# Trig problem, let sinA= -3/5 wiht A in Quad. 3 find Sin2A

1. Jul 17, 2011

### Neopets

1. The problem statement, all variables and given/known data
SinA is = to -3/5 with A in Q3, find sinA?

2. Relevant equations

3. The attempt at a solution

When I did this I set it up like
First find cosA
So to do that I used plus/minus √(1-sin^2A)
The problem is that I really got confused when plugging in the values under the radical correctly, because i thought, -3/5 squared gives you positive 9/25 doesn't it and then the negative sign in the middle should make it 1-9/25, so you'd get 25/25-9/25 = 14/25
and then the answer also says that since it is in quadrant 3 it is negative, so -14/25 ? right?
Then plug that into the next step :
Sin 2A = 2SinACosA
=2(-3/5)(-√14/5)
something is wrong with how I'm doing it and I'd really like to know what. :( :(

2. Jul 17, 2011

### eumyang

It's 16/25, not 14/25.

No. Take the square root first, and then tack on the negative. So it's
$-\sqrt{\frac{16}{25}}=-\frac{4}{5}$

3. Jul 17, 2011

### Neopets

omg thank u, some small silly mistake. ahhhh
thanks for seeing that :)

4. Jul 17, 2011

### GreenPrint

oh wow I thoguht it was sin(a) = -3/5 find a my bad