8.3E Exercises
- Page ID
- 157151
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\(\newcommand{\avec}{\mathbf a}\) \(\newcommand{\bvec}{\mathbf b}\) \(\newcommand{\cvec}{\mathbf c}\) \(\newcommand{\dvec}{\mathbf d}\) \(\newcommand{\dtil}{\widetilde{\mathbf d}}\) \(\newcommand{\evec}{\mathbf e}\) \(\newcommand{\fvec}{\mathbf f}\) \(\newcommand{\nvec}{\mathbf n}\) \(\newcommand{\pvec}{\mathbf p}\) \(\newcommand{\qvec}{\mathbf q}\) \(\newcommand{\svec}{\mathbf s}\) \(\newcommand{\tvec}{\mathbf t}\) \(\newcommand{\uvec}{\mathbf u}\) \(\newcommand{\vvec}{\mathbf v}\) \(\newcommand{\wvec}{\mathbf w}\) \(\newcommand{\xvec}{\mathbf x}\) \(\newcommand{\yvec}{\mathbf y}\) \(\newcommand{\zvec}{\mathbf z}\) \(\newcommand{\rvec}{\mathbf r}\) \(\newcommand{\mvec}{\mathbf m}\) \(\newcommand{\zerovec}{\mathbf 0}\) \(\newcommand{\onevec}{\mathbf 1}\) \(\newcommand{\real}{\mathbb R}\) \(\newcommand{\twovec}[2]{\left[\begin{array}{r}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\ctwovec}[2]{\left[\begin{array}{c}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\threevec}[3]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\cthreevec}[3]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\fourvec}[4]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\cfourvec}[4]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\fivevec}[5]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\cfivevec}[5]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\mattwo}[4]{\left[\begin{array}{rr}#1 \amp #2 \\ #3 \amp #4 \\ \end{array}\right]}\) \(\newcommand{\laspan}[1]{\text{Span}\{#1\}}\) \(\newcommand{\bcal}{\cal B}\) \(\newcommand{\ccal}{\cal C}\) \(\newcommand{\scal}{\cal S}\) \(\newcommand{\wcal}{\cal W}\) \(\newcommand{\ecal}{\cal E}\) \(\newcommand{\coords}[2]{\left\{#1\right\}_{#2}}\) \(\newcommand{\gray}[1]{\color{gray}{#1}}\) \(\newcommand{\lgray}[1]{\color{lightgray}{#1}}\) \(\newcommand{\rank}{\operatorname{rank}}\) \(\newcommand{\row}{\text{Row}}\) \(\newcommand{\col}{\text{Col}}\) \(\renewcommand{\row}{\text{Row}}\) \(\newcommand{\nul}{\text{Nul}}\) \(\newcommand{\var}{\text{Var}}\) \(\newcommand{\corr}{\text{corr}}\) \(\newcommand{\len}[1]{\left|#1\right|}\) \(\newcommand{\bbar}{\overline{\bvec}}\) \(\newcommand{\bhat}{\widehat{\bvec}}\) \(\newcommand{\bperp}{\bvec^\perp}\) \(\newcommand{\xhat}{\widehat{\xvec}}\) \(\newcommand{\vhat}{\widehat{\vvec}}\) \(\newcommand{\uhat}{\widehat{\uvec}}\) \(\newcommand{\what}{\widehat{\wvec}}\) \(\newcommand{\Sighat}{\widehat{\Sigma}}\) \(\newcommand{\lt}{<}\) \(\newcommand{\gt}{>}\) \(\newcommand{\amp}{&}\) \(\definecolor{fillinmathshade}{gray}{0.9}\)We mostly saw angles between \(0\) and \(2\pi\), a full circle, in this section. However, it's possible to make sense of an angle greater than \(2\pi\), by continuing to loop around the origin to achieve more than \(360^\circ\) of rotation. For example, an angle of \(3\pi\) will make a full circle (\(2\pi\)'s worth) and then continue another \(\pi\) radians to arrive at the location shown below.
Notice that this location also matches the angle of just \(\pi\)... We say that \(3\pi\) and \(\pi\) are coterminal angles. Given an angle greater than \(2\pi\), you can find the coterminal angle between \(0\) and \(2\pi\) by subtracting off copies of \(2\pi\) as many times as possible without going negative. In this example, \(3\pi- 2\pi = \pi\) gives me my coterminal angle. The same can be done for angles greater than \(360^\circ\), by subtracting off copies of \(360^\circ\) until you get a coterminal angle smaller than \(360^\circ\). Find the coterminal angles in this preferred range for the angles below.
- \(5\pi\)
- \( 720^\circ\)
- \( \frac{7\pi}{6}\)
- \( 370^\circ\)
- \( \frac{9\pi}{4}\)
- \( \frac{13\pi}{3}\)
- Answer
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- \( \pi\)
- \( 0^\circ \)
- \( \frac{7\pi}{6}\) (this angle is already between \(0\) and \(2\pi\))
- \(10^\circ\)
- \(\frac{\pi}{4}\)
- \( \frac{\pi}{3}\)
Find the positive angle between \(0\) and \(2\pi\) (or between \(0\) and \(360^\circ\)) that is coterminal with the negative angles below. It can often be helpful to draw a picture! Or, you can convert, for degrees for example, by subtracting: \(360 - \) (size of the negative angle) \(=\) (positive coterminal angle).
- \(-45^\circ\)
- \(-180^\circ\)
- \(-\frac{\pi}{6}\)
- \(-\frac{3\pi}{2}\)
- \(-90^\circ\)
- \( -20^\circ\)
- \( - \frac{5\pi}{4}\)
- \(-\frac{11\pi}{6}\)
- Answer
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- \( 315^\circ\)
- \( 180^\circ\)
- \( \frac{11\pi}{6}\)
- \( \frac{\pi}{2}\)
- \( 270^\circ\)
- \( 340^\circ\)
- \( \frac{3\pi}{4}\)
- \( \frac{\pi}{6}\)
Convert from degrees to radians or vice versa. You can use a calculator, but most of these you should be able to eyeball. Try sketching them on a unit circle!
- \( 180^\circ\)
- \( 0\) rad
- \( 30^\circ\)
- \( 45^\circ\)
- \( \frac{\pi}{3} \)
- \( \frac{3\pi}{2}\)
- \( \frac{4\pi}{3}\)
- \( 225^\circ\)
- Answer
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- \(\pi\)
- \(0^\circ\)
- \( \frac{\pi}{6}\)
- \( \frac{\pi}{4}\)
- \( 60^\circ\)
- \( 270^\circ\)
- \( 240^\circ\)
- \( \frac{5\pi}{4}\)
Recall that to find the reference angle of \(\theta\), you take the most direct route from your angle's location straight to the \(x\)-axis. The reference angle \(\theta'\) is the interior angle of the resulting right triangle. For example:
Find the reference angle for each angle below.
- \( \frac{5\pi}{4}\)
- \( 225^\circ\)
- \( \frac{11\pi}{6}\)
- \( \frac{\pi}{3} \)
- \( 45^\circ\)
- \( \frac{5\pi}{3}\)
- \( 120^\circ\)
- \( \frac{7\pi}{6}\)
- Answer
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- \( \frac{\pi}{4}\)
- \(45^\circ\)
- \( \frac{\pi}{6}\)
- \( \frac{\pi}{3}\) (in the first quadrant, the reference angle just is the angle)
- \( 45^\circ\)
- \( \frac{\pi}{3}\)
- \(60^\circ\)
- \( \frac{\pi}{6}\)
Find \( \sin \theta, \cos \theta,\) and \(\tan \theta\) using the triangles (determine any missing side lengths first). Note: figures are not drawn to scale!
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- Answer
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- \( \sin \theta = \frac{3}{5}, \cos \theta = \frac{4}{5}, \tan \theta = \frac{3}{4} \)
- \( \sin \theta = \frac{1}{\sqrt{5}}, \cos \theta = \frac{2}{\sqrt{5}}, \tan \theta = \frac{1}{2} \)
- \( \sin \theta = \frac{1}{2}, \cos \theta = \frac{\sqrt{3}}{2}, \tan \theta = \frac{1}{\sqrt{3}} \)
- \( \sin (45^\circ) = \frac{1}{\sqrt{2}}, \cos (45^\circ)= \frac{1}{\sqrt{2}}, \tan (45^\circ) = 1 \)
Find \( \sin \theta, \cos \theta,\) and \(\tan \theta\) for the given angles. You will probably need to sketch them on a plane to find the reference angles. Watch out for when triangles' sides should be labeled as negative!
- \( \frac{2\pi}{3}\)
- \( \frac{5\pi}{4}\)
- \( -\frac{\pi}{6}\)
- \( - \frac{5\pi}{6} \)
- \( \frac{ 3\pi}{4}\)
- \( \frac{5\pi}{3}\)
- Answer
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- \( \sin \left( \frac{2\pi}{3} \right) = \frac{\sqrt{3}}{2}, \cos \left( \frac{2\pi}{3} \right) = -\frac{1}{2}, \tan \left( \frac{2\pi}{3} \right) = -\sqrt{3} \)
- \( \sin \left( \frac{5\pi}{4}\right) = -\frac{1}{\sqrt{2}}, \cos \left(\frac{5\pi}{4} \right) = -\frac{1}{\sqrt{2}}, \tan \left( \frac{5\pi}{4} \right) = 1 \)
- \( \sin \left( -\frac{\pi}{6}\right) = -\frac{1}{2}, \cos \left( -\frac{\pi}{6} \right) = \frac{\sqrt{3}}{2}, \tan \left( -\frac{\pi}{6} \right) = -\frac{1}{\sqrt{3}} \)
- \( \sin \left( -\frac{5\pi}{6}\right) = -\frac{1}{2}, \cos \left( -\frac{5\pi}{6} \right) = -\frac{\sqrt{3}}{2}, \tan \left( -\frac{5\pi}{6} \right) = \frac{1}{\sqrt{3}} \)
- \( \sin \left( \frac{ 3\pi}{4}\right) = \frac{1}{\sqrt{2}}, \cos \left(\frac{ 3\pi}{4} \right) = -\frac{1}{\sqrt{2}}, \tan \left( \frac{ 3\pi}{4} \right) = -1 \)
- \( \sin \left( \frac{5\pi}{3}\right) = -\frac{\sqrt{3}}{2}, \cos \left(\frac{5\pi}{3} \right) = \frac{1}{2}, \tan \left( \frac{5\pi}{3} \right) = -\sqrt{3} \)
From the given trig ratio and quadrant, find all six trig ratios for that angle.
- \( \sin \theta = \frac{1}{\sqrt{2}}\), Quadrant I
- \( \cos \theta = \frac{\sqrt{3}}{2}\), Quadrant IV
- \( \tan \theta = 1 \), Quadrant III
- \( \sin \theta = \frac{1}{2} \), Quadrant II
- Answer
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- \( \sin \theta = \frac{1}{\sqrt{2}}, \cos\theta = \frac{1}{\sqrt{2}}, \tan \theta = 1, \csc \theta = \sqrt{2}, \sec \theta = \sqrt{2}, \cot \theta =1 \)
- \( \sin \theta = -\frac{1}{2}, \cos\theta = \frac{\sqrt{3}}{2}, \tan \theta =- \frac{1}{\sqrt{3}}, \csc \theta = -2, \sec \theta = \frac{2}{\sqrt{3}}, \cot \theta = -\sqrt{3} \)
- \( \sin \theta = -\frac{1}{\sqrt{2}}, \cos\theta = -\frac{1}{\sqrt{2}}, \tan \theta =1, \csc \theta = -\sqrt{2}, \sec \theta = -\sqrt{2}, \cot \theta =1 \)
- \( \sin \theta = \frac{1}{2}, \cos\theta = -\frac{\sqrt{3}}{2}, \tan \theta = -\frac{1}{\sqrt{3}}, \csc \theta =2 , \sec \theta = -\frac{2}{\sqrt{3}}, \cot \theta = -\sqrt{3} \)
1. Sketch and label the quadrants of the plane.
2. Identify which quadrant each angle lies in: \(\frac{\pi}{3}, \frac{5\pi}{6}, \frac{5\pi}{4}, \frac{5\pi}{3}\)
3. Determine the range of angle values contained in each quadrant. For example, Quadrant I comprises angles between \(0\) and \(\pi/2\) radians.
4. In each quadrant, certain trig ratios always have negative or positive values. For example, for \(\theta\) in Quadrant III, \(\sin \theta\) is always negative, \( \cos \theta\) is always negative, and \( \tan \theta \) is always positive. Give such facts for the other three quadrants.
- Answer
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1.
2. I, II, III, and IV, respectively.
3. Quadrant I: \(0\) to \(\pi/2\). Quadrant II: \(\pi/2\) to \(\pi\). Quadrant III: \(\pi \) to \(3\pi/2\). Quadrant IV: \(3\pi/2 \) to \(2\pi\).
4. Quadrant I: sine, cosine, and tangent all positive. Quadrant II: sine positive, cosine and tangent negative. Quadrant III: sine and cosine negative, tangent positive. Quadrant IV: sine and tangent negative, cosine positive. You can remember this fact with the mnemonic "All Students Take Calculus," which tells you in order, "All positive, Sine positive, Tangent positive, Cosine positive."