Hey Guys,

I'm trying to solve the following problem:

Sin(∆)= 3/5 0 < ∆ < pi/2

Sin(∆/2) = +/- Sqrt[(1-cos∆)/2]

How would I go about that?

I started with

5^2 - 3^2 = sqrt(16) = +/- 4

cos(∆) = 4/5

So --

Sin(∆/2) = sqrt[(1-4/5)/2]

Sqrt[1/5/2] = 1/Sqrt(10) = Sqrt(10)/10

But the book says that this is incorrect. Can someone please assist?

Thank you,

Jason V.