6.2 Trigonometric Functions: Unit Circle Approach/48: Difference between revisions

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\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{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}{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}}\\[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{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}

Revision as of 22:32, 25 August 2022



Failed to parse (Conversion error. Server ("https://en.wikipedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle {\begin{aligned}\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}}}\\[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 {1}{\sqrt {3}}}=-{\frac {1}{\sqrt {3}}}\cdot {\frac {\sqrt {3}}{\sqrt {3}}}={\frac {\sqrt {3}}{3}}\\[2ex]\end{aligned}}}