anonymous
  • anonymous
Ok, verifying an identity question 2!
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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katieb
  • katieb
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anonymous
  • anonymous
\[2\sin B \cos ^{3}B + 2\sin ^{3}B \cos B=\sin 2B\] I can't make heads or tails of this!
TuringTest
  • TuringTest
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
  • TuringTest
so how about try factoring out \(2\sin B\cos B\) from the left and tell me what you get

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anonymous
  • anonymous
That would be something like \[\sin ^{2}B \times \cos ^{2}b\] or would it be sin 2B
TuringTest
  • TuringTest
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
  • anonymous
The last parenthesis with squared sine and cosine
anonymous
  • anonymous
is equal to one.
TuringTest
  • TuringTest
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
  • anonymous
You-re right Turing test. sorry.
TuringTest
  • TuringTest
it's all good buddy ;) emily is not responding anyway
anonymous
  • anonymous
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
  • anonymous
p.s. sorry about the delay, i have a older computer :(
anonymous
  • anonymous
It should be good if Emily take a look at basic trigonometric identities before attempt to solve any of this "complex" identities.
anonymous
  • anonymous
lol just realized that was the answer, sorry !
anonymous
  • anonymous
Thanks both of u, i have helped a lot!
TuringTest
  • TuringTest
anytime :D

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