How do you use the point (-12,3) on the terminal side of the angle to evaluate the six trigonometric functions?

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How do you use the point (-12,3) on the terminal side of the angle to evaluate the six trigonometric functions?

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Answer 1 · 2018-03-04T05:55:16

$tan t = \frac{y}{x} = \frac{3}{- 12} = - \frac{1}{4}$ $tan t = y/x = 3/-12 = - 1/4$
The terminal side of t is in Quadrant 3.
${cos}^{2} t = \frac{1}{1 + {tan}^{2} t} = \frac{1}{1 + \frac{1}{16}} = \frac{16}{17}$ $cos^2 t = 1/(1 + tan^2 t) = 1/(1 + 1/16) = 16/17$
$cos t = - \frac{4}{\sqrt{17}}$ $cos t = - 4/sqrt17$
$sin t = tan t . cos t = (- \frac{1}{4}) (- \frac{4}{\sqrt{17}}) = \frac{1}{\sqrt{17}}$ $sin t = tan t.cos t = (-1/4)(-4/sqrt17) = 1/sqrt17$
$cot t = \frac{1}{tan t} = - 4$ $cot t = 1/(tan t) = - 4$
$sec t = \frac{1}{cos t} = - \frac{\sqrt{17}}{4}$ $sec t = 1/(cos t) = - sqrt17/4$
$csc t = \frac{1}{sin t} = \sqrt{17}$ $csc t = 1/(sin t) = sqrt 17$

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How do you use the point (-12,3) on the terminal side of the angle to evaluate the six trigonometric functions?

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