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Math Help - Trigonometry

  1. #1
    Newbie
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    Sep 2008
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    Stockholm
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    15

    Trigonometry

    Hello!
    The problem is to find A=? B=? and C=?

    (7*cos x +3*sin x)^2 = A*cos 2x+B*sin 2x + C

    Can anyone help me with this problem?
    Thanks
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  2. #2
    Member
    Joined
    Dec 2008
    Posts
    111
    Dear Kluringen,

    you didn't say about x. I suppose the problem is in exact form: find the A, B, C so the equation will be true for all x (element of real).

    Well, we have three parameter --> we have to get three equation.
    I suggest you substitute 3 special value of x in the equation such as x = 0, x=pi/2 and x = pi. Then solve the system of equation for A, B, C.

    And in the end you must contol that these values (A, B, C) satisfy indeed the original equation for all x.
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  3. #3
    Newbie
    Joined
    Dec 2008
    Posts
    3
    (7*cosx+3*sinx)^2 = 49 cos^2 (x) + 9 sin^2 (x) + 2*7*3*Sinx * Cosx

    As we know that
    Sin2x = 2* sinx * cosx
    and
    Cos2x = Cos^2 (x) - Sin^2 (x)
    also we know that cos^2 (x) + sin^2 (x) = 1
    so solving the above 2 equations we will get

    Sin^2 (x) = (1 - cos2x)/2
    Cos^2 (x) = (1+cos2x)/2


    so putting these values in the above expansion

    49 (1 + cos2x)/2 + 9 (1-cos2x)/2 + 21 sin2x
    = 29 + 20 cos2x + 21 sin2x

    So now u can compare and get the answer

    A = 20, B = 21, C = 29
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