Inverse Trigonometric Functions: arcsin, arccos and arctan

Undoing Sine, Cosine and Tangent

If sin θ = 0.5, what is θ? Your calculator says 30°, but 150°, 390° and −210° also have a sine of 0.5. To get a single, well-defined answer, mathematicians restrict each trigonometric function to a range where it takes every value exactly once, and then define its inverse. These inverse trigonometric functions, written sin⁻¹, cos⁻¹ and tan⁻¹ (or arcsin, arccos and arctan), are essential in first-year calculus, physics and engineering. This tutorial explains their domains and principal ranges, their graphs, exact values, important identities, and their derivatives and integrals, with worked examples.

Why the Ranges Must Be Restricted

A function has an inverse only if it is one-to-one: each output comes from exactly one input. The sine function repeats every 360°, so it is not one-to-one. We fix this by keeping only one "branch":

sin⁻¹ x (arcsin): domain −1 ≤ x ≤ 1, range −π/2 ≤ y ≤ π/2 (−90° to 90°).
cos⁻¹ x (arccos): domain −1 ≤ x ≤ 1, range 0 ≤ y ≤ π (0° to 180°).
tan⁻¹ x (arctan): domain all real x, range −π/2 < y < π/2 (strictly between −90° and 90°).

The value given by these functions is called the principal value. Calculators always return principal values; other solutions of an equation such as sin θ = 0.5 must be found using the symmetry of the graphs.

Notation warning: sin⁻¹ x means the inverse function, not 1/sin x. The reciprocal 1/sin x is cosec x. This is different from sin² x, which does mean (sin x)².

Graphs of the Inverse Trigonometric Functions

The graph of an inverse function is the reflection of the original graph in the line y = x.

y = sin⁻¹ x runs from (−1, −π/2) through the origin to (1, π/2), increasing, steepest at the ends and with gradient 1 at the origin.
y = cos⁻¹ x runs from (−1, π) through (0, π/2) to (1, 0), decreasing.
y = tan⁻¹ x passes through the origin and flattens towards the horizontal asymptotes y = π/2 (as x → ∞) and y = −π/2 (as x → −∞).

sin⁻¹ and tan⁻¹ are odd functions: sin⁻¹(−x) = −sin⁻¹ x and tan⁻¹(−x) = −tan⁻¹ x. Arccos is not odd; instead cos⁻¹(−x) = π − cos⁻¹ x.

Exact Values

Use the special angles you already know, and make sure the answer lies in the correct range.

sin⁻¹(1/2) = π/6, sin⁻¹(√3/2) = π/3, sin⁻¹(1) = π/2.
sin⁻¹(−1/2) = −π/6 (not 7π/6 or 11π/6, which are outside the range).
cos⁻¹(1/2) = π/3, cos⁻¹(0) = π/2, cos⁻¹(−1/2) = π − π/3 = 2π/3.
tan⁻¹(1) = π/4, tan⁻¹(√3) = π/3, tan⁻¹(−1) = −π/4.

Example 1. Find cos⁻¹(−√2/2).

cos⁻¹(√2/2) = π/4, so cos⁻¹(−√2/2) = π − π/4 = 3π/4 (135°), which is in the range 0 to π.

Compositions: sin(sin⁻¹ x) and sin⁻¹(sin x)

sin(sin⁻¹ x) = x for every x in [−1, 1]. Applying sine undoes arcsine completely.
sin⁻¹(sin x) = x only when x is in [−π/2, π/2]. Otherwise the answer is the angle in that range with the same sine.

Example 2. Evaluate sin⁻¹(sin(5π/6)).

sin(5π/6) = 1/2, and sin⁻¹(1/2) = π/6, not 5π/6, because 5π/6 is outside the principal range.

Example 3. Find the exact value of cos(sin⁻¹(3/5)).

Let θ = sin⁻¹(3/5), so sin θ = 3/5 and θ is between −π/2 and π/2, where cosine is non-negative. Draw a right-angled triangle with opposite side 3 and hypotenuse 5: the adjacent side is 4. So cos θ = 4/5. In general, cos(sin⁻¹ x) = √(1 − x²).

Example 4. Simplify tan(cos⁻¹ x) for 0 < x ≤ 1.

Let θ = cos⁻¹ x, so cos θ = x = adjacent/hypotenuse. The opposite side is √(1 − x²), so tan θ = √(1 − x²)/x.

Important Identities

sin⁻¹ x + cos⁻¹ x = π/2 for −1 ≤ x ≤ 1. (If sin θ = x, then cos(π/2 − θ) = x.)
tan⁻¹ x + tan⁻¹(1/x) = π/2 for x > 0.
tan⁻¹ a + tan⁻¹ b = tan⁻¹[(a + b)/(1 − ab)] when ab < 1.

Example 5. Show that tan⁻¹(1/2) + tan⁻¹(1/3) = π/4.

Here ab = 1/6 < 1, so the sum is tan⁻¹[(1/2 + 1/3)/(1 − 1/6)] = tan⁻¹[(5/6)/(5/6)] = tan⁻¹(1) = π/4. This kind of identity was used historically to calculate π to many decimal places.

Solving Equations with Inverse Trigonometric Functions

Example 6. Solve sin⁻¹(2x − 1) = π/6.

Take the sine of both sides: 2x − 1 = sin(π/6) = 1/2, so 2x = 3/2 and x = 3/4. Check that 2x − 1 = 1/2 is in [−1, 1] ✓.

Example 7. Solve 2 tan⁻¹ x = π/3.

tan⁻¹ x = π/6, so x = tan(π/6) = 1/√3, or √3/3.

Derivatives of the Inverse Trigonometric Functions

d/dx (sin⁻¹ x) = 1/√(1 − x²)
d/dx (cos⁻¹ x) = −1/√(1 − x²)
d/dx (tan⁻¹ x) = 1/(1 + x²)

Derivation for arcsin. Let y = sin⁻¹ x, so x = sin y. Differentiate with respect to y: dx/dy = cos y. Then dy/dx = 1/cos y = 1/√(1 − sin² y) = 1/√(1 − x²). The positive root is correct because cos y ≥ 0 in the range −π/2 ≤ y ≤ π/2.

Example 8. Differentiate y = tan⁻¹(3x).

By the chain rule, dy/dx = 3/(1 + 9x²).

Example 9. Differentiate y = sin⁻¹(x²).

dy/dx = 2x/√(1 − x⁴).

Integrals That Give Inverse Trigonometric Functions

Reversing the derivatives gives two standard integrals (a > 0):

∫ 1/√(a² − x²) dx = sin⁻¹(x/a) + C
∫ 1/(a² + x²) dx = (1/a) tan⁻¹(x/a) + C

Example 10. Evaluate the integral of 1/(1 + x²) from 0 to 1.

[tan⁻¹ x] from 0 to 1 = π/4 − 0 = π/4.

Example 11. Find ∫ 1/√(9 − x²) dx.

Here a = 3, so the integral is sin⁻¹(x/3) + C.

Example 12. Find ∫ 1/(x² + 4x + 13) dx.

Complete the square: x² + 4x + 13 = (x + 2)² + 9. With u = x + 2 and a = 3, the integral is (1/3) tan⁻¹((x + 2)/3) + C.

Summary

sin⁻¹, cos⁻¹ and tan⁻¹ return principal values in [−π/2, π/2], [0, π] and (−π/2, π/2) respectively.
sin⁻¹ x is the inverse function, not 1/sin x.
sin(sin⁻¹ x) = x always, but sin⁻¹(sin x) = x only when x is in the principal range.
sin⁻¹ x + cos⁻¹ x = π/2; triangles turn expressions like cos(sin⁻¹ x) into √(1 − x²).
The derivatives are 1/√(1 − x²), −1/√(1 − x²) and 1/(1 + x²), giving standard integrals with a² − x² and a² + x².