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Radians and the Unit Circle: Algebra 2 Math Worksheet Answer Key – PDF Download

    This algebra 2 trig worksheet answer key solves all 23 problems of the radians and the unit circle worksheet step by step.

    Each solution names the reference angle or the identity used, then gives the exact answer in bold. Compare your steps as well as your results, and check signs by quadrant.

    Algebra 2Common Core: F-TF.A.1 · F-TF.A.2 · F-TF.C.8

    Algebra 2 Trig Worksheet Answer Key: problems 1–5

    Problem 1: Degrees to radians

    1. \(45^{\circ}\cdot\frac{\pi}{180}=\)\(\frac{\pi}{4}\).
    2. \(120^{\circ}\cdot\frac{\pi}{180}=\)\(\frac{2\pi}{3}\).
    3. \(270^{\circ}\cdot\frac{\pi}{180}=\)\(\frac{3\pi}{2}\).
    4. \(15^{\circ}\cdot\frac{\pi}{180}=\)\(\frac{\pi}{12}\).

    Problem 2: Radians to degrees

    1. \(\frac{\pi}{6}\cdot\frac{180^{\circ}}{\pi}=\)\(30^{\circ}\).
    2. \(\frac{5\pi}{4}\cdot\frac{180^{\circ}}{\pi}=\)\(225^{\circ}\).
    3. \(\frac{7\pi}{3}\cdot\frac{180^{\circ}}{\pi}=\)\(420^{\circ}\).
    4. \(-\frac{\pi}{2}\cdot\frac{180^{\circ}}{\pi}=\)\(-90^{\circ}\).

    Problem 3: Arc length

    1. \(s=r\theta=6\cdot\frac{\pi}{3}=\)\(2\pi\) in, about 6.28 in.
    2. \(s=4(2.5)=\)10 ft.

    Problem 4: Angle from an arc

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

    \(1.5\cdot\frac{180^{\circ}}{\pi}\approx \)\(85.9^{\circ}\).

    Problem 5: Locating the quadrant

    1. \(\frac{5\pi}{6}\) is between \(\frac{\pi}{2}\) and \(\pi\): Quadrant II.
    2. \(\frac{4\pi}{3}\) is between \(\pi\) and \(\frac{3\pi}{2}\): Quadrant III.
    3. \(\frac{7\pi}{4}\) is between \(\frac{3\pi}{2}\) and \(2\pi\): Quadrant IV.
    4. \(\frac{\pi}{3}\) is between \(0\) and \(\frac{\pi}{2}\): Quadrant I.

    Algebra 2 Trig Worksheet Answer Key: problems 6–10

    Problem 6: Special angle values

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

    Problem 7: Unit circle coordinates

    1−11−1(−1/2, √3/2)2π/3(−√2/2, −√2/2)5π/4(√3/2, −1/2)11π/6
    The three points on the unit circle.
    1. Reference angle \(\frac{\pi}{3}\) in Quadrant II (x negative, y positive): \((-\frac{1}{2},\ \frac{\sqrt{3}}{2})\).
    2. Reference angle \(\frac{\pi}{4}\) in Quadrant III (both negative): \((-\frac{\sqrt{2}}{2},\ -\frac{\sqrt{2}}{2})\).
    3. Reference angle \(\frac{\pi}{6}\) in Quadrant IV (x positive, y negative): \((\frac{\sqrt{3}}{2},\ -\frac{1}{2})\).

    Problem 8: Reference angles

    1. \(\frac{5\pi}{6}\) is in Quadrant II: \(\pi-\frac{5\pi}{6}\) \(=\) \(\frac{\pi}{6}\).
    2. \(\frac{7\pi}{6}\) is in Quadrant III: \(\frac{7\pi}{6}-\pi\) \(=\) \(\frac{\pi}{6}\).
    3. \(\frac{5\pi}{3}\) is in Quadrant IV: \(2\pi-\frac{5\pi}{3}\) \(=\) \(\frac{\pi}{3}\).
    4. \(\frac{3\pi}{4}\) is in Quadrant II: \(\pi-\frac{3\pi}{4}\) \(=\) \(\frac{\pi}{4}\).

    Problem 9: Quadrantal angles

    \(\theta\) \(\sin\theta\) \(\cos\theta\) \(\tan\theta\)
    \(0\) \(0\) \(1\) \(0\)
    \(\frac{\pi}{2}\) \(1\) \(0\) undefined
    \(\pi\) \(0\) \(-1\) \(0\)
    \(\frac{3\pi}{2}\) \(-1\) \(0\) undefined

    The points are \((1,0)\), \((0,1)\), \((-1,0)\), \((0,-1)\). Tangent is \(\frac{y}{x}\), so it is undefined when \(x=0\).

    Problem 10: Evaluating in other quadrants

    1. \(\sin\frac{7\pi}{6}\): Quadrant III, reference \(\frac{\pi}{6}\), sine negative: \(-\frac{1}{2}\).
    2. \(\cos\frac{5\pi}{3}\): Quadrant IV, reference \(\frac{\pi}{3}\), cosine positive: \(\frac{1}{2}\).
    3. \(\tan\frac{3\pi}{4}\): Quadrant II, reference \(\frac{\pi}{4}\), tangent negative: \(-1\).

    Algebra 2 Trig Worksheet Answer Key: problems 11–15

    Problem 11: Negative and large angles

    1. \(-\frac{\pi}{3}\) ends in Quadrant IV with reference \(\frac{\pi}{3}\): \(\frac{1}{2}\).
    2. \(-\frac{\pi}{4}\) ends in Quadrant IV: \(-\frac{\sqrt{2}}{2}\).
    3. \(\frac{13\pi}{6}=2\pi+\frac{\pi}{6}\), so it matches \(\frac{\pi}{6}\): \(\frac{1}{2}\).
    4. \(\frac{9\pi}{4}=2\pi+\frac{\pi}{4}\), so it matches \(\frac{\pi}{4}\): \(\frac{\sqrt{2}}{2}\).

    Problem 12: Solving for the angle

    1. Reference angle \(\frac{\pi}{3}\). Sine is positive in Quadrants I and II: \(\theta=\frac{\pi}{3},\ \frac{2\pi}{3}\).
    2. Reference angle \(\frac{\pi}{4}\). Cosine is negative in Quadrants II and III: \(\theta=\frac{3\pi}{4},\ \frac{5\pi}{4}\).

    Problem 13: Checking the identity

    1. \((\frac{1}{2})^{2}+(\frac{\sqrt{3}}{2})^{2}=\frac{1}{4}+\frac{3}{4}=\)\(1\).
    2. \((-\frac{\sqrt{2}}{2})^{2}+(-\frac{\sqrt{2}}{2})^{2}=\frac{2}{4}+\frac{2}{4}=\)\(1\).

    Problem 14: Sine and tangent from cosine

    1−11−1(5/13, −12/13)
    The point \((\frac{5}{13},\ -\frac{12}{13})\) is on the unit circle.

    \(\sin^{2}\theta=1-\frac{25}{169}=\frac{144}{169}\). Quadrant IV has negative sine: \(\sin\theta=-\frac{12}{13}\).

    \(\tan\theta=\frac{-12/13}{5/13}=\)\(-\frac{12}{5}\).

    Problem 15: Sine from cosine

    \(\sin^{2}\theta=1-\frac{25}{169}=\frac{144}{169}\). Sine is negative in Quadrant IV: \(\sin\theta=-\frac{12}{13}\).

    \(\tan\theta=\frac{-12/13}{5/13}=\)\(-\frac{12}{5}\).

    Algebra 2 Trig Worksheet Answer Key: problems 16–20

    Problem 16: Using the tangent

    Write \(\sin\theta=\frac{4}{3}\cos\theta\) and substitute: \(\frac{16}{9}\cos^{2}\theta+\cos^{2}\theta=1\), so \(\frac{25}{9}\cos^{2}\theta=1\) and \(\cos^{2}\theta=\frac{9}{25}\).

    Both are negative in Quadrant III: \(\cos\theta=-\frac{3}{5}\), then \(\sin\theta=\frac{4}{3}(-\frac{3}{5})=-\frac{4}{5}\).

    Problem 17: Radical values

    \(\sin^{2}\theta=1-\frac{5}{9}=\frac{4}{9}\). Sine is negative in Quadrant III: \(\sin\theta=-\frac{2}{3}\).

    \(\tan\theta=\frac{-2/3}{-\sqrt{5}/3}=\frac{2}{\sqrt{5}}=\)\(\frac{2\sqrt{5}}{5}\).

    Problem 18: Ferris wheel height

    1. \(30+25\cdot\frac{1}{2}=\)42.5 ft.
    2. \(30+25(1)=\)55 ft.
    3. \(30+25(-\frac{\sqrt{2}}{2})=30-12.5\sqrt{2}\approx \)12.3 ft.
    4. \(30+25(-1)=\)5 ft.

    Problem 19: Clock hand

    1. \(\frac{1}{3}\cdot 2\pi=\)\(\frac{2\pi}{3}\) radians.
    2. \(s=r\theta=5\cdot\frac{2\pi}{3}=\)\(\frac{10\pi}{3}\) in, about 10.47 in.

    Problem 20: Pizza slice

    1. \(\frac{\pi}{4}\) radians.
    2. \(s=8\cdot\frac{\pi}{4}=2\pi\approx \)6.28 in.

    Algebra 2 Trig Worksheet Answer Key: problems 21–23

    Problem 21: Bicycle wheel

    1. \(3\cdot 2\pi=\)\(6\pi\) radians.
    2. \(s=13\cdot 6\pi=78\pi\approx \)245.04 in\(,\ \frac{245.04}{12}\approx \)20.42 ft.

    Problem 22: Point on the circle

    1−11−1(−12/13, −5/13)
    The point \(P\) in Quadrant III.

    \(x^{2}+y^{2}=1\) gives \(y^{2}=1-\frac{144}{169}=\frac{25}{169}\). In Quadrant III \(y=-\frac{5}{13}\).

    \(\sin\theta=-\frac{5}{13}\), \(\cos\theta=-\frac{12}{13}\), \(\tan\theta=\frac{-5/13}{-12/13}=\)\(\frac{5}{12}\).

    Problem 23: When is the rider low

    \(30+25\sin\theta=17.5\) gives \(25\sin\theta=-12.5\), so \(\sin\theta=-\frac{1}{2}\).

    Reference angle \(\frac{\pi}{6}\); sine is negative in Quadrants III and IV: \(\theta=\frac{7\pi}{6},\ \frac{11\pi}{6}\).

    How to use this algebra 2 trig worksheet answer key

    Read this algebra 2 trig worksheet answer key after trying each problem yourself, and redo any problem you missed.

    Next step after this algebra 2 trig worksheet answer key: Radians and the Unit Circle test

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    Algebra 2 Radians and the Unit Circle worksheet answer key

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