Question

How do you sketch the angle whose terminal side in standard position passes through `(sqrt5,-sqrt15)` and how do you find sin and cos?

Answer 1

`sin a = - sqrt3/2`
`cos a = 1/2`

Explanation:

Call M the point `M(sqrt5, -sqrt15)`
M is in Quadrant IV. On the trig unit circle, M is the terminal point of the arc a, that `tan a = - sqrt15/sqrt5 = - sqrt3`
Trig table gives --> `a = - pi/3 or a = (5pi)/3` (co -terminal arcs).

`sin a = sin (-pi/3) = - sin (pi/3) = - sqrt3/2`
`cos a = cos (-pi/3) = cos (pi/3) = 1/2`