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Math Help - Find the extrema of a trigonometric function

  1. #1
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    Find the extrema of a trigonometric function

    I am to find the extrema of the following function. f(x) = 2 sin x + cos 2x when 0 < x < 2pi. taking the derivative of the function I come up with 2 cos x -2 sin 2x. At this point I need to find the extrema on the interval 0 < x < 2pi. I think the solution is to evaluate 2 cos x = 0. being cos x = -2 and then, -2 sin 2x = 0. that being sin x = 2/2 = sin 1, being pi/2. OR, am I to use the double angle identity in the function and change the derivative of 2 cos x to, 1- 2sin^2 x - 2 sin 2x = 0. I am confused at this point. Can someone please help?
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  2. #2
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    Quote Originally Posted by bosmith View Post
    I am to find the extrema of the following function. f(x) = 2 sin x + cos 2x when 0 < x < 2pi. taking the derivative of the function I come up with 2 cos x -2 sin 2x. At this point I need to find the extrema on the interval 0 < x < 2pi. I think the solution is to evaluate 2 cos x = 0. being cos x = -2 and then, -2 sin 2x = 0. that being sin x = 2/2 = sin 1, being pi/2. OR, am I to use the double angle identity in the function and change the derivative of 2 cos x to, 1- 2sin^2 x - 2 sin 2x = 0. I am confused at this point. Can someone please help?
    You're logic is all wrong when you set 2cos(x)=0 and 2sin(2x)=0. Remember, in algebra, you can only to set factors equal to zero. You can't just set the terms of a sum equal to zero. You could factor the derivative:

    2cos(x)-2sin(2x)=0

    2(cos(x)-sin(2x))=0

    This means that:

    cos(x)-sin(2x)=0

    cos(x)=2cos(x)sin(x)
    ---------------------------------------------------------------------------------
    1=2sin(x) Edit, MY logic was wrong on this step. See the post below.
    ---------------------------------------------------------------------------------
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  3. #3
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    Quote Originally Posted by adkinsjr View Post
    You're logic is all wrong in you set 2cos(x)=0 and 2sin(2x)=0. Remember in algebra, you have to set factors equal to zero. You can't just set the terms of a sum equal to zero. You could factor the derivative:

    2cos(x)-2sin(2x)=0

    2(cos(x)-sin(2x))=0

    This means that:

    cos(x)-sin(2x)=0

    cos(x)=2cos(x)sin(x)

    1=2sin(x)
    You're potentially dividing by 0 here, it would be better to factorise

    cos(x) = 2cos(x)sin(x)

    2cos(x)sin(x)-cos(x) = 0

    cos(x)(2sin(x)-1) = 0

    Either cos(x) = 0 or sin(x) = \frac{1}{2}
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  4. #4
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    Oh yes, thank you. Now MY logic is wrong when I cancel the cosines.
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  5. #5
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    adkinsjr

    Thank you for assisting me. I also want to thank as well as e^(i*pi) . Alright!!!
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