Math  /  Trigonometry

QuestionFind the exact values of the six trigonometric functions of tt for the point P=(223,13)P=\left(\frac{2 \sqrt{2}}{3},-\frac{1}{3}\right) on the unit circle.

Studdy Solution
Substitute the values of sin(t)\sin(t), cos(t)\cos(t), and tan(t)\tan(t) into the formulas for csc(t)\csc(t), sec(t)\sec(t), and cot(t)\cot(t).
csc(t)=113=3\csc(t) = \frac{1}{-\frac{1}{3}} = -3sec(t)=1223=322=324\sec(t) = \frac{1}{\frac{2 \sqrt{2}}{3}} = \frac{3}{2 \sqrt{2}} = \frac{3 \sqrt{2}}{4}cot(t)=124=42=22\cot(t) = \frac{1}{-\frac{\sqrt{2}}{4}} = -\frac{4}{\sqrt{2}} = -2 \sqrt{2}So, the exact values of the six trigonometric functions of tt aresin(t)=13\sin(t) = -\frac{1}{3}cos(t)=223\cos(t) = \frac{2 \sqrt{2}}{3}tan(t)=24\tan(t) = -\frac{\sqrt{2}}{4}csc(t)=3\csc(t) = -3sec(t)=324\sec(t) = \frac{3 \sqrt{2}}{4}cot(t)=22\cot(t) = -2 \sqrt{2}

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