6.2 Trigonometric Functions: Unit Circle Approach/53: Difference between revisions

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\tan{\left(\frac{8\pi}{3}\right)} &= \frac{\frac{\sqrt{3}}{2}}{-\frac{{1}}{2}} = \left(2\right) = \frac{\sqrt{3}}{-1} = -\sqrt{3}  
\tan{\left(\frac{8\pi}{3}\right)} &= \frac{\frac{\sqrt{3}}{2}}{-\frac{{1}}{2}} = \left(2\right) = \frac{\sqrt{3}}{-1} = -\sqrt{3}  


& \cot{\left(\frac{8\pi}{3}\right)} &= \frac{-\frac{1}{2}} {\frac{\sqrt{3}{2} = -\sqrt{3} \\[2ex]  \end{align} </math>
& \cot{\left(\frac{8\pi}{3}\right)} &= \frac{-\frac{1}{2}} {\frac{\sqrt{3}}{2} = -\sqrt{3} \\[2ex]  \end{align} </math>

Revision as of 04:03, 26 August 2022



Failed to parse (unknown function "\begin{align}"): {\displaystyle \begin{align} \sin{\left(\frac{8\pi}{3}\right)} &= \frac{\sqrt{3}}{2} & \csc{\left(\frac{8\pi}{3}\right)} &= {\frac{2}{\sqrt{3}} \cdot \sqrt{3}}= \frac{2\sqrt{3}}{3} \\[2ex] \cos{\left(\frac{8\pi}{3}\right)} &= -\frac{1}{2} & \sec{\left(\frac{8\pi}{3}\right)} &= -\frac{{2}}{1} = -2 \\[2ex] \tan{\left(\frac{8\pi}{3}\right)} &= \frac{\frac{\sqrt{3}}{2}}{-\frac{{1}}{2}} = \left(2\right) = \frac{\sqrt{3}}{-1} = -\sqrt{3} & \cot{\left(\frac{8\pi}{3}\right)} &= \frac{-\frac{1}{2}} {\frac{\sqrt{3}}{2} = -\sqrt{3} \\[2ex] \end{align} }