Verify Algebraically

• August 5th 2008, 12:34 AM
Kitty216
Verify Algebraically
http://i146.photobucket.com/albums/r...6/problem5.jpg

This one really stumped me =[ Does anyone know how to verify it?
• August 5th 2008, 01:19 AM
ticbol
Quote:

Originally Posted by Kitty216
http://i146.photobucket.com/albums/r...6/problem5.jpg

This one really stumped me =[ Does anyone know how to verify it?

I think the (alpha)/3 distracts you. You're used to dealing up to (alpha)/2 only for angles.

Since the alpha/3 is uniform throughout, then let's say angle alpha/ 3 = angle B.

Then the original equation/identity will be
sinBcosB = (1/2)sin(2B)
....which is easy.
• August 5th 2008, 01:21 AM
Gusbob
$\sin \frac{\alpha}{3} \cos \frac{\alpha}{3} = \frac{1}{2} \sin \frac{2\alpha}{3}$



\begin{aligned}
LHS &= \sin \frac{\alpha}{3} \cos \frac{\alpha}{3} \\
&= \frac{1}{2} \left(2\sin \frac{\alpha}{3} \cos \frac{\alpha}{3} \right) \\
\end{aligned}

Using $\sin 2x = 2 \sin x \cos x$

$= \frac{1}{2} \sin \frac{2\alpha}{3}$

$= RHS$
• August 5th 2008, 02:30 AM
Kitty216
Wow.. replacing the alpha/3 with a B really helps! So at the end do I just replace the all the B's back with the alpha/3?
• August 5th 2008, 03:00 AM
ticbol
Quote:

Originally Posted by Kitty216
Wow.. replacing the alpha/3 with a B really helps! So at the end do I just replace the all the B's back with the alpha/3?

Yes.

Later on, in your future Math subjects, you will see that substitutions like that are common.
• August 5th 2008, 11:25 AM
Kitty216
Thanks for the help! =DD