Question

How do you sketch the angle whose terminal side in standard position passes through `(-2,-3sqrt5)` and how do you find sin and cos?

Answer 1

`sin t = - 3sqrt5/7`
`cos t = - 2/7`

Explanation:

Call t the angle whose terminal side passes through
point `(-2, - 3sqrt5)` --> t's terminal side lies in Quadrant 3.
`tan t = y/x = (-3sqrt5)/(-2) = (3sqrt5)/2`
`cos^2 t = 1/(1 + tan^2 t) = 1/(1 + 45/4) = 4/49`
`cos t = - 2/7` (because t lies in Quadrant 3 --> cos t is negative)
`sin^2 t = 1 - cos^2 t = 1 - 4/49 = 45/49`
`sin t = - (3sqrt5)/7` (because t lies in Q.3 --> sin t is negative)