6.2 Trigonometric Functions: Unit Circle Approach/13: Difference between revisions
Jump to navigation
Jump to search
No edit summary |
No edit summary |
||
Line 6: | Line 6: | ||
\sin{(t)} &= \frac{1}{2} & \csc{(t)} &= -\frac{2}{\sqrt{3}}\cdot\frac{\sqrt{3}}{\sqrt{3}}=\frac{2\sqrt{3}}{3}\\[2ex] | \sin{(t)} &= \frac{1}{2} & \csc{(t)} &= -\frac{2}{\sqrt{3}}\cdot\frac{\sqrt{3}}{\sqrt{3}}=\frac{2\sqrt{3}}{3}\\[2ex] | ||
\cos{(t)} &= \frac{\sqrt{3}}{2} & \sec{(t)} &= \frac{2}{1} = 2\\[2ex] | \cos{(t)} &= \frac{\sqrt{3}}{2} & \sec{(t)} &= \frac{2}{1} = 2\\[2ex] | ||
\tan{(t)} &= \frac{\frac{1}{2}}{\frac{\sqrt{3}}{2}} = \frac{1}{\cancel{2}}\cdot\frac{\cancel{2}}{\sqrt{3}} = | \tan{(t)} &= \frac{\frac{1}{2}}{\frac{\sqrt{3}}{2}} = \frac{1}{\cancel{2}}\cdot\frac{\cancel{2}}{\sqrt{3}} = \frac{{1}{\sqrt{3}} \cdot & \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 16:48, 26 August 2022
Failed to parse (unknown function "\begin{align}"): {\displaystyle \begin{align} \sin{(t)} &= \frac{1}{2} & \csc{(t)} &= -\frac{2}{\sqrt{3}}\cdot\frac{\sqrt{3}}{\sqrt{3}}=\frac{2\sqrt{3}}{3}\\[2ex] \cos{(t)} &= \frac{\sqrt{3}}{2} & \sec{(t)} &= \frac{2}{1} = 2\\[2ex] \tan{(t)} &= \frac{\frac{1}{2}}{\frac{\sqrt{3}}{2}} = \frac{1}{\cancel{2}}\cdot\frac{\cancel{2}}{\sqrt{3}} = \frac{{1}{\sqrt{3}} \cdot & \cot{(t)} &= -\frac{1}{\sqrt{3}}=-\frac{1}{\sqrt{3}}\cdot\frac{\sqrt{3}}{\sqrt{3}}=\frac{\sqrt{3}}{3} \\[2ex] \end{align} }