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Trigonometric identity

  1. Nov 27, 2008 #1
    Why is [tex]cos^2 x = \frac{1}{2} + \frac{cos(2x)}{2}[/tex]

    ?
     
  2. jcsd
  3. Nov 27, 2008 #2

    lurflurf

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    Homework Helper

    why is
    [cos(x)]^2=1-[sin(x)]^2
    why is
    [cos(x)]^2=cos(2x)+[sin(x)]^2
     
  4. Nov 27, 2008 #3

    HallsofIvy

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    Starting where? Do you know cos(a+ b)= cos(a)cos(b)- sin(a)sin(b)? If so then take a= b= x. If not then what definition of cos(x) are you using so that you can get that?
     
  5. Nov 27, 2008 #4
    I think this is what you wanted to know:
    cos2x+sin2x=1
    cos2x=1-sin2x
    Since cos(2x)=cos2x-sin2x
    cos2x=1+cos(2x)-cos2x
    2cos2x=1+cos(2x)
    cos2x=1/2 + cos(2x)/2
     
  6. Nov 27, 2008 #5
    RHS:


    [tex]\frac{1}{2} + \frac{cos(2x)}{2}[/tex]

    [tex]=\frac{1+cos(2x)}{2}[/tex]

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

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

    [tex]=cos^{2}x[/tex]

    [tex]=LHS(Shown)[/tex]
     
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