Inverse Trigonometric Functions

In this chapter: why the domain must be restricted, principal value branches of sin⁻¹, cos⁻¹, tan⁻¹, cot⁻¹, sec⁻¹ and cosec⁻¹, their graphs, finding principal values, and the standard identities used to simplify expressions.

Why a principal branch is needed

From Chapter 1, only a bijection has an inverse. The function sin : R → R takes the value 1/2 at π/6, 5π/6, 13π/6 and infinitely many other points, so it is neither one-one nor onto. If we restrict the domain to [−π/2, π/2] and the codomain to [−1, 1], sine becomes a bijection, and its inverse sin⁻¹ : [−1, 1] → [−π/2, π/2] is well defined. Other intervals such as [π/2, 3π/2] would also work, but the one containing the values near zero is chosen as the principal value branch. Unless a question says otherwise, sin⁻¹x always means the value from this branch.

A notation warning: sin⁻¹x is the inverse function (also written arcsin x). It is not (sin x)⁻¹ = 1/sin x.

Domains and principal value ranges

A simple way to remember it: the "sine family" (sin, cosec, tan) takes values between −π/2 and π/2, so negative inputs give negative angles. The "cosine family" (cos, sec, cot) takes values between 0 and π, so negative inputs give obtuse angles. The points removed from cosec⁻¹ and sec⁻¹ are exactly where cosec and sec are undefined, and the open brackets for tan⁻¹ and cot⁻¹ come from the asymptotes.

Finding principal values

Ask "which angle in the principal range has this trig value?" and nothing else. For a negative input to a cosine-family function, find the acute angle first and subtract from π.

Worked example: Find the value of tan⁻¹(1) + cos⁻¹(−1/2) + sin⁻¹(−1/2).
Solution: tan⁻¹(1) = π/4. For cos⁻¹(−1/2), the acute angle with cosine 1/2 is π/3, so the answer in [0, π] is π − π/3 = 2π/3. For sin⁻¹(−1/2), the angle in [−π/2, π/2] is −π/6. Sum = π/4 + 2π/3 − π/6 = (3π + 8π − 2π)/12 = 3π/4.

Standard identities

1. Function of its inverse, and the reverse

sin(sin⁻¹x) = x for every x in [−1, 1]. The other way round needs care: sin⁻¹(sin x) = x only when x lies in [−π/2, π/2]. Outside that interval, first rewrite sin x as the sine of an angle inside the principal range. The same holds for the other five functions with their own ranges.

Worked example: Evaluate sin⁻¹(sin 3π/5).
Solution: 3π/5 = 108° is outside [−π/2, π/2], so the answer is not 3π/5. Since sin(π − θ) = sin θ, sin(3π/5) = sin(2π/5), and 2π/5 = 72° lies in the principal range. So the value is 2π/5.

2. Reciprocal relations

3. Negative arguments

The first line is the sine family (odd functions); the second is the cosine family. This matches the graphs: sin⁻¹ is symmetric about the origin, while cos⁻¹ is symmetric about the point (0, π/2).

4. Complementary pairs

5. Sum and difference of tan⁻¹

The π correction is the one students skip. tan⁻¹2 + tan⁻¹3 is the sum of two angles each greater than π/4, so it must exceed π/2, yet the formula without π gives tan⁻¹(−1) = −π/4. The correct value is π − π/4 = 3π/4.

6. Double angle forms

Similarly, 2 sin⁻¹x = sin⁻¹(2x√(1 − x²)) for |x| ≤ 1/√2, and 3 sin⁻¹x = sin⁻¹(3x − 4x³) for |x| ≤ 1/2. All of these come from putting x = tan θ or x = sin θ and using a double or triple angle formula, which is also the method for "write in simplest form" questions.

Worked example: Write tan⁻¹cos x1 − sin x, −3π/2 < x < π/2, in simplest form.
Solution: Use half angles: cos x = cos²(x/2) − sin²(x/2) and 1 − sin x = (cos(x/2) − sin(x/2))². The fraction becomes (cos(x/2) + sin(x/2))/(cos(x/2) − sin(x/2)) = (1 + tan(x/2))/(1 − tan(x/2)) = tan(π/4 + x/2). For the given x, π/4 + x/2 lies between −π/2 and π/2, which is inside the principal range of tan⁻¹. So the expression equals π/4 + x/2.

That last check (is the angle inside the principal range?) is what separates a correct answer from a nearly correct one. Always write it down.

Common mistakes: (1) Giving cos⁻¹(−1/2) as −π/3. Negative angles are never in the range of cos⁻¹, sec⁻¹ or cot⁻¹. (2) Writing sin⁻¹(sin x) = x without checking that x lies in [−π/2, π/2]. (3) Forgetting the π term in tan⁻¹x + tan⁻¹y when x, y > 0 and xy > 1. (4) Treating sin⁻¹x as 1/sin x. (5) Allowing |x| > 1 inside sin⁻¹ or cos⁻¹; when a question asks for a domain, both conditions −1 ≤ (argument) ≤ 1 must be solved.

JEE and MHT‑CET focus

• The six principal value ranges, including which end points are open and which values are excluded.
• Principal values of standard arguments such as ±1/2, ±√3/2, ±1/√2, ±1, ±√3, ±2.
• sin⁻¹(sin x), cos⁻¹(cos x) and tan⁻¹(tan x) for x outside the principal range.
• The complementary identities and the tan⁻¹ sum formula, used to simplify sums of several terms.
• Substitutions x = tan θ, x = sin θ, x = cos θ to reduce an expression to simplest form.
• Domains of composite expressions like sin⁻¹(2x − 1) or cos⁻¹(x² − 3).

Practice questions