Inverse Trigonometric Functions

A function f: A → B has an inverse f^-1 only when f is bijective — that is, one-one (injective) and onto (surjective). Consider y = sin x on all of ℝ. Because sine repeats every 2π and is symmetric about x = π/2, infinitely many values of x give the same y (for instance sin(π/6) = sin(5π/6) = 1/2). The same many-one behaviour holds for cos x, tan x and the other trigonometric ratios. So none of them is invertible on its full domain.

The fix is to restrict the domain of each trigonometric function to an interval on which it becomes both one-one and onto its range. Such a restricted domain is called a principal value branch. On that branch alone, the function is bijective, and its inverse is well defined. This restricted inverse is written sin^-1x, cos^-1x, tan^-1x, cot^-1x, sec^-1x and cosec^-1x (also read as arcsin, arccos, arctan, and so on).

A crucial notational caution: sin^-1x is not the same as (sin x)^-1. The latter equals 1/sin x = cosec x, while sin^-1x is the angle whose sine is x. Confusing the two is one of the most common errors students make in this chapter.

Worked example. Is f(x) = sin x invertible on the interval [0, π]? On [0, π], sin x increases from 0 to 1 on [0, π/2] and then decreases back to 0 on [π/2, π]; for example sin(π/3) = sin(2π/3) = √3/2. So f is not one-one on [0, π] and has no inverse there. It is one-one only on [-π/2, π/2], which is exactly why that interval is chosen as the principal value branch for sin^-1x.

The six inverse trigonometric functions, together with the standard (principal value) domain and range chosen for each, are:

• sin^-1x: domain x ∈ [-1, 1]; range y ∈ [-π/2, π/2]
• cos^-1x: domain x ∈ [-1, 1]; range y ∈ [0, π]
• tan^-1x: domain x ∈ ℝ; range y ∈ (-π/2, π/2)
• cot^-1x: domain x ∈ ℝ; range y ∈ (0, π)
• sec^-1x: domain |x| ≥ 1; range y ∈ [0, π], y ≠ π/2
• cosec^-1x: domain |x| ≥ 1; range y ∈ [-π/2, π/2], y ≠ 0

Notice that sec^-1x and cosec^-1x are undefined for -1 < x < 1, because sec θ and cosec θ never take values strictly between -1 and 1. Also notice that π/2 is excluded from the range of sec^-1x (since sec θ is undefined at θ = π/2) and 0 is excluded from the range of cosec^-1x (since cosec θ is undefined at θ = 0).

Worked example. Find the principal value of tan^-1(-√3). We need θ ∈ (-π/2, π/2) with tan θ = -√3. Since tan(π/3) = √3 and tangent is an odd function, tan(-π/3) = -√3, and -π/3 lies in (-π/2, π/2). So tan^-1(-√3) = -π/3.

Worked example. Find the principal value of sec^-1(2/√3). We need θ ∈ [0, π], θ ≠ π/2, with sec θ = 2/√3, i.e. cos θ = √3/2. Since cos(π/6) = √3/2 and π/6 lies in [0, π], sec^-1(2/√3) = π/6.

If y = f(x) is a bijection on its principal branch, the graph of y = f^-1(x) is obtained by reflecting the graph of y = f(x) (restricted to the principal branch) in the line y = x. This gives each inverse trigonometric graph a shape that mirrors its parent function.

• y = sin^-1x: an increasing curve from (-1, -π/2) to (1, π/2), passing through the origin, symmetric about the origin (an odd function).
• y = cos^-1x: a decreasing curve from (-1, π) to (1, 0), passing through (0, π/2).
• y = tan^-1x: an increasing curve defined for all real x, passing through the origin, with horizontal asymptotes y = π/2 (as x → ∞) and y = -π/2 (as x → -∞).
• y = cot^-1x: a decreasing curve defined for all real x, with horizontal asymptotes y = 0 (as x → ∞) and y = π (as x → -∞); it passes through (0, π/2).
• y = sec^-1x: defined only for |x| ≥ 1, increasing on each of its two branches, approaching y = π/2 as x → ±∞ but never reaching it.
• y = cosec^-1x: defined only for |x| ≥ 1, decreasing on each of its two branches, approaching y = 0 as x → ±∞ but never reaching it.

A quick way to remember the shape: sin^-1x and tan^-1x are always increasing, while cos^-1x and cot^-1x are always decreasing. Because the graphs are reflections in y = x, the domain of the original restricted function becomes the range of its inverse, and vice versa — this is a useful sanity check whenever you restate a domain/range pair.

Reciprocal identities. Because cosec θ = 1/sin θ, sec θ = 1/cos θ and cot θ = 1/tan θ, we get: sin^-1(1/x) = cosec^-1x for |x| ≥ 1; cos^-1(1/x) = sec^-1x for |x| ≥ 1; and tan^-1(1/x) = cot^-1x for x > 0 (for x < 0 the relation needs a π correction, so NCERT states it only for x > 0).

Odd/even behaviour. sin^-1x, tan^-1x and cosec^-1x are odd functions: sin^-1(-x) = -sin^-1x, tan^-1(-x) = -tan^-1x, cosec^-1(-x) = -cosec^-1x. In contrast, cos^-1x, sec^-1x and cot^-1x are neither odd nor even, but satisfy: cos^-1(-x) = π - cos^-1x, sec^-1(-x) = π - sec^-1x, cot^-1(-x) = π - cot^-1x.

Cancellation identities. For x within the correct domain, sin(sin^-1x) = x, cos(cos^-1x) = x, tan(tan^-1x) = x and similarly for the others. Going the other way, sin^-1(sin x) = x only when x already lies in [-π/2, π/2]; outside that interval you must first bring x into range using the periodicity/symmetry of sine before the cancellation applies. The same caution applies to the other five functions with their own principal ranges.

Worked example. Evaluate cosec^-1(-2). Using the odd-function rule, cosec^-1(-2) = -cosec^-1(2). Since cosec(π/6) = 1/sin(π/6) = 1/(1/2) = 2, cosec^-1(2) = π/6. So cosec^-1(-2) = -π/6, which correctly lies in [-π/2, π/2] excluding 0.

Worked example. Evaluate sec^-1(-√2). Using sec^-1(-x) = π - sec^-1x: sec(π/4) = √2 so sec^-1(√2) = π/4, hence sec^-1(-√2) = π - π/4 = 3π/4.

Complementary identities. Since sin(π/2 - θ) = cos θ, cos(π/2 - θ) = sin θ, and similarly for the other cofunction pairs, we obtain three constant-sum identities: sin^-1x + cos^-1x = π/2 for x ∈ [-1, 1]; tan^-1x + cot^-1x = π/2 for x ∈ ℝ; and sec^-1x + cosec^-1x = π/2 for |x| ≥ 1.

Sum and difference formulas. Combining two arctangents: tan^-1x + tan^-1y = tan^-1((x + y)/(1 - xy)), valid when xy < 1 (a π correction is needed if xy > 1 and x, y are both positive). Similarly tan^-1x - tan^-1y = tan^-1((x - y)/(1 + xy)) when xy > -1. Setting y = x in the sum formula gives the double-angle form 2 tan^-1x = tan^-1(2x/(1 - x²)) for |x| < 1; the same angle can also be written as sin^-1(2x/(1 + x²)) for |x| ≤ 1, or as cos^-1((1 - x²)/(1 + x²)) for x ≥ 0.

Worked example. Prove that tan^-1(1/2) + tan^-1(1/3) = π/4. Here x = 1/2, y = 1/3, so xy = 1/6 < 1, and the sum formula applies directly: (x + y)/(1 - xy) = (1/2 + 1/3)/(1 - 1/6) = (5/6)/(5/6) = 1. So the sum equals tan^-1(1) = π/4, as required.

Worked example. Evaluate 2 tan^-1(1/2) using the double-angle sine form. Here x = 1/2: 2x/(1 + x²) = 1/(1 + 1/4) = 1/(5/4) = 4/5. So 2 tan^-1(1/2) = sin^-1(4/5). As a check, tan of this angle should be 4/3: tan(2θ) = 2(1/2)/(1 - 1/4) = 1/(3/4) = 4/3, and indeed sin^-1(4/5) is the angle in a 3-4-5 right triangle with tan = 4/3.

Many expressions such as sin^-1(2x√(1 - x²)) or tan^-1(√(1 - x²)/x) look complicated but simplify beautifully once you substitute x = sin θ (or cos θ, or tan θ) and recognise a standard double-angle identity underneath.

Worked example (simplification). Express tan^-1(√(1 - x²)/x) in simplest form for 0 < x ≤ 1. Let x = cos θ with θ ∈ [0, π/2), so √(1 - x²) = sin θ. Then tan^-1(sin θ/cos θ) = tan^-1(tan θ) = θ = cos^-1x. Check with x = 1/2: √(1 - 1/4)/(1/2) = (√3/2)/(1/2) = √3, and tan^-1(√3) = π/3, which indeed equals cos^-1(1/2) = π/3.

Worked example (equation with two arctangents). Solve tan^-1(2x) + tan^-1(3x) = π/4. Using the sum formula (valid while 6x² < 1): (2x + 3x)/(1 - 6x²) = tan(π/4) = 1, so 5x = 1 - 6x², giving 6x² + 5x - 1 = 0, which factors as (6x - 1)(x + 1) = 0, so x = 1/6 or x = -1. Checking validity: for x = 1/6, 6x² = 1/6 < 1 (condition satisfied) and both arctangent terms are small positive angles summing to π/4 — valid. For x = -1, 6x² = 6 > 1 (condition fails), and both arctangent terms would be negative, so their sum cannot equal the positive angle π/4 — this root is extraneous. Hence x = 1/6 is the only solution.

Worked example (equation mixing tan and cosec). Solve 2 tan^-1(cos x) = tan^-1(2 cosec x). Using the double-angle form on the left (valid provided cos x ≠ ±1, i.e. x is not an integer multiple of π, so that |cos x| < 1): 2 tan^-1(cos x) = tan^-1(2 cos x/(1 - cos²x)) = tan^-1(2 cos x/sin²x). Equating the arguments with the right-hand side tan^-1(2/sin x): 2 cos x/sin²x = 2/sin x. Cross-multiplying (sin x ≠ 0): 2 cos x sin x = 2 sin²x, and dividing by 2 sin x: cos x = sin x, so tan x = 1, giving x = π/4 (or, more generally, x = nπ + π/4).