6.2 Trigonometric Functions: Unit Circle Approach/48: Difference between revisions
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\begin{align} | \begin{align} | ||
\sin{(\frac{5\pi}{6})} &= -\frac{\sqrt{3}}{2} & \csc{(t)} &= -\frac{2}{\sqrt{3}}\cdot\frac{\sqrt{3}}{\sqrt{3}}=\frac{2\sqrt{3}}{3}\\[2ex] | \sin{\left(\frac{5\pi}{6}\right)} &= -\frac{\sqrt{3}}{2} & \csc{(t)} &= -\frac{2}{\sqrt{3}}\cdot\frac{\sqrt{3}}{\sqrt{3}}=\frac{2\sqrt{3}}{3}\\[2ex] | ||
\cos{(\frac{5\pi}{6})} &= \frac{1}{2} & \sec{(t)} &= \frac{2}{1} = 2\\[2ex] | \cos{\left(\frac{5\pi}{6}\right)} &= \frac{1}{2} & \sec{(t)} &= \frac{2}{1} = 2\\[2ex] | ||
\tan{(\frac{5\pi}{6})} &= \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}}{\sqrt{3}}=\frac{\sqrt{3}}{3} \\[2ex] | \tan{\left(\frac{5\pi}{6}\right)} &= \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}}{\sqrt{3}}=\frac{\sqrt{3}}{3} \\[2ex] | ||
\end{align} | \end{align} | ||
</math> | </math> | ||
Revision as of 22:24, 25 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{5\pi}{6} \Rightarrow \left(\frac{-\sqrt{3}}{2}, \frac{1}{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 {5\pi }{6}}\right)}&=-{\frac {\sqrt {3}}{2}}&\csc {(t)}&=-{\frac {2}{\sqrt {3}}}\cdot {\frac {\sqrt {3}}{\sqrt {3}}}={\frac {2{\sqrt {3}}}{3}}\\[2ex]\cos {\left({\frac {5\pi }{6}}\right)}&={\frac {1}{2}}&\sec {(t)}&={\frac {2}{1}}=2\\[2ex]\tan {\left({\frac {5\pi }{6}}\right)}&={\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}}{\sqrt {3}}}={\frac {\sqrt {3}}{3}}\\[2ex]\end{aligned}}}