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

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<math>
<math>


\frac{8\pi}{3}\Rightarrow \left(\frac{\sqrt{2}}{2} ,\frac{\sqrt{2}}{2}\right)</math><br><br> </math>
\frac{8\pi}{3}\Rightarrow \left(\frac{\sqrt{2}}{2} ,\frac{\sqrt{2}}{2}\right)</math><br><br>


<math>
<math>
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\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{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}


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

Revision as of 03:37, 26 August 2022



Failed to parse (unknown function "\begin{align}"): {\displaystyle \begin{align} \sin{\left(\frac{5\pi}{6}\right)} &= \frac{1}{2} & \csc{\left(\frac{5\pi}{6}\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] \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} & \cot{\left(\frac{5\pi}{6}\right)} &= -\frac{\sqrt{3}}{1}= -\sqrt{3} \\[2ex] }