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Thread: trigonometry problem with sin(sqrt(B-x^2))

  1. #1
    Aug 2008

    trigonometry problem with sin(sqrt(B-x^2))

    If B is a constant, for what value(s) of x will the graph of the function f(x) = sin(sqrt(B-x^2)) have a maximum.

    i know sin has a max when it equals 1 so i set sin(sqrt(B-x^2)) = 1 and therefore sqrt(B-x^2) = pi/2 + 2(pi)n where n is an integer. however my teacher seemed to have done the problem a different way. He did:

    sin(sqrt(B-x^2 + 2(pi)r)) = 1 so therefore pi/2 = sqrt(B - x^2 + 2(pi)r). i don't understand why a 2(pi) r term was added under the radical as opposed to what i did which was add the 2(pi)n term after the pi/2. Was my way correct?
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  2. #2
    MHF Contributor
    Oct 2009
    I agree with your approach, and I also don't understand your teacher's solution.
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