Based on the unit circle shown, Josiah claims that \( \sin \left(\frac{5 \pi}{6}\right)=-\frac{\sqrt{3}}{2} \). Is Josiah correct? Use the drop-down menus to explain.
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The angle \( \frac{5\pi}{6} \) is in the second quadrant. Show more…
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Explain the mistake that is made. Use the unit circle to evaluate $\tan \left(\frac{5 \pi}{6}\right)$ exactly. $\begin{array}{l}\text { Tangent is the } \\ \text { ratio of sine } \\ \text { to cosine. }\end{array} \tan \left(\frac{5 \pi}{6}\right)=\frac{\sin \left(\frac{5 \pi}{6}\right)}{\cos \left(\frac{5 \pi}{6}\right)}$ Use the unit circle to $\begin{array}{l}\text { identify sine } \\ \text { and cosine. }\end{array} \quad \sin \left(\frac{5 \pi}{6}\right)=-\frac{\sqrt{3}}{2}$ and $\cos \left(\frac{5 \pi}{6}\right)=\frac{1}{2}$ Substitute values for $\begin{array}{l}\text { sine and } \\ \text { cosine. }\end{array} \quad \tan \left(\frac{5 \pi}{6}\right)=\frac{-(\sqrt{3} / 2)}{1 / 2}$ Simplify. $\tan \left(\frac{5 \pi}{6}\right)=-\sqrt{3}$ This is incorrect. What mistake was made?
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If $\sin ^{-1}\left(\sin \frac{\pi}{3}\right)=\frac{\pi}{3},$ is $\sin ^{-1}\left(\sin \frac{5 \pi}{6}\right)=\frac{5 \pi}{6} ?$ Explain your answer.
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In Exercises 69 and $70,$ explain the mistake that is made. Use the unit circle to evaluate $\tan \left(\frac{5 \pi}{6}\right)$ exactly. Solution: $\begin{array}{l}\text { Tangent is the ratio } \\ \text { of sine to cosine. }\end{array} \tan \left(\frac{5 \pi}{6}\right)=\frac{\sin \left(\frac{5 \pi}{6}\right)}{\cos \left(\frac{5 \pi}{6}\right)}$ Use the unit circle to $\begin{array}{l}\text { identify sine } \\ \text { and cosine. }\end{array} \quad \sin \left(\frac{5 \pi}{6}\right)=-\frac{\sqrt{3}}{2}$ and $\cos \left(\frac{5 \pi}{6}\right)=\frac{1}{2}$ Substitute values for ${\begin{array}{l}\text { sine and } \\ \text { cosine. }\end{array}} \quad \tan \left(\frac{5 \pi}{6}\right)=\frac{-\frac{\sqrt{3}}{2}}{\frac{1}{2}}$ Simplify. $\tan \left(\frac{5 \pi}{6}\right)=-\sqrt{3}$ This is incorrect. What mistake was made?
DEFINITION 3 OF TRIGONOMETRIC FUNCTIONS: UNIT CIRCLE APPROACH
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