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]
Last edited by stapel; 03-09-2018 at 03:25 PM. Reason: Typing out the text in the graphics; creating useful subject line.
You would do well to find the Double Angle formula for sine. Hint: [tex]\sin(2x) = ...[/tex]
"Unique Answers Don't Care How You Find Them." - Many may have said it, but I hear it most from me.
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?
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!
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