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Prove identity sin4x=(4sinxcosx)(1-2sin^2x)

  1. May 30, 2012 #1
    1. The problem statement, all variables and given/known data
    sin4x=(4sinxcosx)(1-2sin^2x)


    2. Relevant equations
    Trig identities.


    3. The attempt at a solution
    sin4x=(4sinxcosx)(1-2sin^2x)
    (4sinxcosx)(cos2x)
    stuck right here...
     
  2. jcsd
  3. May 30, 2012 #2
    Try starting with sin(4x) = sin(2x + 2x), and applying the formula for the sine of the sum of two angles.
     
  4. May 30, 2012 #3
    so it would be 2sinxcosx+2sinxcosx--->4sin2xcos2x-->is this right so far? i think im wrong on the last part...btw im horrible at precalc
     
  5. May 30, 2012 #4
    Check the boldfaced items in the above, but the final result is correct. Just substitute the double angle formulas for sin2x and cos2x to complete the proof of the identity.
     
  6. May 30, 2012 #5
    ok i get this-->4sin2xcos2x-->4(2sinxcosx)(1-2sin^2x)--> was I supposed to do something to the bolded 4? as the final thing is (4sinxcosx)(1-2sin^2x)
     
  7. May 31, 2012 #6
    never mind i get it. I was over thinking it lol
     
  8. May 31, 2012 #7
    You can also do it from RHS to LHS

    [tex]sin(4x) = (4sin\: x.cos\: x)(1-2sin^2x)[/tex]

    Write both the brackets in terms of sin2x and cos2x.
     
  9. May 31, 2012 #8
    I thought I had been right but I was still lost. can you please look over my work and tell me where I went wrong. thanks
    sin4x=(4sinxcosx)(1-sin^2x)--->sin4x= sin(2x+2x)--->2sinxcosx+2sinxcosx= 4sin2xcos2x-->4(2sinxcosx)(cos2x)----> I don't know how 4(2sinxcosx) is equal to (4sinxcosx).
     
  10. May 31, 2012 #9
    This part is wrong.

    Whats the formula for sin(2x) in terms of x?

    Once you answer that, your equation is [itex]sin2(2x)[/itex], just expand it.
     
  11. May 31, 2012 #10
    I did that because isn't sin4x the same thing as sin2x + sin 2x? and according to my double angle formula sheet sin2x= 2sinxcosx, so this is why i did 2sinxcosx +2sinxcosx = 4sin2xcos2x . I don't really understand what you mean by "sin(2x) in terms of x" I am not the greatest in math -.- .
     
  12. May 31, 2012 #11
    You already have the answer! Use that bold relation!

    And no, [itex]sin4x\neq sin 2x+sin2x[/itex]

    But,

    [tex]sin4x=sin2(2x)[/tex]
     
  13. May 31, 2012 #12
    ok I think I understand it now. Can I ask you one last favor and please write it down as if you had just gotten the question on a quiz or something? I want to see every step by step.
    so 1) Prove : sin4x=(4sinxcosx)(1-2sin^2x)
    please Its late where I live and I need some sleep.
     
  14. May 31, 2012 #13
    No.

    By the homework guidelines, I cannot. And I preferably wouldn't. You won't enjoy math unless you do it by yourself! :smile:


    As you say you understand it, why don't you write it down here? If there's any mistakes, I can point them out.
     
  15. May 31, 2012 #14
    ok well here it goes... sin4x=(4sinxcosx)(1-2sin^2x)-->sin4x--> sin2(2x)--> 2(2sinxcosx)-->4sin2xcos2x-->...the confusing part-->4(2sinxcosx)(1-sin^2x)-->(4sinxcosx)(1-sin^2x)

    on the italic part am i supposed to distribute the 2 out to 2sinxcosx so i can get the 4sin2xcos2x so that i can factor out the 4 and then have 4(sin2xcos2x) and then use double angle formula and get 4(2sinxcosx)(cos2x, which becomes1-sin^2x)?
     
    Last edited: May 31, 2012
  16. May 31, 2012 #15
    Umm, no :redface:

    [tex]sin4x = sin2(2x) = sin2y[/tex]

    here y=2x. Now using the sin2y formula...

    [tex]sin2y = 2siny.cosy[/tex]

    Buy you already have y=2x. So,

    [tex]sin4x=2(sin2x.cos2x)[/tex]

    Now proceed by writing sin2x and cos2x in terms of sinx and cosx.
     
  17. May 31, 2012 #16
    ok makes more sense now, but now that i have 2(2sinxcosx)(cos2x)...does the 2 way in front get distrubuted like this--> (4sinxcosx)(cos2x) why not like this (4sin2xcos2x)(cos2x)?
     
  18. May 31, 2012 #17
    Yes, you get (4sinx.cosx)(cos2x).

    (4sin2x.cos2x)(cos2x) doesn't make sense, where did you get it from?? There wasn't a 2 along with a sin2x.cos2x to begin with.
     
  19. May 31, 2012 #18

    I guess this response is useless now but I would use the exponential forms of sine and cosine.
     
  20. May 31, 2012 #19
    thank you all who helped or tried to. I finally understand. The problem I was having was thinking 2(2sinxcosx)-->(4sin2xcos2x), but now i understand it just becomes 4sinxcosx. Really appreciate the help
     
  21. Jun 1, 2012 #20
    Glad you figured it out!! :biggrin:
     
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