Angles, Radians and the Unit Circle: Precalculus Math Worksheet Answer Key – PDF Download

Precalculus Unit Circle Worksheet Answer Key: problems 1–5

Problem 1: Degrees to radians

1. \(30^{\circ}\cdot\frac{\pi}{180}=\)\(\frac{\pi}{6}\)
2. \(135^{\circ}\cdot\frac{\pi}{180}=\)\(\frac{3\pi}{4}\)
3. \(270^{\circ}\cdot\frac{\pi}{180}=\)\(\frac{3\pi}{2}\)
4. \(-210^{\circ}\cdot\frac{\pi}{180}=\)\(-\frac{7\pi}{6}\)
5. \(15^{\circ}\cdot\frac{\pi}{180}=\)\(\frac{\pi}{12}\)

Problem 2: Radians to degrees

1. \(\frac{5\pi}{6}\cdot\frac{180^{\circ}}{\pi}=\)\(150^{\circ}\)
2. \(\frac{7\pi}{4}\cdot\frac{180^{\circ}}{\pi}=\)\(315^{\circ}\)
3. \(\frac{2\pi}{3}\cdot\frac{180^{\circ}}{\pi}=\)\(120^{\circ}\)
4. \(\frac{11\pi}{6}\cdot\frac{180^{\circ}}{\pi}=\)\(330^{\circ}\)
5. \(3\cdot\frac{180^{\circ}}{\pi}\approx\) \(171.89^{\circ}\)

Problem 3: Coterminal angles

1. \(\frac{17\pi}{6}-2\pi=\)\(\frac{5\pi}{6}\)
2. \(-\frac{\pi}{3}+2\pi=\)\(\frac{5\pi}{3}\)
3. \(\frac{25\pi}{4}-6\pi=\)\(\frac{\pi}{4}\)
4. \(-\frac{11\pi}{4}+4\pi=\)\(\frac{5\pi}{4}\)

Problem 4: Reference angles

1. Quadrant II, reference angle \(\frac{\pi}{6}\)
2. Quadrant IV, reference angle \(\frac{\pi}{4}\)
3. Quadrant IV, reference angle \(\frac{\pi}{6}\)
4. Coterminal angle \(\frac{\pi}{3}\); Quadrant I, reference angle \(\frac{\pi}{3}\)
5. Coterminal angle \(\frac{4\pi}{3}\); Quadrant III, reference angle \(\frac{\pi}{3}\)

Problem 5: Special angle values

Read the coordinates \((\cos\theta,\sin\theta)\) and divide for the tangent.

\(\theta\)	\(\sin\theta\)	\(\cos\theta\)	\(\tan\theta\)
\(\frac{\pi}{6}\)	\(\frac{1}{2}\)	\(\frac{\sqrt{3}}{2}\)	\(\frac{\sqrt{3}}{3}\)
\(\frac{\pi}{4}\)	\(\frac{\sqrt{2}}{2}\)	\(\frac{\sqrt{2}}{2}\)	\(1\)
\(\frac{\pi}{3}\)	\(\frac{\sqrt{3}}{2}\)	\(\frac{1}{2}\)	\(\sqrt{3}\)

Precalculus Unit Circle Worksheet Answer Key: problems 6–10

Problem 6: Point on the unit circle

1. The reference angle satisfies \(\sin\frac{\pi}{6}=\frac{1}{2}\). In Quadrant II, \(\theta=\pi-\frac{\pi}{6}=\)\(\frac{5\pi}{6}\)
2. \(\sin\theta=\)\(\frac{1}{2}\), \(\cos\theta=\)\(-\frac{\sqrt{3}}{2}\), \(\tan\theta=\frac{1/2}{-\sqrt{3}/2}=\)\(-\frac{\sqrt{3}}{3}\)

Problem 7: Exact values by symmetry

1. \(\sin(\frac{5\pi}{6})=\)\(\frac{1}{2}\). Reference angle \(\frac{\pi}{6}\), Quadrant II, sine is positive there.
2. \(\cos(\frac{7\pi}{4})=\)\(\frac{\sqrt{2}}{2}\). Reference angle \(\frac{\pi}{4}\), Quadrant IV, cosine is positive there.
3. \(\tan(\frac{2\pi}{3})=\)\(-\sqrt{3}\). Reference angle \(\frac{\pi}{3}\), Quadrant II, tangent is negative there.
4. \(\sin(\frac{11\pi}{6})=\)\(-\frac{1}{2}\). Reference angle \(\frac{\pi}{6}\), Quadrant IV, sine is negative there.
5. \(\cos(\frac{4\pi}{3})=\)\(-\frac{1}{2}\). Reference angle \(\frac{\pi}{3}\), Quadrant III, cosine is negative there.

Problem 8: Reading a point from an angle

1. The terminal side is \(\frac{5\pi}{4}-\pi=\)\(\frac{\pi}{4}\) past the negative x-axis.
2. Both coordinates have magnitude \(\frac{\sqrt{2}}{2}\) and are negative in Quadrant III: \(P(-\frac{\sqrt{2}}{2},-\frac{\sqrt{2}}{2})\)
3. \(\tan\frac{5\pi}{4}=\frac{-\sqrt{2}/2}{-\sqrt{2}/2}=\)1

Problem 9: Even and odd identities

1. \(\sin(-\frac{\pi}{3})=-\sin(\frac{\pi}{3})=\)\(-\frac{\sqrt{3}}{2}\).
2. \(\cos(-\frac{\pi}{4})=\cos(\frac{\pi}{4})=\)\(\frac{\sqrt{2}}{2}\).
3. \(\tan(-\frac{\pi}{6})=-\tan(\frac{\pi}{6})=\)\(-\frac{\sqrt{3}}{3}\).
4. \(\sin(-\frac{5\pi}{6})=-\sin(\frac{5\pi}{6})=\)\(-\frac{1}{2}\).
5. \(\cos(-\frac{2\pi}{3})=\cos(\frac{2\pi}{3})=\)\(-\frac{1}{2}\).

Problem 10: Coordinates on the unit circle

1. \(\sin^{2}\theta=1-\frac{9}{25}=\frac{16}{25}\), and sine is negative in Quadrant IV, so \(\sin\theta=-\frac{4}{5}\). Then \(\tan\theta=\frac{-4/5}{3/5}=-\frac{4}{3}\)
2. \(\frac{25}{169}+y^{2}=1\) gives \(y^{2}=\frac{144}{169}\). In Quadrant III, \(y=-\frac{12}{13}\)

Precalculus Unit Circle Worksheet Answer Key: problems 11–15

Problem 11: Periodicity

1. \(\sin(\frac{25\pi}{6})=\sin(\frac{25\pi}{6}-4\pi)=\sin(\frac{\pi}{6})=\)\(\frac{1}{2}\).
2. \(\cos(-\frac{13\pi}{3})=\cos(-\frac{13\pi}{3}+6\pi)=\cos(\frac{5\pi}{3})=\)\(\frac{1}{2}\).
3. \(\tan(\frac{9\pi}{4})=\tan(\frac{9\pi}{4}-2\pi)=\tan(\frac{\pi}{4})=\)\(1\).
4. \(\sin(-\frac{17\pi}{4})=\sin(-\frac{17\pi}{4}+6\pi)=\sin(\frac{7\pi}{4})=\)\(-\frac{\sqrt{2}}{2}\).
5. \(\tan(-\frac{11\pi}{6})=\tan(-\frac{11\pi}{6}+2\pi)=\tan(\frac{\pi}{6})=\)\(\frac{\sqrt{3}}{3}\).

Problem 12: Solving trigonometric equations

1. The reference angle is \(\frac{\pi}{6}\). The value is positive, so the terminal side is in Quadrant I or II: \(\theta=\)\(\frac{\pi}{6}\) and \(\theta=\pi-\frac{\pi}{6}=\)\(\frac{5\pi}{6}\).
2. The reference angle is \(\frac{\pi}{4}\). The value is negative, so the terminal side is in Quadrant II or III: \(\theta=\pi-\frac{\pi}{4}=\)\(\frac{3\pi}{4}\) and \(\theta=\pi+\frac{\pi}{4}=\)\(\frac{5\pi}{4}\).
3. The reference angle is \(\frac{\pi}{4}\). The value is negative, so the terminal side is in Quadrant II or IV: \(\theta=\pi-\frac{\pi}{4}=\)\(\frac{3\pi}{4}\) and \(\theta=2\pi-\frac{\pi}{4}=\)\(\frac{7\pi}{4}\).
4. The reference angle is \(\frac{\pi}{3}\). The value is negative, so the terminal side is in Quadrant III or IV: \(\theta=\pi+\frac{\pi}{3}=\)\(\frac{4\pi}{3}\) and \(\theta=2\pi-\frac{\pi}{3}=\)\(\frac{5\pi}{3}\).

Problem 13: Even or odd functions

1. \(f(-x)=(-x)\sin(-x)=(-x)(-\sin x)=x\sin x=f(x)\), so even
2. \(g(-x)=(-x)\cos(-x)=-x\cos x=-g(x)\), so odd
3. \(h(\frac{\pi}{4})=\sqrt{2}\) but \(h(-\frac{\pi}{4})=-\frac{\sqrt{2}}{2}+\frac{\sqrt{2}}{2}=0\), which equals neither \(h(\frac{\pi}{4})\) nor \(-h(\frac{\pi}{4})\), so neither

Problem 14: Solutions from a graph

On \([0,2\pi)\), \(\sin x=\frac{1}{2}\) at \(x=\frac{\pi}{6}\) and \(x=\frac{5\pi}{6}\). Sine has period \(2\pi\), so add \(2\pi\): \(\frac{\pi}{6}+2\pi=\frac{13\pi}{6}\) and \(\frac{5\pi}{6}+2\pi=\frac{17\pi}{6}\).

\(x=\frac{\pi}{6},\ \frac{5\pi}{6},\ \frac{13\pi}{6},\ \frac{17\pi}{6}\)

Problem 15: Arc length

1. \(150^{\circ}=\)\(\frac{5\pi}{6}\)
2. \(s=r\theta=12\cdot\frac{5\pi}{6}=\)\(10\pi\approx 31.42\) in
3. \(s=8\cdot 2.5=\)\(20\text{ ft}\)

Precalculus Unit Circle Worksheet Answer Key: problems 16–20

Problem 16: Central angle from arc length

\(\theta=\frac{s}{r}=\frac{9}{6}=\)1.5 radians.

\(1.5\cdot\frac{180^{\circ}}{\pi}\approx\) \(85.94^{\circ}\).

Problem 17: Ferris wheel height

\(\sin\frac{4\pi}{3}=-\frac{\sqrt{3}}{2}\) (reference angle \(\frac{\pi}{3}\), Quadrant III).

\(h=35+30(-\frac{\sqrt{3}}{2})=\)\(35-15\sqrt{3}\approx 9.02\) ft.

Problem 18: Bicycle wheel speed

\(\omega=3\cdot 2\pi=6\pi\approx 18.85\) rad/s.

The radius is 13 in, so \(v=r\omega=13\cdot 6\pi=78\pi\approx\) 245.04 in/s.

\(245.04\cdot\frac{3600}{12\cdot 5280}\approx\) 13.92 mph.

Problem 19: Turntable speed

\(\omega=\frac{100}{3}\cdot 2\pi=\frac{200\pi}{3}\approx\) 209.44 rad/min.

\(v=5\cdot\frac{200\pi}{3}=\frac{1000\pi}{3}\approx\) 1047.20 in/min for a point 5 in from the center.

Problem 20: Windshield wiper

\(140^{\circ}=140\cdot\frac{\pi}{180}=\frac{7\pi}{9}\).

\(s=r\theta=18\cdot\frac{7\pi}{9}=\)\(14\pi\approx 43.98\) in.

Precalculus Unit Circle Worksheet Answer Key: problems 21–23

Problem 21: Lighthouse beam

\(\omega=2\cdot 2\pi=4\pi\approx 12.57\) rad/min.

\(v=5\cdot 4\pi=20\pi\approx 62.83\) mi/min, and \(62.83\cdot 60\approx\) 3769.91 mph.

Problem 22: Running track curve

\(\theta=\frac{s}{r}=\frac{300}{100}=\)3 radians.

\(3\cdot\frac{180^{\circ}}{\pi}\approx\) \(171.89^{\circ}\).

Problem 23: Minute hand travel

Twenty minutes is one third of a turn, so \(\theta=\frac{2\pi}{3}\).

\(s=6\cdot\frac{2\pi}{3}=\)\(4\pi\approx 12.57\) in.

Average speed: \(\frac{4\pi}{20}=\frac{\pi}{5}\approx\) 0.63 in/min.

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