Given the point (6,-8) on the terminal side of ø in standard position, find the exact value of csc ø?

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Given the point (6,-8) on the terminal side of ø in standard position, find the exact value of csc ø?

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Answer 1 · 2018-05-18T04:54:16

$csc t = - \frac{5}{4}$ $csc t = - 5/4$

Explanation:

Call t the arc (angle), that its terminal side passes at point (6, -8).
t is in Quadrant 4.
$tan t = \frac{y}{x} = - \frac{8}{6} = - \frac{4}{3}$ $tan t = y/x = -8/6 = - 4/3$
$cot = \frac{1}{tan t} = - \frac{3}{4}$ $cot = 1/(tan t) = - 3/4$
${sin}^{2} t = \frac{1}{1 + {cot}^{2} t} = \frac{1}{1 + \frac{9}{16}} = \frac{16}{25}$ $sin^2 t = 1/(1 + cot^2 t) = 1/(1 + 9/16) = 16/25$
$sin t = \pm \frac{4}{5}$ $sin t = +- 4/5$
Since t is in Quadrant 4, then, sin t is negative
$sin t = - \frac{4}{5}$ $sin t = - 4/5$
$csc t = \frac{1}{sin t} = - \frac{5}{4}$ $csc t = 1/(sin t) = - 5/4$

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Given the point (6,-8) on the terminal side of ø in standard position, find the exact value of csc ø?

1 Answer

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