Prove 2πx≤sinx≤x for x∈[0,π2].
I tried to do this in two ways, I'm not sure about CMVT and I have a problem with the other way.
Using Cauchy's MVT:
RHS:
sinx≤x⟹sinxx≤1
So define:
f(x)=sinx, g(x)=x then from CMVT:
f(π2)g(π2)=f′(c)g′(c)=cosc
and from the fact that c is between 0 and π/2⟹cosc≤1.
LHS: In the same manner but here I run into some trouble:
2πx≤sinx⟹2xπsinx≤1
So:
f(π2)g(π2)=f′(c)g′(c)⟹1sinπ2=2πcosc
Here actually
1sinπ2=1 so it's also ≤1
Is it correct to use CMVT like this ?
The other way:
We want to show: f(x)=sinx−x<0 and g(x)=2xπ−sinx<0 by deriving both it's easy to show that the inequality stands for f but for g it isn't so obvious that g′(x)=2π−cosx is negative. In fact for x=π2 it's positive. Please help figure this out.
This is the same The sine inequality 2πx≤sinx≤x for $0
Note: I can't use integrals.
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