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} | \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 \frac{2}{1} \\[2ex] | ||
\end{align} | \end{align} | ||
</math> | </math> | ||
Revision as of 17:13, 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 (Conversion error. Server ("https://en.wikipedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle {\begin{aligned}\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 {\frac {2}{1}}\\[2ex]\end{aligned}}}