Question

Given that sin t= (-2/3) and that P(t) is a point in the 3rd quadrant, what is the cos of t?

Answer 1

See explanation.

Explanation:

If we have `sin alpha` given, thento calculate `cos alpha` we use the identity:

`sin^2alpha+cos^2alpha=1`

If we substitute the given value we get:

`(-2/3)^2+cos^2alpha=1`

`4/9+cos^2alpha=1`

`cos^2alpha=1-4/9`

`cos^2alpha=5/9`

`cos alpha=-sqrt(5)/3 vv cos alpha=sqrt(5)/3`

The given equation has 2 solutions. To choose one we have to use the given information about the quadrant, in which the angle is located.

The angle is located in `Q3`. This means that its sine and cosine are both negative. Finally we can write the answer:

`cos alpha=-sqrt(5)/3`