Question

If sinx = -3/5, x is in quadrant III, then how do you find sin2x?

Answer 1

`sin2x=24/25`

Explanation:

`sin2x=2sinxcosx`
`cosx=-sqrt(1-sin^2x` (cosx <0 in quadrant III)
so `cosx=-sqrt(1-9/25)`
that's `cosx=-4/5`
and `sin2x=2(-3/5)(-4/5)`
that's `24/25`