need some help with trig function: 4 sin(a + pi/6) cos(a - pi/6)

[chaNio]

So i basically hate to solve this thing:

. . . . .$\displaystyle 4\, \sin\left(\alpha\, +\, \dfrac{\pi}{6}\right)\, \cos\left(\alpha\, -\, \dfrac{\pi}{6}\right)$

 

[tkhunny]

You would do well to find the Double Angle formula for sine. Hint: $\displaystyle \sin(2x) = ...$

 

[Dr.Peterson]

tkhunny's suggestion is particularly apt if the expression is

4 sin(alpha + pi/6) cos(alpha + pi/6)

But it looks to me more like

4 sin(alpha + pi/6) cos(alpha - pi/6)

Am I right about that? (I'm also not sure I'm reading the "alpha" correctly.)

If so, then there are some identities called "product-to-sum" identities you can use; if you haven't seen them, they can be derived from the angle-sum and angle-difference identities.

Does any of that sound familiar?

 

[stapel]

So i basically hate to solve this thing:

. . . . .$\displaystyle 4\, \sin\left(\alpha\, +\, \dfrac{\pi}{6}\right)\, \cos\left(\alpha\, -\, \dfrac{\pi}{6}\right)$

There is no "equals", so this is not an "equation" and cannot be "solved". What were the actual instructions?

When you reply, please include a clear listing of your thoughts and efforts so far. Thank you!