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

From Burton Tech. Points Wiki
Jump to navigation Jump to search
No edit summary
No edit summary
Line 2: Line 2:


\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{\sqrt{2}}{2} ,\frac{\sqrt{2}}{2}\right)</math><br><br> </math>
<math>
\begin{align}
\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]

Revision as of 03:36, 26 August 2022



</math>

<math> \begin{align}

\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)} &= \fracTemplate: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]