How do you find the five remaining trigonometric functions of this problem?

#theta# if sec#theta# =3/2 and sin#theta#< 0

1 Answer
Apr 29, 2018

Too long for this space, see the exolplanation below.

Explanation:

First get the sine and the cosine. The cosine follows immediately.

#sec(\theta)=1/{\cos(theta)}=3/2#; thus #\cos(theta)=2/3#.

To get the sine use the Pythagorean identity #sin^2(\theta)+\cos^2(\theta)=1#. Since the sine is given as less than zero we put a negative sign on the square root that comes up when we solve for the sine:

#\sin(\theta)=-\sqrt{1-\cos^2(\theta)}=-\sqrt{1-4/9}=-{\sqrt{5}}/3#.

Now just use the quotient definitions for the rest:

#\tan(\theta)={\sin(\theta}}/{\cos(\theta}}={-{\sqrt{5}}/3}/{2/3}=-{\sqrt{5}}/2#.

#\cot(\theta)={\cos(\theta}}/{\sin(\theta}}={2/3}/{-\sqrt{5}}/3=-2{\sqrt{5}}/5#.

#\sec(\theta)=1/{\cos(\theta}}=1/{2/3}=3/2# (given).

#\csc(\theta)=1/{\sin(\theta}}=1/{{-\sqrt{5}}/3}=-{3\sqrt{5}}/5#.