranges: sin⁻¹x ∈ [−π/2,π/2], cos⁻¹x ∈ [0,π], tan⁻¹x ∈ (−π/2,π/2).
• sin⁻¹x + cos⁻¹x = π/2; tan⁻¹x + cot⁻¹x = π/2; sec⁻¹x + cosec⁻¹x = π/2.
• tan⁻¹x + tan⁻¹y = tan⁻¹((x+y)/(1−xy)), xy<1.
• 2tan⁻¹x = sin⁻¹(2x/(1+x²)) = tan⁻¹(2x/(1−x²)).
• sin⁻¹(−x) = −sin⁻¹x; cos⁻¹(−x) = π − cos⁻¹x.

Usage: keep results in principal ranges; adjust negatives with the (−x) identities.

Worked Example

Problem: Evaluate (a) sin⁻¹(1/2) + cos⁻¹(1/2), (b) tan⁻¹(1) + tan⁻¹(2) + tan⁻¹(3), giving principal values.

Solution:
Inverse trig functions return the PRINCIPAL value in a fixed range: sin⁻¹ ∈ [−π/2, π/2], cos⁻¹ ∈ [0, π], tan⁻¹ ∈ (−π/2, π/2).

(a) sin⁻¹(1/2) = the angle in [−π/2, π/2] whose sine is 1/2 = π/6.
cos⁻¹(1/2) = the angle in [0, π] whose cosine is 1/2 = π/3.
Sum = π/6 + π/3 = π/6 + 2π/6 = 3π/6 = π/2.
(This illustrates the identity sin⁻¹x + cos⁻¹x = π/2 for all x in [−1, 1].)

(b) tan⁻¹(1) = π/4 (since tan(π/4) = 1).
For tan⁻¹(2) + tan⁻¹(3), use the addition formula tan⁻¹a + tan⁻¹b = π + tan⁻¹((a+b)/(1−ab)) when ab > 1 (here ab = 6 > 1, so add π):
(a + b)/(1 − ab) = (2 + 3)/(1 − 6) = 5/(−5) = −1.
So tan⁻¹(2) + tan⁻¹(3) = π + tan⁻¹(−1) = π − π/4 = 3π/4.
Now total = tan⁻¹(1) + [tan⁻¹(2) + tan⁻¹(3)] = π/4 + 3π/4 = 4π/4 = π.
(A well-known result: tan⁻¹1 + tan⁻¹2 + tan⁻¹3 = π.)

Answer: (a) sin⁻¹(1/2) + cos⁻¹(1/2) = π/6 + π/3 = π/2 (the identity sin⁻¹x + cos⁻¹x = π/2). (b) tan⁻¹1 + tan⁻¹2 + tan⁻¹3 = π/4 + 3π/4 = π.