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Trigonometric identity

  1. Jan 14, 2007 #1
    In my book, (cos4x)^2 is written 1+cos8x without refering to any formula. Which trig. identity is used here?
     
  2. jcsd
  3. Jan 14, 2007 #2

    cristo

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    Try looking at the identity for cos(2x)
     
  4. Jan 14, 2007 #3
    The correct identity is (cos4x)^2 = (1+cos8x)/2 .
     
  5. Jan 14, 2007 #4
    You mean cos(2x) = (cosx)^2 - (sinx)^2 ?
     
  6. Jan 14, 2007 #5

    cristo

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    Yes, and as arunbg says, there is a factor of 1/2 missing from your given identity.
     
  7. Jan 14, 2007 #6
    the identity is cos^2x = (1 + cos2x)/2 is it not?
     
  8. Jan 14, 2007 #7
    Yes, my mistake.
     
  9. Jan 14, 2007 #8

    cristo

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    One can derive this from the double angle identity for cos(2x) using further the identity that cos2x+sin2x=1
     
  10. Jan 15, 2007 #9
    ..........
     
    Last edited: Jan 15, 2007
  11. Jan 15, 2007 #10

    dextercioby

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    Nope.

    [tex] \sin^{2} x=\frac{1-\cos 2x}{2} [/tex]

    Daniel.
     
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