Question

the point (3,-6) lies on the terminal side of angle 0. Find the exact value of the six trigonometric functions of 0. Sin cos tan csc sec cot

Answer 1

ANSWER

See explanation

EXPLANATION

The point (3,-6) is the fourth quadrant.

In this quadrant only the cosine ratio and the secant ratio are positive.

The remaining four trigonometric ratios are negative.

The diagram is shown in the attachment.

We use the Pythagoras Theorem to find the hypotenuse, h.

${h}^{2} = {6}^{2} + {3}^{2}$

${h}^{2} = 36 + 9$

${h}^{2} = 45$

$h = \sqrt{45}$

$h = 3 \sqrt{5}$

$\sin( \theta) = - \frac{opp}{hyp}$

$\sin( \theta) = - \frac{6}{3 \sqrt{5} }$

$\sin( \theta) = - \frac{2 \sqrt{5} }{ 5 }$

$\sin( \theta) = - \frac{2}{ \sqrt{5} }$

$\ cos(\theta) = \frac{adj}{hyp}$

$\ cos(\theta) = \frac{3}{3 \sqrt{5} }$

$\ cos(\theta) = \frac{1}{ \sqrt{5} }$

$\ cos(\theta) = \frac{ \sqrt{5} }{5}$

$\tan(\theta)= - \frac{opp}{hyp}$

$\tan(\theta)= - \frac{6}{3} = - 2$

$\csc(\theta)= - \frac{hyp}{opp}$

$\csc(\theta)= - \frac{3 \sqrt{5} }{6} = - \frac{\sqrt{5} }{2}$

$\sec( \theta) = \frac{hyp}{adj}$

$\sec( \theta) = \frac{3 \sqrt{5} }{3} = \sqrt{5}$

$\cot( \theta) = - \frac{adj}{opp}$

$\cot( \theta) = - \frac{3}{6} = - \frac{1}{2}$