6.2 Trigonometric Functions: Unit Circle Approach/13: Difference between revisions
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\sin{(t)} &= \frac{1}{2} & \csc{(t)} &= \frac{1}{\frac{1}{2}}=2\\[2ex] | \sin{(t)} &= \frac{1}{2} & \csc{(t)} &= \frac{1}{\frac{1}{2}}=2\\[2ex] | ||
\cos{(t)} &= \frac{\sqrt{3}}{2} & \sec{(t)} &= \frac{1}{\frac{\sqrt{3}}{2}}=\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{1}{\frac{\sqrt{3}}{2}}=\frac{2}{\sqrt{3}} \cdot \frac{\sqrt{3}}{\sqrt{3}} = \frac{2\sqrt{3}}{3}\\[2ex] | ||
\tan{(t)} &= \frac{\frac{1}{2}}{\frac{\sqrt{3}}{2}} = \frac{1}{\cancel{2}}\cdot\frac{\cancel{2}}{\sqrt{3}} = \frac{1}{\cancel{\sqrt{3}}} \cdot \frac{\cancel{\sqrt{3}}}{\sqrt{3}}=\frac{\sqrt{3}}{3} & \cot{(t)} &= \frac{\frac{\sqrt{3}}{2}}{\frac{1}{2}}=\frac{\sqrt{3}}{2}/cdot | \tan{(t)} &= \frac{\frac{1}{2}}{\frac{\sqrt{3}}{2}} = \frac{1}{\cancel{2}}\cdot\frac{\cancel{2}}{\sqrt{3}} = \frac{1}{\cancel{\sqrt{3}}} \cdot \frac{\cancel{\sqrt{3}}}{\sqrt{3}}=\frac{\sqrt{3}}{3} & \cot{(t)} &= \frac{\frac{\sqrt{3}}{2}}{\frac{1}{2}}=\frac{\sqrt{3}}{2} /cdot {2} \\[2ex] | ||
\end{align} | \end{align} | ||
</math> | </math> | ||
Revision as of 17:12, 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 \left(\frac{\sqrt{3}}{2}, \frac{1}{2}\right)}
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 \begin{align} \sin{(t)} &= \frac{1}{2} & \csc{(t)} &= \frac{1}{\frac{1}{2}}=2\\[2ex] \cos{(t)} &= \frac{\sqrt{3}}{2} & \sec{(t)} &= \frac{1}{\frac{\sqrt{3}}{2}}=\frac{2}{\sqrt{3}} \cdot \frac{\sqrt{3}}{\sqrt{3}} = \frac{2\sqrt{3}}{3}\\[2ex] \tan{(t)} &= \frac{\frac{1}{2}}{\frac{\sqrt{3}}{2}} = \frac{1}{\cancel{2}}\cdot\frac{\cancel{2}}{\sqrt{3}} = \frac{1}{\cancel{\sqrt{3}}} \cdot \frac{\cancel{\sqrt{3}}}{\sqrt{3}}=\frac{\sqrt{3}}{3} & \cot{(t)} &= \frac{\frac{\sqrt{3}}{2}}{\frac{1}{2}}=\frac{\sqrt{3}}{2} /cdot {2} \\[2ex] \end{align} }