http://i146.photobucket.com/albums/r...6/problem5.jpg

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

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- Aug 5th 2008, 12:34 AMKitty216Verify Algebraically
http://i146.photobucket.com/albums/r...6/problem5.jpg

This one really stumped me =[ Does anyone know how to verify it? - Aug 5th 2008, 01:19 AMticbol
- Aug 5th 2008, 01:21 AMGusbob
$\displaystyle \sin \frac{\alpha}{3} \cos \frac{\alpha}{3} = \frac{1}{2} \sin \frac{2\alpha}{3}$

$\displaystyle

\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 $\displaystyle \sin 2x = 2 \sin x \cos x $

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

$\displaystyle = RHS$ - Aug 5th 2008, 02:30 AMKitty216
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?

- Aug 5th 2008, 03:00 AMticbol
- Aug 5th 2008, 11:25 AMKitty216
Thanks for the help! =DD