From Burton Tech. Points Wiki
Jump to navigation
Jump to search
|
|
| Line 8: |
Line 8: |
| \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}{1} = 2\\[2ex] |
|
| |
|
| \tan{\left(\frac{5\pi}{6}\right)} &= \frac{\frac{1}{2}}{-\frac{\sqrt{3}}{2}} = \frac{1}{2}\cdot\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)\cdot\left(-\frac{2}{\sqrt{3}}\right) = -\frac{1}{\sqrt{3}}\cdot\frac{\sqrt{3}}{\sqrt{3}} = -\frac{\sqrt{3}}{3} |
|
| |
|
|
| |
|
Revision as of 22:30, 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}{\sqrt {3}}}\cdot {\frac {\sqrt {3}}{\sqrt {3}}}={\frac {2{\sqrt {3}}}{3}}\\[2ex]\cos {\left({\frac {5\pi }{6}}\right)}&={\frac {-{\sqrt {3}}}{2}}&\sec {\left({\frac {5\pi }{6}}\right)}&={\frac {2}{1}}=2\\[2ex]\tan {\left({\frac {5\pi }{6}}\right)}&={\frac {\frac {1}{2}}{-{\frac {\sqrt {3}}{2}}}}=\left({\frac {1}{2}}\right)\cdot \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}}}](https://en.wikipedia.org/api/rest_v1/media/math/render/svg/21296c01118d2966e031e04816441bfdfab7438e)