Thursday, April 16, 2009

Sine Half-Angle Formula Discussion Post

Hint #1:
Identify which line segments on the diagram are equal to sin(alpha/2) and cos(alpha/2), and note how sin(alpha) and cos(alpha) show up in the diagram.

Hint #2:
What is the value of angle ECB? Are there similar triangle possibilities here?

Hint #3:
CE=CD-AB or AB=CD-CE

Hint #4:
Get an expression for cos(alpha) from triangle ECB. Every term in your expression should be convertible to sines or cosines of alpha and (alpha/2).

Hint #5:
From here it's just algebra. Solve for sin(alpha/2), remembering that you can convert a cosine to a sine using the pythagorean theorem.

5 comments:

Vranian said...

Ok, lets see where everyone has gotten...I've begun to write out the formula for the cosine of a/2... (a-alpha.) Where are y'all?

Scott said...

From hint #1, it says to note how sin(alpha) and cos(alpha) show up in the diagram. I don't see this anywhere. I've found that angle BCE equals alpha, but I don't see what the hint is referring to. I've found ratios for sin and cos(alpha/2), but not for alpha. Any ideas?

sam said...

i'm in the same boat as you scott

Scott said...

Use the formulas we found on Wednesday: sin(a)=opp. side/hypotenuse, cos(a)=adj. side/hypotenuse.

Using those, and having established that angle BCE equals alpha, you can do hint #4. I've written everything in terms of sines and cosines of alpha and alpha/2, so I think it's just figuring out the algebra from there.

sam said...

Has anyone proven that CD bisects OB?