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Math Help - Range of cos(x)

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    Range of cos(x)

    Hi! I need to find the range of the function y = cos(x) on the interval -pi/6<x<pi/4. Help would be appreciated!
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    Re: Range of cos(x)

    The correct answer is something like: [p,q]. I know that q = 1, but cannot find p! Any ideas?
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    Re: Range of cos(x)

    Quote Originally Posted by Dmiller93 View Post
    The correct answer is something like: [p,q]. I know that q = 1, but cannot find p! Any ideas?
    p=\frac{\sqrt2}{2}
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  4. #4
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    Re: Range of cos(x)

    If you want a kind of graphical intepretation of the problem then draw the goniometric circle and also an angle \alpha=\frac{\pi}{4} and an angle \beta=\frac{-\pi}{6}.
    You can see that \cos\left(\alpha)<\cos\left(\beta).

    If \beta \rightarrow 0 then \cos(\beta) \rightarrow 1
    If \alpha \rightarrow 0 then \cos(\alpha) \rightarrow 1

    And offcourse if \alpha=\beta=0 then \cos(\alpha)=\cos(\beta)=1

    That means that \cos(\alpha)=\cos\left(\frac{\pi}{4}\right)=\frac{  \sqrt{2}}{2}
    is the lower limit of the range and 1 the upper limit of the range, therefore the range is:
    \[\frac{\sqrt{2}}{2},1\]
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    Re: Range of cos(x)

    Cos(x) decreases, from 1 to 0, for x from 0 to \pi/2. \pi/4 is in that range so one end point is cos(\pi/4). One way to remember that is to remember that \pi/4+ \pi/4= \pi/2 so if a right triangle has one angle \pi/4 radians, the other is also. Of course, if two angles are equal, the sides opposite them are also equal. That is, we have an isosceles right triangle. If one leg has length s, so has the other and the length of the hypotenuse, c, is given by c^2= s^2+ s^2= 2s^2 so c= s\sqrt{2}. The cosine is "opposite side over they hypotenuse" so cos(\pi/4)= \frac{s}{s\sqrt{2}}= \frac{1}{\sqrt{2}}= \frac{\sqrt{2}}{2}.
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