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Math Help - range of trigonometric function

  1. #1
    ohm
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    range of trigonometric function

    i have confusion in finding range of sin^4x + cos^2x.
    I have replace cos^2x as sin^2x -1 and let sin^2x as t.
    It becomes then t^2-t+1. Now how can i proceed?
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  2. #2
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    Re: range of trigonometric function

    Quote Originally Posted by ohm View Post
    i have confusion in finding range of sin^4x + cos^2x.
    I have replace cos^2x as sin^2x -1 and let sin^2x as t.
    It becomes then t^2-t+1. Now how can i proceed?
    That is a very good start. I suggest that you continue by completing the square, to get (t-\tfrac12)^2 + \tfrac34. You know that t must lie between 0 and 1, so from that you can work out the range of the function.
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    ohm
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    Re: range of trigonometric function

    Quote Originally Posted by Opalg View Post
    That is a very good start. I suggest that you continue by completing the square, to get (t-\tfrac12)^2 + \tfrac34. You know that t must lie between 0 and 1, so from that you can work out the range of the function.
    Sorry! I have done so, but not understanding whether i should put value of t here. If i do so answer is 1 in both cases. Help
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    Re: range of trigonometric function

    Ok, do you know the range of this function?

    (t-\tfrac12)^2 + \tfrac34.
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    ohm
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    Re: range of trigonometric function

    Quote Originally Posted by pickslides View Post
    Ok, do you know the range of this function?

    (t-\tfrac12)^2 + \tfrac34.
    no!
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    Re: range of trigonometric function

    do you know the range of t?
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    Re: range of trigonometric function

    Well I would say the range of t is (0.75, \infty)

    And as Opalg already suggested the range between 0 and 1 then the original functions range is the intersection of the 2.

    I.e \displaystyle [0,1] \cap \left[\frac{3}{4}, \infty\right) = \dots
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    Re: range of trigonometric function

    Quote Originally Posted by ohm View Post
    Quote Originally Posted by Opalg View Post
    That is a very good start. I suggest that you continue by completing the square, to get (t-\tfrac12)^2 + \tfrac34. You know that t must lie between 0 and 1, so from that you can work out the range of the function.
    Sorry! I have done so, but not understanding whether i should put value of t here. If i do so answer is 1 in both cases. Help
    You have to think about what happens as t goes from 0 to 1. As t goes from 0 to 1/2, (t-\tfrac12)^2 goes from 1/4 down to 0. And as t goes from 1/2 to 1, (t-\tfrac12)^2 goes from 0 back up to 1/4. So the range of (t-\tfrac12)^2 will be the interval [0,1/4]. It follows that the range of (t-\tfrac12)^2 + \tfrac34 is the interval [3/4,1].
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