Precalculus Inverse Trigonometric Functions and Equations Worksheet Solutions (PDF)

Precalculus math worksheet solutions: Inverse Trigonometric Functions and Equations – Maths PDF

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This inverse trig answers for Precalculus covers inverse trigonometric functions and equations (Common Core F-TF.B.6, F-TF.B.7), with a free PDF to print.

Use this key to check each inverse trig value and equation. Solutions show the reference angle, the quadrants and the period, with final answers in bold.

Precalculus · Common Core State Standards: F-TF.B.6, F-TF.B.7

Inverse trig answers: exercises 1 to 10

Exercise 1 – Exact values of inverse trig

\(\sin\frac{\pi}{6}=\frac12\). \(\cos\frac{\pi}{6}=\frac{\sqrt3}{2}\). \(\tan\frac{\pi}{4}=1\).

\(\frac{\pi}{6},\ \frac{\pi}{6},\ \frac{\pi}{4}\)

Exercise 2 – Negative inputs

Arcsin and arctan give angles in \((-\frac{\pi}{2},\frac{\pi}{2})\): \(-\frac{\pi}{4}\) and \(-\frac{\pi}{3}\). Arccos gives an angle in \([0,\pi]\), here quadrant II: \(\frac{2\pi}{3}\).

\(-\frac{\pi}{4},\ \frac{2\pi}{3},\ -\frac{\pi}{3}\)

Exercise 3 – Domain and range

The graph runs from \(x=-1\) to \(x=1\), and \(y\) from \(-\frac{\pi}{2}\) to \(\frac{\pi}{2}\).

Domain \([-1,1]\); range \(\left[-\frac{\pi}{2},\frac{\pi}{2}\right]\)

Exercise 4 – Calculator values

\(\arcsin0.4\approx 0.4115\). \(\arctan2.5\approx 1.1903\).

\(0.4115\) and \(1.1903\)

Exercise 5 – Inverse properties

Each function undoes its inverse, and all the inputs are in the right domain or range.

\(0.3,\ -0.8,\ \frac{\pi}{6}\)

Exercise 6 – Outside the principal range

\(\sin\frac{7\pi}{6}=-\frac12\) and \(\arcsin\left(-\frac12\right)=-\frac{\pi}{6}\). \(\cos\left(-\frac{\pi}{3}\right)=\frac12\) and \(\arccos\frac12=\frac{\pi}{3}\).

\(-\frac{\pi}{6}\) and \(\frac{\pi}{3}\)

Exercise 7 – Cosine of an arcsine

Let \(\theta=\arcsin\frac35\), an angle in quadrant I. The triangle has opposite 3, hypotenuse 5, so adjacent 4.

\(\frac45\)

Exercise 8 – Tangent of an arccosine

Adjacent 5, hypotenuse 13, so opposite \(\sqrt{169-25}=12\). \(\tan\theta=\frac{12}{5}\).

\(\frac{12}{5}\)

Exercise 9 – Sine of an arctangent

Opposite 3, adjacent 4, hypotenuse 5.

\(\frac35\)

Exercise 10 – Solve sin x = 1/2

Quadrants I and II.

The reference angle is \(\frac{\pi}{6}\). Sine is positive in quadrants I and II: \(\frac{\pi}{6}\) and \(\pi-\frac{\pi}{6}\).

\(x=\frac{\pi}{6},\ \frac{5\pi}{6}\)

Exercises 11 and up

Exercise 11 – Solve cos x

Reference angle \(\frac{\pi}{4}\). Cosine is negative in quadrants II and III: \(\pi-\frac{\pi}{4}\) and \(\pi+\frac{\pi}{4}\).

\(x=\frac{3\pi}{4},\ \frac{5\pi}{4}\)

Exercise 12 – Solve tan x

Reference angle \(\frac{\pi}{3}\). Tangent is negative in quadrants II and IV: \(\pi-\frac{\pi}{3}\) and \(2\pi-\frac{\pi}{3}\).

\(x=\frac{2\pi}{3},\ \frac{5\pi}{3}\)

Exercise 13 – General solutions

In one period: \(\frac{\pi}{3}\) and \(\frac{2\pi}{3}\). Add \(2\pi k\).

\(x=\frac{\pi}{3}+2\pi k\) or \(x=\frac{2\pi}{3}+2\pi k\), \(k\) an integer

Exercise 14 – Rearrange first

\(\sin x=\frac{\sqrt2}{2}\). Reference angle \(\frac{\pi}{4}\), quadrants I and II.

\(x=\frac{\pi}{4},\ \frac{3\pi}{4}\)

Exercise 15 – Quadratic in sine

Factor: \((2\cos x-1)(\cos x+1)=0\). \(\cos x=\frac12\) gives \(\frac{\pi}{3}\) and \(\frac{5\pi}{3}\). \(\cos x=-1\) gives \(\pi\).

\(x=\frac{\pi}{3},\ \pi,\ \frac{5\pi}{3}\)

Exercise 16 – A non-special value

Two intersections.

\(x=\arccos0.3\approx 1.266\). The other solution is \(2\pi-1.266\approx 5.017\).

\(x\approx 1.266\) and \(5.017\)

Exercise 17 – Double angle inside

Let \(u=2x\) on \([0,4\pi)\). \(\sin u=\frac{\sqrt3}{2}\) gives \(u=\frac{\pi}{3},\frac{2\pi}{3},\frac{7\pi}{3},\frac{8\pi}{3}\). Divide by 2.

\(x=\frac{\pi}{6},\ \frac{\pi}{3},\ \frac{7\pi}{6},\ \frac{4\pi}{3}\)

Exercise 18 – Ramp angle

sin θ = 2/12.

\(\sin\theta=\frac{2}{12}=\frac16\), so \(\theta=\arcsin\frac16\approx 9.6^{\circ}\).

\(9.6^{\circ}\)

Exercise 19 – Ladder

cos θ = 4/15.

\(\cos\theta=\frac{4}{15}\), so \(\theta=\arccos\frac{4}{15}\approx 74.5^{\circ}\).

\(74.5^{\circ}\)

Exercise 20 – Ferris wheel timing

\(30-25\cos u=40\), so \(\cos u=-0.4\) where \(u=\frac{\pi t}{15}\). \(u=\arccos(-0.4)\approx 1.9823\), so \(t\approx 9.46\). The other solution in a cycle is \(u=2\pi-1.9823\), so \(t=30-9.46=20.54\).

\(t\approx 9.46\) s and \(20.54\) s

Key facts behind this inverse trig answers

Key facts

arcsin and arctan have range (or closed range) centered on 0; arccos has range [0, pi].
sin(arcsin x) = x on [-1, 1], but arcsin(sin x) = x only for x in [-pi/2, pi/2].
sin^-1 x means the inverse function, not 1/sin x.
Sine and cosine equations usually have two solutions per period; tangent has one.
Factor quadratic-type trig equations before solving.

How to use this inverse trig answers

Cover each solution, solve the exercise again, then uncover it and compare every line. If your answer differs, find the first step that changed and redo the exercise from there.

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