Chapter 2.7 Practice

Worked Out Questions

  1. Calculate \(\angle pOq\) where \(p = (\tfrac{1}{6}, \tfrac{\sqrt{35}}{6}) \quad{\rm and} \quad q = (\tfrac{1}{2}, \tfrac{\sqrt{3}}{2}).\)

    Answer

    The definition of the angle \(\angle pOq\) is \(p^{-1}\star q.\) So \[\begin{align*} \angle pOq&=\left(\tfrac{1}{6}, -\tfrac{\sqrt{35}}{6}\right)\star\left(\tfrac{1}{2}, \tfrac{\sqrt{3}}{2}\right)\\ &=\left(\frac{1+3\sqrt{25}}{12},\frac{\sqrt{3}-\sqrt{35}}{12}\right). \end{align*}\]

  2. Find three radian measures that correspond to the point on the unit circle with principle angle measure equal to \(\frac{3\pi}{8}.\)

    Answer

    Recall that one full rotation is \(2\pi\) in radians. So here are three radian measures that correspond to the point on the unit circle with principle angle measure equal to \(\frac{3\pi}{8}:\)

    • \(\frac{3\pi}{8}+2\pi=\frac{19\pi}{8}\)
    • \(\frac{3\pi}{8}+4\pi=\frac{35\pi}{8}\)
    • \(\frac{3\pi}{8}-2\pi=-\frac{13\pi}{8}\)

  3. Convert \(222^{\circ}\) into radians.

    Answer

    Recall that a full circle is measured in \(2\pi\) in radians and \(360^{\circ}\) in degrees. So \[2\pi=360^{\circ}=\text{full fraction of a circle}\] implies that \[\frac{2\pi}{360^{\circ}}=1.\] Thus, \(222^{\circ}\) can be converted to radians like this: \[\begin{align*} 222^{\circ}&=222^{\circ}\cdot \frac{2\pi}{360^{\circ}}\\ &=\frac{37\pi}{30}. \end{align*}\]

  4. Convert \(\frac{2\pi}{3}\) into degrees.

    Answer

    Recall that a full circle is measured in \(2\pi\) in radians and \(360^{\circ}\) in degrees. So \[2\pi=360^{\circ}=\text{full fraction of a circle}\] implies that \[\frac{360^{\circ}}{2\pi}=1.\] Thus, \(\frac{2\pi}{3}\) can be converted to degrees like this: \[\begin{align*} \frac{2\pi}{3}&=\frac{2\pi}{3}\cdot \frac{360^{\circ}}{2\pi}\\ &=120^{\circ}. \end{align*}\]

  5. There are \(105\) \(C\) radians in a circle. Determine the \(C\) radian measure of angle that is \(\frac{20}{30}\) radians. Also determine the degree measure of an angle that is 25 \(C\) radians.

    Answer

    Recall that a full circle is measured in \(2\pi\) in radians and it is given that there are \(150\) \(C\) radians in a circle. So \[2\pi=150\,C=\text{full fraction of a circle}\] implies that \[\frac{150\,C}{2\pi}=1.\] Thus, \(\frac{20}{30}\) radians can be converted to \(C\) radians like this: \[\begin{align*} \frac{20}{30}&=\frac{20}{30}\cdot \frac{105\,C}{2\pi}\\ &=\frac{35}{\pi}\,C. \end{align*}\] The degree measure of an angle that is \(25\,C\) radians can be determined like this: \[\begin{align*} 25\,C&=25\,C\cdot \frac{360^{\circ}}{105\,C}\\ &=\frac{600}{7}^{\circ}. \end{align*}\]

  6. There are angles \(A\) in quadrant II and \(B\) in quadrant III so that \(\sin\left(A\right)=\frac{3}{11}\) and \(\cos(B)=-\frac{1}{10}.\) Determine \(\cos(A+B)\) and \(\tan(A+B).\)

    Answer

    Any angle \(\theta\) can be represented by point \((\cos(\theta),\sin(\theta))\) where \[\cos^2(\theta)+\sin^2(\theta)=1.\] This means that \[\begin{align*} \sin^2(A)+\cos^2(A)&=1\\ \left(\frac{3}{11}\right)^2+\cos^2(A)&=1 \end{align*}\] and \[\begin{align*} \sin^2(B)+\cos^2(B)&=1\\ \sin^2(A)+\left(-\frac{1}{10}\right)^2&=1. \end{align*}\] Solve each equation to get \[\cos(A)=-\frac{4\sqrt{7}}{11}\quad\text{and}\quad\sin(B)=-\frac{3\sqrt{11}}{10}\] where \(\cos(A)\) is negative because \(A\) is in quadrant II and \(\sin(B)\) is negative because \(B\) is in quadrant III. Now use the angle addition formula to obtain: \[\begin{align*} \cos(A+B)&=\cos(A)\cos(B)-\sin(A)\sin(B)\\ &=\left(-\frac{4\sqrt{7}}{11}\right)\cdot\left(-\frac{1}{10}\right)-\left(\frac{3}{11}\right)\cdot\left(-\frac{3\sqrt{11}}{10}\right)\\ &=\frac{4\sqrt{7}+9\sqrt{11}}{110}. \end{align*}\] For \(\tan(A+B)\), calculate it by using \[\tan(A+B)=\frac{\sin(A+B)}{\cos(A+B)}\] , which means calculating \(\sin(A+B)\) first then using the formula, or calculate it by using the formula \[\tan(A+B)=\frac{\tan(A)+\tan(B)}{1-\tan(A)\tan(B)}.\] Calculate to get \[\begin{align*}\tan(A+B)&=\frac{\sin(A+B)}{\cos(A+B)}\\ &=\frac{\frac{12\sqrt{77}-3}{110}}{\frac{4\sqrt{7}+9\sqrt{11}}{110}}\\ &=\frac{12\sqrt{77}-3}{4\sqrt{7}+9\sqrt{11}}. \end{align*}\]

  7. Calculate sine, cosine, and tangent at the following angles \(\theta=135^{\circ}\), \(\theta=\frac{5\pi}{4}.\)

    Answer

    \[\sin(135^{\circ})=\frac{\sqrt{2}}{2},\quad \cos(135^{\circ})=\frac{\sqrt{2}}{2},\quad \tan(135^{\circ})=-1\] and \[\sin\left(\frac{5\pi}{4}\right)=-\frac{\sqrt{2}}{2},\quad \cos\left(\frac{5\pi}{4}\right)=-\frac{\sqrt{2}}{2},\quad \tan\left(\frac{5\pi}{4}\right)=1.\]

  8. Use angle addition formula to calculate \(\sin(165^{\circ})\) by looking at the unit circle standard angles and seeing which angles add up to \(165\) degrees.

    Answer

    The angle \(165^{\circ}\) can decompose as \[165^{\circ}=120^{\circ}+45^{\circ}\quad\text{or}\quad 165^{\circ}=135^{\circ}+30^{\circ}.\] Use either decomposition and \[\sin(A+B)=\cos(A)\sin(B)+\cos(B)\sin(A)\] to get that \[\sin(165^{\circ})=\frac{\sqrt{6}-\sqrt{2}}{4}.\]

  9. Use the half angle formula to determine \(\sin(45^{\circ})\) and \(\sin(22.5^{\circ}).\)

    Answer

    The half angle formula states that \[\sin\left(\frac{\theta}{2}\right)=\begin{cases}\sqrt{\frac{1-\cos(\theta)}{2}}&\text{if }\frac{\theta}{2}\text{ in I or II}\\-\sqrt{\frac{1-\cos(\theta)}{2}}&\text{if }\frac{\theta}{2}\text{ in III or IV}\end{cases}.\] Because \(\frac{90^{\circ}}{2}=45^{\circ}\), we have that \[\begin{align*} \sin\left(45^{\circ}\right)&=\sin\left(\frac{90^{\circ}}{2}\right)\\ &=\sqrt{\frac{1-\cos(90^{\circ})}{2}}\\ &=\sqrt{\frac{1}{2}}\\ &=\frac{1}{\sqrt{2}}. \end{align*}\] Because \(\frac{45^{\circ}}{2}=22.5^{\circ}\), we have that \[\begin{align*} \sin\left(22.5^{\circ}\right)&=\sin\left(\frac{45^{\circ}}{2}\right)\\ &=\sqrt{\frac{1-\cos(45^{\circ})}{2}}\\ &=\sqrt{\frac{1-\frac{1}{\sqrt{2}}}{2}}\\ &=\sqrt{\frac{\sqrt{2}-1}{2\sqrt{2}}}\\ &=\frac{\sqrt{\sqrt{2}-1}}{\sqrt{2\sqrt{2}}}. \end{align*}\]

  10. What is \(R_\theta(3,5)?\)

    Answer

    \[R_{\theta}(3,5)=(\cos(\theta),\sin(\theta))\star(3,5)=(3\cos(\theta)-5\sin(\theta),5\cos(\theta)+3\sin(\theta)),\] which produces the point obtained after rotating the point \((3,5)\) about \((0,0)\) by the angle \(\theta.\)

Questions

  1. Convert the following degree measures of angles into radian measures of the angles. Make sure your answers are in \([0,2\pi)\).
    1. \(268^{\circ}\)
    2. \(15^{\circ}\)
    3. \(-55^{\circ}\)
    4. \(-450^{\circ}\)
  2. Convert the following radian measures of angles into degree measures of the angles. Make sure your answers are in \([0,360).\)
    1. \(\frac{7\pi}{8}\) radians
    2. \(\frac{8\pi}{3}\) radians
    3. \(\frac{241\pi}{240}\) radians
    4. \(3\) radians
  3. For each of these points, determine all values \(t\) so that the point is in the unit circle.
    1. \(\left(t,1\right)\)
    2. \(\left(-1,t\right)\)
    3. \(\left(\frac{1}{6},t\right)\)
    4. \(\left(t,-\frac{5}{7}\right)\)
    5. \(\left(\frac{\sqrt{3}}{11},t\right)\)
    6. \(\left(t,-\frac{2\sqrt{5}}{9}\right)\)
  4. For each angle \(A\) and \(B\) given, calculate \(\sin(A+B),\sin(A-B),\cos(A+B),\cos(A-B),\tan(A+B),\) and \(\tan(A-B).\)
    1. angle \(A\) is in quadrant I and angle \(B\) is in quadrant I so that \(\sin(A)=\frac{3}{7}\) and \(\cos(B)=\frac{1}{9}\)
    2. angle \(A\) is in quadrant II and angle \(B\) is in quadrant IV so that \(\sin(A)=\frac{2}{5}\) and \(\cos(B)=\frac{4}{5}\)
    3. angle \(A\) is in quadrant III and angle \(B\) is in quadrant IV so that \(\cos(A)=-\frac{\sqrt{5}}{11}\) and \(\sin(B)=-\frac{4}{5}\)

Answers

    1. \(\frac{67\pi}{45}\)
    2. \(\frac{\pi}{12}\)
    3. \(-\frac{61}{36}\)
    4. \(\frac{3\pi}{2}\)
    1. \(\left(157.5\right)^{\circ}\)
    2. \(480^{\circ}\)
    3. \(180.75^{\circ}\)
    4. \(\frac{540}{\pi}\)
    1. \(\left(0,1\right)\)
    2. \(\left(-1,0\right)\)
    3. \(\left(\frac{1}{6},\frac{\sqrt{35}}{6}\right)\) and \(\left(\frac{1}{6},-\frac{\sqrt{35}}{6}\right)\)
    4. \(\left(\frac{2\sqrt{6}}{7},-\frac{5}{7}\right)\) and \(\left(-\frac{2\sqrt{6}}{7},-\frac{5}{7}\right)\)
    5. \(\left(\frac{\sqrt{3}}{11},\frac{\sqrt{118}}{11}\right)\) and \(\left(\frac{\sqrt{3}}{11},-\frac{\sqrt{118}}{11}\right)\)
    6. \(\left(\frac{\sqrt{61}}{9},-\frac{2\sqrt{5}}{9},\right)\) and \(\left(-\frac{\sqrt{61}}{9},-\frac{2\sqrt{5}}{9}\right)\)
    1. \[\sin(A+B)=\frac{3+\sqrt{3200}}{63},\quad \cos(A+B)=\frac{\sqrt{40}-3\sqrt{80}}{63}\quad \text{and}\quad \tan(A+B)=\frac{3+\sqrt{3200}}{\sqrt{40}-3\sqrt{80}}\] \[\sin(A-B)=\frac{\sqrt{3200}-3}{63} ,\quad \cos(A-B)=\frac{\sqrt{40}+3\sqrt{80}}{63}\quad\text{and}\quad \tan(A-B)=\frac{\sqrt{3200}-3}{\sqrt{40}+3\sqrt{80}}\]
    2. \[\sin(A+B)=\frac{8+3\sqrt{21}}{25},\quad \cos(A+B)=\frac{-4\sqrt{21}+6}{25}\quad \text{and}\quad \tan(A+B)=\frac{8+3\sqrt{21}}{-4\sqrt{21}+6}\] \[\sin(A-B)=\frac{3\sqrt{21}-8}{25},\quad \cos(A-B)=\frac{-4\sqrt{21}-6}{25}\quad \text{and}\quad \tan(A-B)=\frac{3\sqrt{21}-8}{-4\sqrt{21}-6}\]
    3. \[\sin(A+B)=\frac{4\sqrt{5}+3\sqrt{116}}{55},\quad \cos(A+B)=\frac{-3\sqrt{5}+4\sqrt{116}}{55}\quad \text{and}\quad \tan(A+B)=\frac{4\sqrt{5}+3\sqrt{116}}{-3\sqrt{5}+4\sqrt{116}}\] \[\sin(A-B)=\frac{4\sqrt{5}-3\sqrt{116}}{55},\quad \cos(A+B)=\frac{-3\sqrt{5}-4\sqrt{116}}{55}\quad \text{and}\quad \tan(A-B)=\frac{4\sqrt{5}-3\sqrt{116}}{-3\sqrt{5}-4\sqrt{116}}\]

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