So i basically hate to solve this thing:
. . . . .[tex]4\, \sin\left(\alpha\, +\, \dfrac{\pi}{6}\right)\, \cos\left(\alpha\, -\, \dfrac{\pi}{6}\right)[/tex]
So i basically hate to solve this thing:
. . . . .[tex]4\, \sin\left(\alpha\, +\, \dfrac{\pi}{6}\right)\, \cos\left(\alpha\, -\, \dfrac{\pi}{6}\right)[/tex]
You would do well to find the Double Angle formula for sine. Hint: [tex]\sin(2x) = ...[/tex]
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?