Ok, verifying an identity question 2!

Question

anonymous · Answer

$$2\sin B \cos ^{3}B + 2\sin ^{3}B \cos B=\sin 2B$$
I can't make heads or tails of this!

turingtest · Answer

The way to do this is not obvious, so I'm going to show you what I will work from to do this problem: http://tutorial.math.lamar.edu/pdf/Trig_Cheat_Sheet.pdf

turingtest · Answer

so how about try factoring out $2\sin B\cos B$ from the left and tell me what you get

anonymous · Answer

That would be something like $$\sin ^{2}B \times \cos ^{2}b$$ or would it be sin 2B

turingtest · Answer

no, I meant$$2\sin B \cos ^{3}B + 2\sin ^{3}B \cos B=2\sin B\cos B(\cos^2B+\sin^2B)$$now how does this simplify?

anonymous · Answer

The last parenthesis with squared sine and cosine

anonymous · Answer

is equal to one.

turingtest · Answer

I was really hoping to let emily realize that on her own juancarlos, but yeah...
so why don't you guys go ahead and finish the proof

anonymous · Answer

You-re right Turing test. sorry.

turingtest · Answer

it's all good buddy ;)
emily is not responding anyway

anonymous · Answer

ok, it fine guys 
so$$2\sin B cosB \times(\cos ^{2}B+\sin ^{2}B) or 2\sin B cosB(1)  $$ but how would I get rid of the cos

anonymous · Answer

p.s. sorry about the delay, i have a older computer :(

anonymous · Answer

It should be good if Emily take a look at basic trigonometric identities before attempt to solve any of this "complex" identities.

anonymous · Answer

lol just realized that was the answer, sorry !

anonymous · Answer

Thanks both of u, i have helped a lot!

turingtest · Answer

anytime :D