How do you use the point (-1,5) on the terminal side of the angle to evaluate the six trigonometric functions?

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How do you use the point (-1,5) on the terminal side of the angle to evaluate the six trigonometric functions?

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Answer 1 · 2017-03-14T13:50:27

Call t the angle that is located in Quadrant 2, with
$tan t = \frac{y}{x} = \frac{5}{- 1} = - 5$ $tan t = y/x = 5/-1 = -5$
${cos}^{2} t = \frac{1}{1 + {tan}^{2} t} = \frac{1}{1 + 25} = \frac{1}{26}$ $cos ^2 t = 1/(1 + tan^2 t) = 1/(1 + 25) = 1/26$
$cos t = - \frac{1}{\sqrt{26}} = - \frac{\sqrt{26}}{26}$ $cos t = - 1/(sqrt26) = - sqrt26/26$
(because t is in Quadrant 2, cos t < 0)
${sin}^{2} t = \frac{1}{1 + {cot}^{2} t} = \frac{1}{1 + \frac{1}{25}} = \frac{25}{26}$ $sin^2 t = 1/(1 + cot^2 t) = 1/(1 + 1/25) = 25/26$
$sin t = + \frac{5}{\sqrt{26}}$ $sin t = + 5/sqrt26$ (t in Q. 2 --> sin t > 0)
$tan t = \frac{sin}{cos} = (\frac{5}{\sqrt{26}}) (\frac{- \sqrt{26}}{1}) = - 5$ $tan t = sin/(cos) = (5/(sqrt26))((- sqrt26)/1) = - 5$ OK
$sec t = \frac{1}{cos} = - \sqrt{26}$ $sec t = 1/(cos) = - sqrt26$
$csc t = \frac{1}{sin} = \frac{\sqrt{26}}{5}$ $csc t = 1/(sin) = sqrt26/5$

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How do you use the point (-1,5) on the terminal side of the angle to evaluate the six trigonometric functions?

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