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

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\sin{\left(\frac{8\pi}{3}\right)} &= \frac{\sqrt{3}}{2} & \csc{\left(\frac{8\pi}{3}\right)} &= {\frac{2}{\sqrt{3}} \cdot \sqrt{3}}= \frac{2\sqrt{3}}{3}  \\[2ex]
\sin{\left(\frac{8\pi}{3}\right)} &= \frac{\sqrt{3}}{2} & \csc{\left(\frac{8\pi}{3}\right)} &= {\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{{2}}{1}\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{{2}}{1} = -2 \\[2ex]  


\tan{\left(\frac{8\pi}{3}\right)} &= \frac{\frac{\sqrt{3}}{2}}{-\frac{{1}}{2}} = \left(2\right) = \frac{\sqrt{3}}{-1} = -\sqrt{3}  
\tan{\left(\frac{8\pi}{3}\right)} &= \frac{\frac{\sqrt{3}}{2}}{-\frac{{1}}{2}} = \left(2\right) = \frac{\sqrt{3}}{-1} = -\sqrt{3}  


& \cot{\left(\frac{8\pi}{3}\right)} &= -\frac{\sqrt{3}}{1}= -\sqrt{3} \\[2ex]  \end{align} </math>
& \cot{\left(\frac{8\pi}{3}\right)} &= -\frac{\sqrt{3}}{1}= -\sqrt{3} \\[2ex]  \end{align} </math>

Revision as of 03:59, 26 August 2022

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://en.wikipedia.org/api/rest_v1/":): {\displaystyle \frac{8\pi}{3}\Rightarrow \left(-\frac{1}{2} ,\frac{\sqrt{3}}{2}\right)}

Failed to parse (Conversion error. Server ("https://en.wikipedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle {\begin{aligned}\sin {\left({\frac {8\pi }{3}}\right)}&={\frac {\sqrt {3}}{2}}&\csc {\left({\frac {8\pi }{3}}\right)}&={{\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 {2}{1}}=-2\\[2ex]\tan {\left({\frac {8\pi }{3}}\right)}&={\frac {\frac {\sqrt {3}}{2}}{-{\frac {1}{2}}}}=\left(2\right)={\frac {\sqrt {3}}{-1}}=-{\sqrt {3}}&\cot {\left({\frac {8\pi }{3}}\right)}&=-{\frac {\sqrt {3}}{1}}=-{\sqrt {3}}\\[2ex]\end{aligned}}}