Thread: Half-Angle Formula Help

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.

Originally Posted by JasonV

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.

you are correct ... books have been wrong before.

Originally Posted by skeeter

you are correct ... books have been wrong before.

Thanks man.