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.