How Do You Solve for Sin(theta/2) Using the Half Angle Formula?

[daliberataaa!]

Homework Statement

given that sin theta = 3/5, 0 < theta < pi/2 find sin theta/2

Homework Equations

http://www.intmath.com/Analytic-trigonometry/sinalphaon2.gif

The Attempt at a Solution

I appologize that I don't know how to use all the symbols on a keyboard (square roots, etc)
But I have so far gotten it to :

square root of 1-3/5 all over 2. I have a sample of this same problem in my book, but they don't explain the next step. Suddenly this problem from here just turns into square root of 1/5...can someone please explain how they got to there?

Thank you all.

 

[daliberataaa!]

ok i feel like the biggest idiot in the world...i got it.

I blame sleep deprivation...haha.

 

[Architeuthis Dux]

Dear student,

Thank you for your question. The solution to this problem involves using the half angle formula for sine:

sin(theta/2) = ±√((1-cos(theta))/2)

In your attempt, you have correctly substituted the given value of sin(theta) into the formula and simplified it to the square root of (1-3/5)/2. However, to complete the solution, we need to find the value of cos(theta).

To do this, we can use the Pythagorean identity: sin^2(theta) + cos^2(theta) = 1. Since we know that sin(theta) = 3/5, we can substitute this into the identity and solve for cos(theta):

(3/5)^2 + cos^2(theta) = 1
9/25 + cos^2(theta) = 1
cos^2(theta) = 16/25
cos(theta) = ±4/5

Now we can substitute this value into our original equation for sin(theta/2):

sin(theta/2) = ±√((1-4/5)/2)
sin(theta/2) = ±√(1/5)
sin(theta/2) = ±√(1)/√(5)
sin(theta/2) = ±1/√(5)

Since 0 < theta < pi/2, we know that theta is in the first quadrant and therefore sin(theta/2) must be positive. Therefore, we can drop the ± and our final answer is:

sin(theta/2) = 1/√(5)

I hope this helps clarify the next step in the solution. Good luck with your studies!

Best regards,