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Are these equal?

  1. Aug 18, 2012 #1
    I'm verify some trigonometry equations and am confused about a couple of things. (This is self-study, I'm not in school)

    The equation cos4x = 1+8cos^4-8cos^2 can be solved by re-writing as 2(cos2x)^2 -1 and factoring out which yields the correct answer, however based on what i've seen in other double angle identity equations one can re-write cos4x as cos(2x + 2x).

    I believe this can be written as [(2cos^2 - 1)(2cos^2-1)] + [(2cos^2-1)(2cos^2-1)], however the answer then comes out to 8cos^4-8cos^2+2. So it's off by "+1".

    What about my thinking is flawed? They can't both be correct!

    Thanks for your help.

    Joe
     
  2. jcsd
  3. Aug 18, 2012 #2
    cos(a+b)=cosacosb-sinasinb ,what are you doing with second one in cos(2x+2x)
     
  4. Aug 18, 2012 #3
    I'm obviously not writing the formula out correctly since the answer is not the same. I was just solving how I'v written it out [(2cos^2 - 1)(2cos^2-1)] + [(2cos^2-1)(2cos^2-1)]

    Is there another way to write out cos(4x) as cos(2x+2x) instead of cos 2[(cos2x)^2-1] and have the answer come out to be 1+ 8cos^2x - 8cos^2x?

    How do you add cos(2x + 2x)? or does cos4x not equal cos(2x+2x)?

    Thanks,
    Joe
     
    Last edited: Aug 18, 2012
  5. Aug 18, 2012 #4
    This section of PF library would be of help to you. It lists all the identities you need.
     
  6. Aug 18, 2012 #5

    SammyS

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    cos(2x + 2x) ≠ cos(2x) + cos(2x)

    You appear to be assuming that they are equal !
     
  7. Aug 18, 2012 #6

    CAF123

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    Yes, and to expand do what user andrien suggested.
    [tex] cos(2x + 2x) = cos2xcos2x -sin2xsin2x = cos^2 2x - sin^2 2x[/tex] and use known trigonometric formula to simplify the expression down to one involving only [itex] cos, [/itex] as required.
     
  8. Aug 18, 2012 #7

    Mentallic

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    [tex]\cos(4x)\neq \cos\left(2(\cos^2(2x)-1)\right)[/tex]

    Because you already know that [itex]\cos(2x)=2cos^2(x)-1[/itex] hence [itex]2\cos^2(2x)-1=\cos(4x)[/itex] (do you see how that works?) and so finally, if we plug this expression into
    [tex]\cos\left(2(\cos^2(2x)-1)\right)[/tex]
    we will have that equivalent to
    [tex]\cos\left(2\cos^2(2x)-2)\right)[/tex][tex]=\cos\left(2\cos^2(2x)-1+1)\right)[/tex][tex]=\cos\left(\cos(4x)+1)\right)[/tex]
     
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