Question

Find the exact value of cos(theta) for an angle (theta) with tan (theta)= -2/3 and with its terminal side in Quadrant II.


`a.(2√(13) )/(13) \\\\b. - (3√(13) )/(13) \\\\c. -(2)/(5) \\\\d. (3)/(√(13) )`

Answer 1

b

Explanation:

1 + tan²theta = sec²theta

1 + (-2/3)² = sec²theta

1 + 4/9 = sec²theta

13/9 = sec²theta

sec theta = -sqrt(13)/3

Because Quadrant 2

cos theta = 1 ÷ sec theta

Cos there = -3/sqrt(13)

-3sqrt(13)/13

After rationalization

Answer 2

`-(3√(13))/(13)`.

Explanation:

Since we are in quadrant two, cosine value is negative while sine value is positive.

We are going to use the Pythagorean Identity:
`1+\tan^2(\theta)=\sec^2(\theta)`.

`1+((-2)/(3))^2=\sec^2(\theta)`

`1+(4)/(9)=\sec^2(\theta)`

`(9+4)/(9)=\sec^2(\theta)`

`(13)/(9)=\sec^2(\theta)`

`\pm \sqrt{(13)/(9)}=\sec(\theta)`

`\pm (√(13))/(3)=\sec(\theta)`

Since cosine and secant are reciprocals then they will have the same sign as along as they both exist.

`\sec(\theta)=-(√(13))/(3)`

`\cos(\theta)=-(3)/(√(13))`.

I don't see this answer as I'm going to rationalize the denominator.

`\cos(\theta)=-(3)/(√(13)) \cdot (√(13))/(√(13))`.

`\cos(\theta)=-(3√(13))/(13)`.