Question

If `sinx=(4/5)`, how do you find `sin2x`?

Answer 1

`sin2x = 2*4/5*3/5=24/25`

Explanation:

Since `sinx=4/5` we have a (3, 4, 5) triangle and we can totally define all sines and cosines, `cosx=3/5`

Now we use the double angle identities:
`sin2x=2sinx*cosx`
`sin2x = 2*4/5*3/5=24/25`