6.2 Trigonometric Functions: Unit Circle Approach/19: Difference between revisions
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\begin{align} | \begin{align} | ||
\sin{(t)} &= -\frac{1}{3} & \csc{(t)} &= | \sin{(t)} &= -\frac{1}{3} & \csc{(t)} &= \frac{1}{-\frac{1}{3}} = \frac{1}{1}\cdot-\frac{3}{1} = -3\\[2ex] | ||
\cos{(t)} &= \frac{2\sqrt{2}}{3} & \sec{(t)} &= \frac{2}{1} = 2\\[2ex] | \cos{(t)} &= \frac{2\sqrt{2}}{3} & \sec{(t)} &= \frac{1}{\frac{2\sqrt{2}}{3}} = \frac{1}{1}\cdot\frac{3}{2\sqrt{2}} = \frac{3}{2\sqrt{2}}\cdot\frac{\sqrt{2}}{\sqrt{2}}=\frac{3\sqrt{2}}{4}\\[2ex] | ||
\tan{(t)} &= \frac{-\frac{1}{3}}{\frac{2\sqrt{2}}{3}} = -\frac{1}{3}\cdot\frac{3}{2\sqrt{2}} = \frac{1}{2\sqrt{2}}\cdot\frac{\sqrt{2}}{\sqrt{2}} = \frac{\sqrt{2}}{4} & \cot{(t)} &= -\frac{1}{\sqrt{3}}=-\frac{1}{\sqrt{3}}\cdot\frac{\sqrt{3}}{\sqrt{3}}=\frac{\sqrt{3}}{3} \\[2ex] | \tan{(t)} &= \frac{-\frac{1}{3}}{\frac{2\sqrt{2}}{3}} = -\frac{1}{3}\cdot\frac{3}{2\sqrt{2}} = \frac{1}{2\sqrt{2}}\cdot\frac{\sqrt{2}}{\sqrt{2}} = \frac{\sqrt{2}}{4} & \cot{(t)} &= -\frac{1}{\sqrt{3}}=-\frac{1}{\sqrt{3}}\cdot\frac{\sqrt{3}}{\sqrt{3}}=\frac{\sqrt{3}}{3} \\[2ex] | ||
Revision as of 17:22, 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{2\sqrt{2}}{3}, -\frac{1}{3}\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}{3}}&\csc {(t)}&={\frac {1}{-{\frac {1}{3}}}}={\frac {1}{1}}\cdot -{\frac {3}{1}}=-3\\[2ex]\cos {(t)}&={\frac {2{\sqrt {2}}}{3}}&\sec {(t)}&={\frac {1}{\frac {2{\sqrt {2}}}{3}}}={\frac {1}{1}}\cdot {\frac {3}{2{\sqrt {2}}}}={\frac {3}{2{\sqrt {2}}}}\cdot {\frac {\sqrt {2}}{\sqrt {2}}}={\frac {3{\sqrt {2}}}{4}}\\[2ex]\tan {(t)}&={\frac {-{\frac {1}{3}}}{\frac {2{\sqrt {2}}}{3}}}=-{\frac {1}{3}}\cdot {\frac {3}{2{\sqrt {2}}}}={\frac {1}{2{\sqrt {2}}}}\cdot {\frac {\sqrt {2}}{\sqrt {2}}}={\frac {\sqrt {2}}{4}}&\cot {(t)}&=-{\frac {1}{\sqrt {3}}}=-{\frac {1}{\sqrt {3}}}\cdot {\frac {\sqrt {3}}{\sqrt {3}}}={\frac {\sqrt {3}}{3}}\\[2ex]\end{aligned}}}