Half-Angle Formula Help

JasonV · #1 $Old$ November 4th, 2009, 05:04 PM

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.

skeeter · #2 $Old$ November 4th, 2009, 05:16 PM

Quote:

Originally Posted by JasonV $View Post$

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.

JasonV · #3 $Old$ November 4th, 2009, 05:18 PM

Quote:

Originally Posted by skeeter $View Post$

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

Thanks man.