From Burton Tech. Points Wiki
Jump to navigation
Jump to search
|
|
| (79 intermediate revisions by the same user not shown) |
| Line 1: |
Line 1: |
| <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{1}{2} ,\frac{\sqrt{3}}{2}\right)</math><br><br> |
|
| |
|
| <math> | | <math> |
| \begin{align} | | \begin{align} |
|
| |
|
| \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{8\pi}{3}\right)} &= \frac{\sqrt{3}}{2} & \csc{\left(\frac{8\pi}{3}\right)} &= \frac{1}{\frac{\sqrt{3}}{\cancel2}} \cdot \cancel{2} = \frac{2}{\sqrt{3}} \cdot \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}}{-\sqrt{3}}\cdot\frac{\sqrt{3}}{\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{1}{-\frac{1}{\cancel{2}}} \cdot \cancel{2} = -\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)\left(-\frac{2}{\sqrt{3}}\right) = -\frac{1}{\sqrt{3}}\cdot\frac{\sqrt{3}}{\sqrt{3}} = -\frac{\sqrt{3}}{3} | | \tan{\left(\frac{8\pi}{3}\right)} &= \frac{\frac{\sqrt{3}}{\cancel 2}}{-\frac{{1}}{\cancel 2}} \cdot \cancel{2} = \frac{\sqrt{3}}{-1} = -\sqrt{3} |
|
| |
|
| & \cot{\left(\frac{5\pi}{6}\right)} &= -\frac{\sqrt{3}}{1}= -\sqrt{3} \\[2ex] | | & \cot{\left(\frac{8\pi}{3}\right)} &= \frac{-\frac{1}{\cancel 2}} {\frac{\sqrt{3}}{\cancel 2}} \cdot \cancel 2 = \frac{-1}{\sqrt{3}} \cdot \sqrt{3}= \frac{-\sqrt{3}}{3} \\[2ex] \end{align} </math> |
| | |
| </math> | |
Latest revision as of 09:45, 27 August 2022
![{\displaystyle {\begin{aligned}\sin {\left({\frac {8\pi }{3}}\right)}&={\frac {\sqrt {3}}{2}}&\csc {\left({\frac {8\pi }{3}}\right)}&={\frac {1}{\frac {\sqrt {3}}{\cancel {2}}}}\cdot {\cancel {2}}={\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 {1}{-{\frac {1}{\cancel {2}}}}}\cdot {\cancel {2}}=-{\frac {2}{1}}=-2\\[2ex]\tan {\left({\frac {8\pi }{3}}\right)}&={\frac {\frac {\sqrt {3}}{\cancel {2}}}{-{\frac {1}{\cancel {2}}}}}\cdot {\cancel {2}}={\frac {\sqrt {3}}{-1}}=-{\sqrt {3}}&\cot {\left({\frac {8\pi }{3}}\right)}&={\frac {-{\frac {1}{\cancel {2}}}}{\frac {\sqrt {3}}{\cancel {2}}}}\cdot {\cancel {2}}={\frac {-1}{\sqrt {3}}}\cdot {\sqrt {3}}={\frac {-{\sqrt {3}}}{3}}\\[2ex]\end{aligned}}}](https://en.wikipedia.org/api/rest_v1/media/math/render/svg/e84bee6192feaf91f5efed7e735f9ff95013fdeb)