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Math Help - Sinusodial functions

  1. #1
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    Exclamation Sinusodial functions

    Find the regions where sinx < x
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  2. #2
    No one in Particular VonNemo19's Avatar
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    Quote Originally Posted by math123456 View Post
    Find the regions where sinx < x

    Maybe this will help. sinx\leq{x}\Rightarrow{sinx}-x\leq0







    I took this image from http://webgraphing.com/
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  3. #3
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    Exclamation sinusodial functions

    i'm still confused, i would you solve the function to find those regions.
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  4. #4
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    Quote Originally Posted by math123456 View Post
    Find the regions where sinx < x
    Are exact values of the endpoints of the intervals required?
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  5. #5
    No one in Particular VonNemo19's Avatar
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    Quote Originally Posted by math123456 View Post
    i'm still confused, i would you solve the function to find those regions.
    sinx is always between -1 and 1. It follows then, that when |x|>1, the graph will either become more and more negative, or more and more positive. Know that, all you must do is analyze the behavior of the graph for -1<x<1 and determine the behavior of the function for those values.
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  6. #6
    Senior Member pankaj's Avatar
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    Let f(x)=\sin x-x

    f'(x)=\cos x-1=-2\sin^2\frac{x}{2}\leq 0

    Thus f(x) is a decreasing funtion(i.e. if x_{1}<x_{2} ,then f(x_{1})>f(x_{2}))

    Thus for x\geq 0

    f(x)\leq f(0)

    f(x)\leq 0

    \sin x-x\leq 0


    \sin x\leq x for x\in [0,\infty)
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  7. #7
    No one in Particular VonNemo19's Avatar
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    Quote Originally Posted by pankaj View Post
    Let f(x)=\sin x-x

    f'(x)=\cos x-1=-2\sin^2\frac{x}{2}\leq 0

    Thus f(x) is a decreasing funtion(i.e. if x_{1}<x_{2} ,then f(x_{1})>f(x_{2}))

    Thus for x\geq 0

    f(x)\leq f(0)

    f(x)\leq 0

    \sin x-x\leq 0


    \sin x\leq x for x\in [0,\infty)
    This was my initial mode of attack as well, but this is the pre-calculus subforum, so I disregarded the use of the first derivative test.
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  8. #8
    Senior Member pankaj's Avatar
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    Quote Originally Posted by VonNemo19 View Post
    This was my initial mode of attack as well, but this is the pre-calculus subforum, so I disregarded the use of the first derivative test.
    O.K. then.

    Now consider acircle of radius 1 unit.Let a chord AB of this circle subtend an angle x at the centre O of the circle.

    Clearly,Ar( \triangle AOB)<Area(\sector AOB).

    Ar( \triangle AOB)= \frac{1}{2}\sin x

    Area of sector AOB= \frac{1}{2}.1^2.x=\frac{1}{2}.x

    \frac{1}{2}\sin x\leq \frac{1}{2}x

     <br />
\sin x\leq x<br />

    which is true for all non-negative x.If x<0,then \sin x>x
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  9. #9
    Super Member dhiab's Avatar
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    Hello : this graphic -solution
    Attached Files Attached Files
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