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Math Help cos^6 x

  1. Dec 5, 2004 #1
    Rewrite the expression in terms of the first power of the cosine:

    cos^6 x

    My guess would be that it would be cos(x)*cos(x)*cos(x)*cos(x)*cos(x)*cos(x)

    I'm not sure if that is it, or any way to start the problem even

    Please help me :cry:

    edit :

    cos^6(x)=cos^2(x)*cos^2(x)*cos^2(x)
    =.5(1+cos2x)*.5(1+cos2x)*.5(1+cos2x)
    then ???
    ?
     
    Last edited: Dec 5, 2004
  2. jcsd
  3. Dec 5, 2004 #2
    Maybe [tex]\cos^2{x}=\frac{1+\cos{2x}}{2}[/tex] will help?
     
  4. Dec 5, 2004 #3

    Isn't that what i have?

    cos^6(x)=cos^2(x)*cos^2(x)*cos^2(x)
    =.5(1+cos2x)*.5(1+cos2x)*.5(1+cos2x)

    i am lost again :bugeye:
     
    Last edited: Dec 5, 2004
  5. Dec 5, 2004 #4

    Hurkyl

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    Gold Member

    Do it again.
     
  6. Dec 5, 2004 #5
    what do you mean do it again
     
  7. Dec 5, 2004 #6
    Sorry, I must have posted right before you edited your original post.
     
  8. Dec 5, 2004 #7
    what would i need for the next step
     
  9. Dec 5, 2004 #8
    I really really need help someone
     
  10. Dec 5, 2004 #9
    Hmmm.. maybe it's

    [tex]
    \cos^6 x = (\cos^2 x)^3
    [/tex]

    so then ...

    [tex]
    \cos^6 x = ( \frac { 1 + cos 2x } { 2 } )^3
    [/tex]
     
  11. Dec 5, 2004 #10
    how is cos2x = 1 + cos2x / 2 ???

    isnt cos2x = 1-sin2x ??????
     
  12. Dec 5, 2004 #11
    [tex]cos(2x)=cos^2(x)-sin^2(x)[/tex]
    [tex]cos(2x)=cos^2(x)-(1-cos^2(x))[/tex]
    [tex]cos(2x)=2cos^2(x)-1[/tex]
    [tex]cos^2(x)= (\frac { 1 + cos 2x } { 2 } )[/tex]
     
  13. Dec 5, 2004 #12
    Since we can express cos(2x) and cos(3x) in terms of cos(x) it is a good idea to do so right away when you are asked such a question because you cannot break it up like (5x+x) because that would require a polynomial for cos(5x) in terms of cos(x) which you are not likely to remember or even derive easily. So the sensible thing to do would be to break it up as cos(3x+3x) or cos(2(3x)) and proceed further.
     
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