Question

How do you find the exact value of the five remaining trigonometric function given `sintheta=1/3` and the angle is in standard position in quadrant II?

Answer 1

`sin(theta)=1/3`
`cos(theta)=-(2sqrt(2))/3`
`tan(theta)=-sqrt(2)/4`
`csc(theta)=3`
`sec(theta)=-(3sqrt(2))/4`
`cot(theta)=-2sqrt(2)`

Explanation:

First, `csc(theta)=1/sin(theta)=1/(1/3)=3`.

We know that `sin^2(theta)+cos^2(theta)=1`, or `cos(theta)=+-sqrt(1-sin^2(theta))=+-sqrt(1-(1/3)^2)=+-(2sqrt(2))/3`. Now, the angle is in quadrant 2. This means that its cosine value is negative. Thus, `cos(theta)=-(2sqrt(2))/3`.

Then, `sec(theta)=1/cos(theta)=1/(-(2sqrt(2))/3)=-3/(2sqrt(2))=-(3sqrt(2))/4`.

We know that `tan(theta)=sin(theta)/cos(theta)=(1/3)/(-(2sqrt(2))/3)=-1/(2sqrt(2))=-sqrt(2)/4`.

Then, `cot(theta)=1/tan(theta)=1/(-sqrt(2)/4)=-4/sqrt(2)=-2sqrt(2)`.