Inverse Trigonometric Functions and Trigonometric Equations: Precalculus Math Worksheet Answer Key – PDF Download

Precalculus Inverse Trig Worksheet Answer Key: problems 1–5

Problem 1: Evaluate arcsin

The answer must lie in \( [-\frac{\pi}{2},\frac{\pi}{2}] \).

1. \(\frac{\pi}{6}\)
2. \(-\frac{\pi}{6}\)
3. \(\frac{\pi}{3}\)
4. \(-\frac{\pi}{2}\)

Problem 2: Evaluate arccos

The answer must lie in \( [0,\pi] \).

1. \(\frac{\pi}{4}\)
2. \(\frac{2\pi}{3}\) (cosine is negative in Quadrant II)
3. \(\frac{\pi}{2}\)
4. \(\pi\)

Problem 3: Evaluate arctan

The answer must lie in \( (-\frac{\pi}{2},\frac{\pi}{2}) \).

1. \(\frac{\pi}{4}\)
2. \(-\frac{\pi}{4}\)
3. \(\frac{\pi}{3}\)
4. \(-\frac{\pi}{6}\)

Problem 4: Graph of arctan

Every real number can be an input, and the curve stays between the dashed lines at \(y=\pm\frac{\pi}{2}\).

Domain: all real numbers. Range: \( (-\frac{\pi}{2},\frac{\pi}{2}) \). Asymptotes: \(y=\frac{\pi}{2}\) and \(y=-\frac{\pi}{2}\). \(\tan^{-1}(0)=0\).

Problem 5: Restricted cosine

The thick piece is \( [0,\pi] \).

1. \(\frac{\pi}{3}\) (already in \( [0,\pi] \))
2. \(\cos(-\frac{\pi}{3})=\frac{1}{2}\), and \(\cos^{-1}\frac{1}{2}=\)\(\frac{\pi}{3}\)
3. \(\cos\frac{4\pi}{3}=-\frac{1}{2}\), and \(\cos^{-1}(-\frac{1}{2})=\)\(\frac{2\pi}{3}\)

Precalculus Inverse Trig Worksheet Answer Key: problems 6–10

Problem 6: Compositions that cancel

1. 0.3
2. -0.8
3. 25 (every real number is allowed for \(\tan^{-1}\))
4. undefined, because \(1.2\) is not in \( [-1,1] \)

Problem 7: Arcsine of a sine

1. \(\frac{\pi}{4}\) (already in the range)
2. \(\sin\frac{3\pi}{4}=\frac{\sqrt{2}}{2}\), so \(\frac{\pi}{4}\)
3. \(\sin\frac{7\pi}{6}=-\frac{1}{2}\), so \(-\frac{\pi}{6}\)
4. \(\sin(-\frac{2\pi}{3})=-\frac{\sqrt{3}}{2}\), so \(-\frac{\pi}{3}\)

Problem 8: Triangle compositions

1. The missing leg is \(\sqrt{5^{2}-3^{2}}=4\), so \(\sin\theta=\)\(\frac{4}{5}\)
2. With opposite 5 and hypotenuse 13 the adjacent side is 12, so \(\frac{12}{13}\)
3. Adjacent 4, hypotenuse 5, opposite 3: \(\frac{3}{4}\)
4. The angle is in Quadrant IV where cosine is positive: opposite \(-3\), adjacent 4, so \(-\frac{3}{4}\)

Problem 9: Calculator practice

1. 0.36
2. 1.37
3. 1.33
4. 2.28

Problem 10: Solve a sine equation

The reference angle is \(\sin^{-1}\frac{\sqrt{3}}{2}=\frac{\pi}{3}\). Sine is positive in Quadrants I and II, so the second solution is \(\pi-\frac{\pi}{3}=\frac{2\pi}{3}\).

\(x=\frac{\pi}{3}\) or \(x=\frac{2\pi}{3}\)

Precalculus Inverse Trig Worksheet Answer Key: problems 11–15

Problem 11: Solve a cosine equation

The reference angle is \(\frac{\pi}{4}\). Cosine is negative in Quadrants II and III: \(\pi-\frac{\pi}{4}=\frac{3\pi}{4}\) and \(\pi+\frac{\pi}{4}=\frac{5\pi}{4}\).

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

Problem 12: Solve a tangent equation

\(\tan^{-1}(-1)=-\frac{\pi}{4}\). Tangent repeats every \(\pi\), so add \(\pi\) and \(2\pi\) to reach the interval: \(-\frac{\pi}{4}+\pi=\frac{3\pi}{4}\) and \(-\frac{\pi}{4}+2\pi=\frac{7\pi}{4}\).

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

Problem 13: Equation with a unit circle

Subtract and divide: \(\cos x=-\frac{1}{2}\). The line \(x=-\frac{1}{2}\) meets the circle in Quadrants II and III. \(\cos^{-1}(-\frac{1}{2})=\frac{2\pi}{3}\) and the other angle is \(2\pi-\frac{2\pi}{3}=\frac{4\pi}{3}\).

\(x=\frac{2\pi}{3}\) or \(x=\frac{4\pi}{3}\)

Problem 14: Sine equation by calculator

\(\sin x=0.4\). \(x=\sin^{-1}(0.4)\approx 0.41\). The second solution is \(\pi-0.4115\approx 2.73\).

\(x\approx 0.41\) or \(x\approx 2.73\)

Problem 15: Cosine equation by calculator

\(x=\cos^{-1}(0.3)\approx 1.27\). Cosine is also positive in Quadrant IV: \(2\pi-1.2661\approx 5.02\).

\(x\approx 1.27\) or \(x\approx 5.02\)

Precalculus Inverse Trig Worksheet Answer Key: problems 16–20

Problem 16: General solutions

1. \(x=\frac{\pi}{6}+2\pi k\) or \(x=\frac{5\pi}{6}+2\pi k\)
2. \(x=\frac{\pi}{2}+\pi k\) (both \(\frac{\pi}{2}\) and \(\frac{3\pi}{2}\) are included)
3. \(x=\frac{\pi}{3}+\pi k\)

Problem 17: Quadratic in sine

Factor: \((2\sin x+1)(\sin x-1)=0\). If \(\sin x=1\) then \(x=\frac{\pi}{2}\). If \(\sin x=-\frac{1}{2}\) then \(x=\frac{7\pi}{6}\) or \(x=\frac{11\pi}{6}\).

\(x=\frac{\pi}{2}\), \(\frac{7\pi}{6}\), \(\frac{11\pi}{6}\)

Problem 18: Multiple-angle equation

Let \(u=2x\), so \(0\le u< 4\pi\). \(\sin u=\frac{\sqrt{2}}{2}\) gives \(u=\frac{\pi}{4},\frac{3\pi}{4},\frac{9\pi}{4},\frac{11\pi}{4}\). Divide by 2.

\(x=\frac{\pi}{8},\frac{3\pi}{8},\frac{9\pi}{8},\frac{11\pi}{8}\)

Problem 19: Wheelchair ramp

\(\sin\theta=\frac{1}{12}\), so \(\theta=\sin^{-1}(\frac{1}{12})\approx 0.0834\text{ rad}\).

\(\theta\approx 4.78^{\circ}\)

Problem 20: Ladder angle

1. \(\cos\theta=\frac{4}{16}=0.25\), so \(\theta=\cos^{-1}(0.25)\approx\)\(75.5^{\circ}\)
2. \(16\cos 70^{\circ}\approx 5.47\), so about 5.5 ft

Precalculus Inverse Trig Worksheet Answer Key: problems 21–24

Problem 21: Daylight hours

\(4\sin(\frac{2\pi}{365}(t-80))=2\), so \(\sin u=\frac{1}{2}\) with \(u=\frac{2\pi}{365}(t-80)\). Then \(u=\frac{\pi}{6}\) or \(\frac{5\pi}{6}\). \(t-80=\frac{365}{12}\approx 30.4\) gives \(t\approx 110.4\); \(t-80=\frac{5\cdot 365}{12}\approx 152.1\) gives \(t\approx 232.1\).

About day 110 (April 20) to day 232 (August 20) has more than 14 hours of daylight.

Problem 22: Spring height

Height 23: \(\cos\frac{2\pi t}{3}=\frac{1}{2}\), so \(\frac{2\pi t}{3}=\frac{\pi}{3}\) or \(\frac{5\pi}{3}\), giving \(t=0.5\) and \(t=2.5\).

Height 16: \(\cos\frac{2\pi t}{3}=-\frac{2}{3}\). \(\cos^{-1}(-\frac{2}{3})\approx 2.3005\), so \(t=\frac{3(2.3005)}{2\pi}\approx 1.10\); the other angle is \(2\pi-2.3005\), giving \(t\approx 1.90\).

23 in at \(t=0.5\) and \(t=2.5\) s; 16 in at \(t\approx 1.10\) and \(t\approx 1.90\) s

Problem 23: Household voltage

\(\sin(120\pi t)=\frac{100}{170}\approx 0.5882\). \(\sin^{-1}(0.5882)\approx 0.6288\). So \(t=\frac{0.6288}{120\pi}\approx 0.001668\) s and \(t=\frac{\pi-0.6288}{120\pi}\approx 0.006665\) s.

\(t\approx 1.67\) ms and \(t\approx 6.67\) ms; one cycle is \(\frac{2\pi}{120\pi}=\frac{1}{60}\) s, about 16.67 ms

Problem 24: Sun angle

1. \(\tan\theta=\frac{40}{55}\), so \(\theta=\tan^{-1}(\frac{40}{55})\approx\)\(36.0^{\circ}\)
2. \(\text{shadow}=\frac{40}{\tan 60^{\circ}}=\frac{40}{\sqrt{3}}\approx\)23.1 ft

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