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Another trig question

  1. Apr 11, 2014 #1
    1. The problem statement, all variables and given/known data
    show that sin4x + 2sin2x = 8sinxcos^3x



    2. Relevant equations
    sin2x =2sinxcosx


    3. The attempt at a solution
    I started out by letting sin4x = sin(2x*2) so that I could plugg in sin2x = 2sinxcosx in the equation.
    sin(2x*2) +2sin2x =
    sin2*sin2x +2sin2x =
    sin2 * (2sinxcosx) + 2*sin2x =
    sin2* (2sinxcosx) + 2* (2sinxcosx)
    and I end up with 8sinxcosx
    which is wrong!
    ??
     
  2. jcsd
  3. Apr 11, 2014 #2
    sin(ab) ≠ (sina)(sinb) i.e sin4x≠(sin2)(sin2x)

    sin4x = 2sin2xcos2x
     
  4. Apr 11, 2014 #3
    Ok, did you derive that from sin2x = 2sinxcosx?
    In that case, I don´t understand why sin4x = 2sin2xcos2x and not 4sin2xcos2x (i.e twice as much as sin2x = 2sinxcosx?)
     
  5. Apr 11, 2014 #4
    Yes

    Please answer this - What is sin2A ?
     
    Last edited: Apr 11, 2014
  6. Apr 11, 2014 #5
    2sinAcosA?

    Or is this some sort of a trick question?
     
  7. Apr 11, 2014 #6
    Not at all . Just trying to make you understand the formula :smile:

    sin2A = 2sinAcosA . Now replace A with 2x on both the sides .What do you get ?
     
  8. Apr 11, 2014 #7
    Ah!! I see. Thanks!
    sin2*2x= 2sin2xcos2x.
    I´ll carry on now and see if I get anywhere.
     
  9. Apr 11, 2014 #8
    Somehow it keeps on getting more and more complicated. Do you have any idea on how I could keep it "simple"?

    sin4x + 2sin2x = 8sinxcos^3x
    on the left hand side:
    2sin2xcos2x + 2sin2x =
    2*2sinxcosx(cos^2x - sin^2x) + 2*2sinxcosx =
    4sinxcos^3x - 2sin^3xcosx + 4sinxcosx

    It really doesn´t feel right...
     
  10. Apr 11, 2014 #9
    Is there something common in the two terms ? If yes ,take out the common factor .What do you get ?
     
  11. Apr 11, 2014 #10
    2sin2x(cos2x + 1)?
     
  12. Apr 11, 2014 #11
    Good...

    1+cos2x = ?
     
  13. Apr 11, 2014 #12
    1 + cos^2x - sin^2x
    ?
     
  14. Apr 11, 2014 #13
    Correct...Is there some other way of expressing this expression ? Can a couple of terms be combined ?
     
  15. Apr 11, 2014 #14
    There are two options:
    1) 1 + cos^2x -sin^2x =
    1+cos^2x - (1-cos^2x)=2cos^2x

    2) 1+(1-sin^2x)-sin^2x=
    1+1 - sin^2x -sin^2x = 2-2sin^2x = 2(1-sin^2x)

    ?
     
  16. Apr 11, 2014 #15
    Right...

    Now put sin2x = 2sinxcosx and (1+cos2x) = 2cos2x in the expression you have in post#10.
     
    Last edited: Apr 11, 2014
  17. Apr 11, 2014 #16
    Perfect! Now I got it.
    Thanks a lot!! you´ve been a massive help.
    Do you have any general tips on how to solve such problems?
     
  18. Apr 11, 2014 #17
    Remembering trigonometric formulas and practicing variety of problems of increasing complexity is the key .If you remember the formulas then it will help you in manipulating the trigonometric expressions .

    You should be comfortable with double ,triple angle formulas .How to convert from one form to other .

    Cos2x = cos2x-sin2x = 1-2sin2x = 2cos2x-1 .

    sinx = √[(1-cos2x)/2] ; cosx = √[(1+cos2x)/2]

    All the above are various ways of writing the same thing .With practice you will recognize which form to use .
     
  19. Apr 11, 2014 #18
    Ok. I´ll do my best. Thanks once again!
     
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