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Homework Help: Request for help proving this identity!

  1. Dec 8, 2005 #1
    I've wracked my brains out trying to prove this identity. If anyone could offer some suggestions, I'd greatly appreciate it!

    (sin(x))^4+(cos(x))^4 = 0.25∙cos(4x)+0.75
     
  2. jcsd
  3. Dec 8, 2005 #2

    HallsofIvy

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    cos(2x)= cos2(x)- sin2(x)
    sin(2x)= 2sin(x)cos(x)

    So cos(4x)= ?
     
  4. Dec 8, 2005 #3
    Cos(4x)=Cos(2(2x))=
    ...So does that mean Cos(4x)=2((cosx)^2-(sinx)^2)?
     
  5. Dec 8, 2005 #4

    HallsofIvy

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    Only if x2 is the same as 2x.
     
  6. Dec 8, 2005 #5
    Hmmm...in that case...I'm not really sure how to go about reducing cos(4x) to a single angle using the double-angle cosine formula. :confused:

    EDIT: Ahh...I now see that cos(4x)=cos(2x+2x)=cos(2x)cos(2x)-sin(2x)sin(2x)...
    more work underway!

    EDIT2: Ohhh, here we go: cos(4x)=((cos(x))^2- (sin(x))^2)((cos(x))^2- (sin(x))^2)-(2sin(x)cos(x))(2sin(x)cos(x))!
     
    Last edited: Dec 8, 2005
  7. Dec 8, 2005 #6
    Hmm...this was all I could narrow it down to:

    cos(4x)= (cos(x))^4 + (sin(x))^4 - 6(sin(x))^2(cos(x))^2
     
  8. Dec 9, 2005 #7

    HallsofIvy

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    Good! That's exactly right.

    Now go back to the original identity. Because I dislike fractions, multiply through by 4. Then put formula you have above into the equation and move (almost) everything over to the left side.
     
  9. Dec 9, 2005 #8
    Hmm...after fiddling around some more following your suggestions I have:

    (sin(x))^4+(cos(x))^4 = = -2(sin(x))^2(cos(x))^2
     
  10. Dec 9, 2005 #9

    HallsofIvy

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    What happened to the "3/4" in the orginal equation??
     
  11. Dec 9, 2005 #10
    The 3/4 turned into a 3 after I multiplied everything by 4. Afterwards, the 4sin and 4cos on the left became 3sin and 3cose when moving the stuff on the right to the left. I then divided everything by 3.
     
    Last edited: Dec 9, 2005
  12. Dec 10, 2005 #11

    VietDao29

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    When divided by 3, shouldn't it become '1'?
    [tex]3 \cos ^ 4 x + 3 \sin ^ 4 x = -6 \sin ^ 2 x \cos ^ 2 x + 3[/tex]
    Devide both sides by 3 gives:
    [tex]\Leftrightarrow \cos ^ 4 x + \sin ^ 4 x = -2 \sin ^ 2 x \cos ^ 2 x + 1[/tex]
     
  13. Dec 10, 2005 #12

    HallsofIvy

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    Yes! Now where is the 1 you get when you divide by 3?

    Look at VietDao's post.
    After moving (almost) everything to the left you should have:
    [tex]cos^4(x)+ 2sin^2(x)cos^2(x)+ sin^4(x)= 1[/tex]

    Does that give you any ideas? How can you factor the left hand side?
     
    Last edited by a moderator: Dec 10, 2005
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