6.2 Trigonometric Functions: Unit Circle Approach/48: Difference between revisions
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(Created page with "<math>\frac{5\pi}{6} \Rightarrow (\frac{-\sqrt{3}}{2}, \frac{1}{2})</math> <math> \begin{align} \sin{(t)} &= -\frac{\sqrt{3}}{2} & \csc{(t)} &= -\frac{2}{\sqrt{3}}\cdot\frac{\sqrt{3}}{\sqrt{3}}=\frac{2\sqrt{3}}{3}\\[2ex] \cos{(t)} &= \frac{1}{2} & \sec{(t)} &= \frac{2}{1} = 2\\[2ex] \tan{(t)} &= \frac{-\frac{\sqrt{3}}{2}}{\frac{1}{2}} = -\frac{\sqrt{3}}{2}\cdot\frac{2}{1} = -\sqrt{3} & \cot{(t)} &= -\frac{1}{\sqrt{3}}=-\frac{1}{\sqrt{3}}\cdot\frac{\sqrt{3}}{\s...") |
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<math>\frac{5\pi}{6} \Rightarrow (\frac{-\sqrt{3}}{2}, \frac{1}{2})</math> | <math>\frac{5\pi}{6} \Rightarrow \left(\frac{-\sqrt{3}}{2}, \frac{1}{2}\right)</math><br><br> | ||
<math> | <math> | ||
\begin{align} | \begin{align} | ||
\sin{( | \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] | |||
\end{align} | \end{align} | ||
</math> | </math> | ||
Latest revision as of 22:36, 25 August 2022
![{\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}}}}\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]\end{aligned}}}](https://en.wikipedia.org/api/rest_v1/media/math/render/svg/e5f387867e199026604dc845372bee26c4d5ce65)