You can factor out 2sinxcosx from that entire expression. Multiply it back in and maybe you'll get what they're doing.:
Hopefully you'll recognize the expressions for sin2x and cos2x.
Okay, I'm trying to simplify a trigonometric thing. I say "thing" because it's not an equation, and that has me a little confused.
Beginning:
So, I'm not given what that is equal to, but told to factor and simplify. I looked up the answer, and they gave me the last two steps.
Second-to-last:
Last:
Okay, getting from the second-to-last step to the last is easy, using the double-angle formulae. I just don't know how to get from the beginning to the second-to-last step! I checked it out in MS Excel, and
does in fact equal
... I just don't know how they got there. Is it that there is some common factor in the first two terms of the original thing? Can I divide each side of a subtraction operation in half or something?
Oh thanks! Hop David has supplied the steps I was missing.
Let me reiterate to see if I have it right. First, I can pull out (2 sin^{2}x) from 4 sin^{3}x cosx. That leaves me with 2 sinx cosx on both sides of my subtraction sign. I missed that at first.
Next, the point I REALLY missed, was that A - AB = A(1-B). It would be so easy to go the other way, but my reverse gears are rusty. Thus I can go from
2 sinx cosx - 2 sinx cosx (2 sin^{2}x)
to
2 sinx cosx (1 - 2 sin^{2}x).
THEN I can use my double-angle identities. Geez, I've done like twenty problems in my book after that one ... they're all cakewalks by comparison.