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Math Help - Trig questions help

  1. #1
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    Trig questions help

    Hiya, i'm kind of stuck on this problem:

    Using cos2A = cos(squared)A - sin(squared) A or otherwise

    prove cos2a = 1/2(1+cos2A)


    any help appreciated, thanks!
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  2. #2
    MHF Contributor
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    prove cos2a = 1/2(1+cos2A) ?

    Is cos2a the same as cos2A ?
    if yes,
    cos2a = 1/2(1+cos2A)
    Multiply both sides by 2,
    2cos2a = 1 +cos2A
    2cos2a -cos2A = 1
    cos2a = 1 ----------------No, that is not true.

    That means cos2a is not the same as cos2A.

    I think your cos2a should have been cos^2(A), or (cosA)^2.
    That is the same as your "cos(squared)A"

    Let us see,
    cos^2(A) = (1/2)[1 +cos(2A)]
    cos^2(A) = (1/2)[{sin^2(A) +cos^2(A)} +{cos^2(A) -sin^2(A)}]
    cos^2(A) = (1/2)[sin^2(A) +cos^2(A) +cos^2(A) -sin^2(A)]
    cos^2(A) = (1/2)[2cos^2(A)]
    cos^2(A) = cos^2(A) -----------***
    Proven.

    ---------
    We used also the trig identity
    sin^2(A) +cos^2(A) = 1
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  3. #3
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    Sorry! Big oops, I pasted the wrong part of the question, god I feel like such an eejit!

    Yes that was a cos(squared)A on the left. I proved that fine, it was the next part that got me:

    hence or otherwise solve the equation -
    1+cos2x=cosx where 0 degrees </= x </= 360
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  4. #4
    MHF Contributor
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    hence or otherwise solve the equation -
    1+cos2x=cosx where 0 degrees </= x </= 360

    Don't feel too bad. Everyone commits mistakes.

    1 +cos(2X) = cosX
    1 +cos^2(X) -sin^2(X) = cosX
    We make them all in cosX,
    1 +cos^2(X) -[1 -cos^2(X)] = cosX
    1 +cos^2(X) -1 +cos^2(X) = cosX
    2cos^2(X) = cosX
    2cos^2(X) -cosX = 0
    (cosX)(2cosX -1) = 0

    cosX = 0
    X = arccos(0)
    x = 90deg or 270deg -------***

    2cosX -1 = 0
    2cosX = 1
    cosX = 1/2 --------positive. Cosine is positive in the 1st and 4th quadrants.
    X = arccos(1/2)
    X = 60deg ---in the 1st quadrant.
    X = 360 -60 = 300deg ---in the 4th quadrant.

    Therefore, X = 60, 90, 270, or 300 degrees. --------answer.
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