# How do you use the point (-9,14) on the terminal side of the angle to evaluate the six trigonometric functions?

Then teach the underlying concepts
Don't copy without citing sources
preview
?

#### Explanation

Explain in detail...

#### Explanation:

I want someone to double check my answer

1
Nghi N. Share
Mar 19, 2017

The terminal side, containing point (-9, 14), is in Quadrant 3.
Call t the angle.
$\tan t = - \frac{14}{9}$
${\cos}^{2} t = \frac{1}{1 + {\tan}^{2} t} = \frac{1}{1 + \frac{196}{81}} = \frac{81}{277}$
$\cos t = \pm \frac{9}{\sqrt{277}}$
Because t is in Quadrant 3, cos t is negative -->
$\cos t = - \frac{9}{\sqrt{277}}$
$\sin t = \tan t . \cos t = \left(- \frac{14}{9}\right) \left(- \frac{9}{\sqrt{277}}\right) = \frac{14}{\sqrt{277}}$
$\tan t = - \frac{14}{9}$
$\cot t = - \frac{9}{14}$
$\sec t = \frac{1}{\cos} = - \frac{\sqrt{277}}{9}$
$\csc t = \frac{1}{\sin} = \frac{\sqrt{277}}{14}$

• 7 minutes ago
• 9 minutes ago
• 9 minutes ago
• 10 minutes ago
• 44 seconds ago
• 2 minutes ago
• 5 minutes ago
• 5 minutes ago
• 6 minutes ago
• 6 minutes ago
• 7 minutes ago
• 9 minutes ago
• 9 minutes ago
• 10 minutes ago