1. ## Verify Algebraically

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

2. Originally Posted by Kitty216

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.

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

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

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

$\displaystyle = RHS$

4. 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?

5. 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.

6. Thanks for the help! =DD