6.2 Trigonometric Functions: Unit Circle Approach/63: Difference between revisions

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\begin{align}
\begin{align}


\sin{\left(\frac{5\pi}{6}\right)} &= \frac{1}{2} & \csc{\left(\frac{5\pi}{6}\right)} &= \frac{2}{1}=2\\[2ex]
\sin{\left(\frac{-14\pi}{3}\right)} &= \frac{1}{2} & \csc{\left(\frac{-14\pi}{3}\right)} &= \frac{2}{1}=2\\[2ex]


\cos{\left(\frac{5\pi}{6}\right)} &= \frac{-\sqrt{3}}{2} & \sec{\left(\frac{5\pi}{6}\right)} &= \frac{{2}}{-\sqrt{3}}\cdot\frac{\sqrt{3}}{\sqrt{3}} = -\frac{2\sqrt{3}}{3}\\[2ex]  
\cos{\left(\frac{-14\pi}{3}\right)} &= \frac{-\sqrt{3}}{2} & \sec{\left(\frac{-14\pi}{3}\right)} &= \frac{{2}}{-\sqrt{3}}\cdot\frac{\sqrt{3}}{\sqrt{3}} = -\frac{2\sqrt{3}}{3}\\[2ex]  


\tan{\left(\frac{5\pi}{6}\right)} &= \frac{\frac{1}{2}}{-\frac{\sqrt{3}}{2}} = \left(\frac{1}{2}\right)\left(-\frac{2}{\sqrt{3}}\right) = -\frac{1}{\sqrt{3}}\cdot\frac{\sqrt{3}}{\sqrt{3}} = -\frac{\sqrt{3}}{3}
\tan{\left(\frac{-14\pi}{3}\right)} &= \frac{\frac{1}{2}}{-\frac{\sqrt{3}}{2}} = \left(\frac{1}{2}\right)\left(-\frac{2}{\sqrt{3}}\right) = -\frac{1}{\sqrt{3}}\cdot\frac{\sqrt{3}}{\sqrt{3}} = -\frac{\sqrt{3}}{3}


& \cot{\left(\frac{5\pi}{6}\right)} &= -\frac{\sqrt{3}}{1}= -\sqrt{3} \\[2ex]
& \cot{\left(\frac{-14\pi}{3}\right)} &= -\frac{\sqrt{3}}{1}= -\sqrt{3} \\[2ex]


\end{align}
\end{align}
</math>
</math>

Revision as of 15:57, 29 August 2022