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Math Help - Finding crossover points of 2 trig functions

  1. #1
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    Finding crossover points of 2 trig functions

    Hi.

    Can anyone help me find the angles (values of t) where:

    25*sin(t) = 22*sin(2t) - 18*cos(2t)


    I know the general way to do it is to get all one trig function and form a quadratic then solve it, but I'm not too hot on the identities and I can't seem to get rid of the double angle bits and make everything sine or anything like that!


    Help!


    Thanks!
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  2. #2
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    Quote Originally Posted by TheOnlyNameLeft View Post
    Hi.

    Can anyone help me find the angles (values of t) where:

    25*sin(t) = 22*sin(2t) - 18*cos(2t)


    I know the general way to do it is to get all one trig function and form a quadratic then solve it, but I'm not too hot on the identities and I can't seem to get rid of the double angle bits and make everything sine or anything like that!


    Help!


    Thanks!
    25*sin(t) = 22*sin(2t) - 18*cos(2t)

    sin(2t) = 2*sin(t)*cos(t)
    cos(2t) = 1 - 2*sin^2(t)

    So
    25*sin(t) = 44*sin(t)*cos(t) - 18 + 36*sin^2(t)

    Now, the problem is that we need to get rid of the cos(t) term somehow. There is only one reasonable way to do this: we know that sin^2(t) + cos^2(t) = 1, so...
    cos(t) = (+/-)sqrt{1 - sin^2(t)}

    Thus:
    25*sin(t) = (+/-)44*sin(t)*sqrt{1 - sin^2(t)} - 18 + 36*sin^2(t)

    Do yourself a favor at this point, let x = sin(t). Then:
    25x = (+/-)44x*sqrt{1 - x^2} - 18 + 36x^2

    Isolate the square root term:
    (+/-)44x*sqrt{1 - x^2} = -36x^2 + 25x + 18

    Square both sides:
    1936x^2*(1 - x^2) = (-36x^2 + 25x + 18)^2

    1936x^2 - 1936x^4 = 1296x^4 - 1800x^3 - 671x^2 + 900x + 324

    3232x^4 -1800x^3 - 2607x^2 + 900x + 324 = 0

    There's nothing easy about solving this, AND we need to check each x back in the original problem:
    25x = (+/-)44x*sqrt{1 - x^2} - 18 + 36x^2

    Once that is done, the final solution is
    t = asn(x)

    I have attached a graph of the polynomial function. I'll let you do the approximations and get the solutions for t.

    -Dan
    Attached Thumbnails Attached Thumbnails Finding crossover points of 2 trig functions-polynomial.jpg  
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  3. #3
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    Thanks a lot.


    Had some attempts but didn't think of getting rid of the cos(t) using the sin^2 + cos^2 = 1 identity!



    That's actually a very small part of a dynamics exam question about finding required flywheel sizes. I think I'll have to go visit the lecturer and see if he's expecting us to do things like that in the exam or use the graphs given to estimate the angles where the functions cross!


    Thanks again.
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