Real Analysis Prove that 2/π ≤(sinx)/x ≤ 1 for all |x|≤ π/2 ? Just need the 2/π greater than part.
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Hint: convexity – Wojowu Apr 07 '18 at 15:56
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2https://math.stackexchange.com/questions/980551/how-to-show-sin-x-geq-frac2x-pi-x-in-0-frac-pi2?noredirect=1&lq=1 https://math.stackexchange.com/questions/1126449/proof-for-forall-x-in-0-frac-pi2-quad-sinx-ge-fracx2?noredirect=1&lq=1 https://math.stackexchange.com/questions/407517/the-sine-inequality-frac2-pi-x-le-sin-x-le-x-for-0x-frac-pi2?noredirect=1&lq=1 https://math.stackexchange.com/questions/213382/prove-an-inequality-with-a-sin-function-sinx-frac2-pi-x-for-0x-fr?noredirect=1&lq=1https://math.stackexchange.com/questions/2029743/show-that-x-sin-frac-pi-x2-forall-x-in-0-1?noredirect=1 – Clement C. Apr 07 '18 at 15:58
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Hint:
On the interval $\bigl[0,\frac\pi2\bigr]$, the sine function is concave. As a consequence, the slopes of the chords joining a point of the curve to the origin are decreasing, and $\frac2\pi$ is the slope of the chord joining the local maximum $\bigl(\frac\pi2,1\bigr)$ to the origin.
Bernard
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You mean that in the interval the slope of any point on sine curve (that is cosx ) is greater than the chord with slope 2/π.How are you then getting it in sinx / x form ? Integration of cosx < 2/π ? Sounds correct but I am looking for a solution based on monotonicity of functions. Thank you. – DEEPANSHU KAUL PHILIP Apr 07 '18 at 18:13
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The slope of the chord joining the origin to another point (or any fixed point to another point on its right), not the slope of the tangent. – Bernard Apr 07 '18 at 18:18
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That is fine but how is that leading up to 2/π ≤ sin(x) / x ? – DEEPANSHU KAUL PHILIP Apr 07 '18 at 18:37
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$\sin x/x$ is the slope of the chord joining the point with abscissa $x$ to the origin. It decreases down to its value at $\frac\pi2$. – Bernard Apr 07 '18 at 18:40
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I did it using g(x)=xcos(x)- sin(x) being decreasing and g(0)=0. – DEEPANSHU KAUL PHILIP Apr 07 '18 at 19:00
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Perfect! Actually this would be valid for any concave function (even non-differentiable) – as far as I remember, it is an exercise in Michael Spivak's Calculus. – Bernard Apr 07 '18 at 19:29
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We can assume that $$0<x\le \frac{\pi}{2}$$ since $$\frac{\sin(x)}{x}$$ is even. We multiply by $x>0$ and we have to prove that $$\frac{2}{\pi}x\le\sin(x)\le x$$. Defining $$f(x)=x-\sin(x)$$ then we get $$f'(x)=1-\cos(x)>0$$ and let $$g(x)=\sin(x)-\frac{2}{\pi}x$$ then we get $$g'(x)=\cos(x)-\frac{2}{\pi}$$ and $$g''(x)=-\sin(x)<0$$ Can you finish?
Dr. Sonnhard Graubner
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Sir, I did something similar. Unlike the f(x) function, g(x) is not strictly monotonic. It is changing slope at x= cos–¹(2/π). Also, please explain a bit more on how to get 2/π xsinx ≤ x ? – DEEPANSHU KAUL PHILIP Apr 07 '18 at 18:31