# Prove that 2/π ≤(sinx)/x ≤ 1 for all |x|≤ π/2. [duplicate]

Real Analysis Prove that 2/π ≤(sinx)/x ≤ 1 for all |x|≤ π/2 ? Just need the 2/π greater than part.

## marked as duplicate by Clement C., John Doe, Leucippus, Misha Lavrov, Jack D'Aurizio real-analysis StackExchange.ready(function() { if (StackExchange.options.isMobile) return; $('.dupe-hammer-message-hover:not(.hover-bound)').each(function() { var$hover = $(this).addClass('hover-bound'),$msg = $hover.siblings('.dupe-hammer-message');$hover.hover( function() { $hover.showInfoMessage('', { messageElement:$msg.clone().show(), transient: false, position: { my: 'bottom left', at: 'top center', offsetTop: -7 }, dismissable: false, relativeToBody: true }); }, function() { StackExchange.helpers.removeMessages(); } ); }); }); Apr 7 '18 at 16:21

• Hint: convexity – Wojowu Apr 7 '18 at 15:56
• – Clement C. Apr 7 '18 at 15:58

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.

• 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 7 '18 at 18:13
• 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 7 '18 at 18:18
• That is fine but how is that leading up to 2/π ≤ sin(x) / x ? – DEEPANSHU KAUL PHILIP Apr 7 '18 at 18:37
• $\sin x/x$ 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 7 '18 at 18:40
• Alright got it ! Thank you. – DEEPANSHU KAUL PHILIP Apr 7 '18 at 18:47

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?

• 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 7 '18 at 18:31