Inverse Trigonometric Functions Worksheet Answer Key – Precalculus

Problem 1 Evaluate arcsin

The answer must lie in [−π/2, π/2].

(a) π/3. (b) −π/6. (c) 0. (d) −π/2.

Problem 2 Evaluate arccos

The answer must lie in [0, π].

(a) π/3. (b) 5π/6. (c) π. (d) π/2.

Problem 3 Evaluate arctan

The answer must lie in (−π/2, π/2).

(a) π/4. (b) −π/3. (c) 0. (d) π/6.

Problem 4 Domain and Range from a Graph

The curve begins at x = −1 and ends at x = 1: domain [−1, 1].

Its lowest height is −π/2 and its highest is π/2: range [−π/2, π/2]. At x = 0 the output is 0.

Problem 5 Why Restrict the Domain?

sin(π/6) = 1/2 and sin(5π/6) = 1/2. One output comes from two inputs, so sine is not one-to-one on [0, 2π] and a single inverse cannot choose between them.

Restrict to [−π/2, π/2], where sine is increasing.

Problem 6 Solve Three Basic Equations

(a) Reference angle π/6, sine positive in Quadrants I and II: x = π/6, 5π/6.

(b) Cosine is negative in Quadrants II and III with reference π/4: x = 3π/4, 5π/4.

(c) Tangent is 1 at π/4 and every π after: x = π/4, 5π/4.

Problem 7 Compositions with Triangles

(a) Let θ = arccos(8/17); the adjacent side is 8, the hypotenuse 17, so the opposite side is 15: 15/17.

(b) Opposite 7, hypotenuse 25, adjacent 24: 24/25.

(c) θ is in Quadrant IV (arcsin range), cosine is positive 12/13, so tan = (−5/13) ÷ (12/13) = −5/12.

(d) Opposite 12, adjacent 5, hypotenuse 13: 12/13.

Problem 8 Careful with arcsin(sin x)

(a) sin(5π/6) = 1/2 and arcsin(1/2) = π/6.

(b) cos(−π/3) = 1/2 and arccos(1/2) = π/3.

(c) tan(3π/4) = −1 and arctan(−1) = −π/4.

(d) cos(7π/4) = √2/2 and arccos(√2/2) = π/4.

Problem 9 Equation Solved in General

sin x = √3/2, reference angle π/3, sine positive in Quadrants I and II.

On [0, 2π): x = π/3, 2π/3. General: x = π/3 + 2πk or 2π/3 + 2πk, k an integer.

Problem 10 A Quadratic in Cosine

cos² x = 3/4, so cos x = ±√3/2.

cos x = √3/2 gives x = π/6, 11π/6; cos x = −√3/2 gives x = 5π/6, 7π/6.

x = π/6, 5π/6, 7π/6, 11π/6.

Problem 11 Equation with a Multiple Angle Input

sin u = √3/2 when u = π/3 + 2πk or 2π/3 + 2πk. With u = 3x: x = π/9 + 2πk/3 or 2π/9 + 2πk/3.

On [0, 2π): x = π/9, 2π/9, 7π/9, 8π/9, 13π/9, 14π/9. General: x = π/9 + 2πk/3 or 2π/9 + 2πk/3.

Problem 12 Use a Calculator for cos x = 0.3

In radian mode, arccos 0.3 ≈ 1.2661. By symmetry the second solution is 2π − 1.2661 ≈ 5.0171.

Problem 13 Negative Sine Value

arcsin(−0.45) ≈ −0.4668, which is in Quadrant IV but negative, so convert. The solutions lie in Quadrants III and IV.

Quadrant III: π + 0.4668 ≈ 3.6084. Quadrant IV: 2π − 0.4668 ≈ 5.8164.

Problem 14 Tangent Equation with Technology

tan x = −5/3. arctan(−5/3) ≈ −1.0304, which is outside [0, 2π), so add π and 2π.

x ≈ 2.1112 and 5.2528.

Problem 15 Factor Instead of Dividing

Move terms: cos x(2 sin x − 1) = 0. Then cos x = 0 gives x = π/2, 3π/2, and sin x = 1/2 gives x = π/6, 5π/6.

x = π/6, π/2, 5π/6, 3π/2. Dividing by cos x would remove the solutions where cos x = 0, namely π/2 and 3π/2.

Problem 16 Ferris Wheel at 50 Feet

Solve 30 − 25cos(πt/20) = 50: cos(πt/20) = −0.8. Then πt/20 = arccos(−0.8) ≈ 2.4981, so t ≈ 15.9 s.

By symmetry the second time is 40 − 15.9 ≈ 24.1 s.

Problem 17 Tide Window

Solve 3.5 sin(πt/6) ≥ 2: sin(πt/6) ≥ 0.5714. Then arcsin 0.5714 ≈ 0.6082, so the first crossing is πt/6 = 0.6082, giving t ≈ 1.16. The second is πt/6 = π − 0.6082, giving t ≈ 4.84.

The depth is at least 6.5 ft for 1.16 ≤ t ≤ 4.84, about 3.7 hours.

Problem 18 Wheelchair Ramp Angles

(a) sin θ = 1.1/14, so θ = arcsin(0.0786) ≈ 4.51°.

(b) tan θ = 1/12, so θ = arctan(1/12) ≈ 4.76°.

(c) h = 20 sin(4.76°) ≈ 1.66 feet.

Problem 19 Swinging Pendulum

Period = 2π ÷ (5π/4) = 1.6 s. Solve cos(5πt/4) = 0.2/0.35 ≈ 0.5714: 5πt/4 = arccos(0.5714) ≈ 0.9626, so t ≈ 0.245 s.

The other solution is 5πt/4 = 2π − 0.9626, so t ≈ 1.355 s.

Problem 20 Quadratic in Sine with an Impossible Factor

Factor: (2 sin x − 1)(sin x + 2) = 0.

sin x = −2 is impossible because sine never goes below −1. sin x = 1/2 gives x = π/6, 5π/6.

General: x = π/6 + 2πk or 5π/6 + 2πk.