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

Mar 14, 2017

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