Show that for \( 0 < x < \frac{\pi}{2} \), we have

\[ \sin x < x < \tan x .\]

This is a list of Calculus proof based problems that I like. Please avoid posting complete solutions, so that others can work on it.

Show that for \( 0 < x < \frac{\pi}{2} \), we have

\[ \sin x < x < \tan x .\]

This is a list of Calculus proof based problems that I like. Please avoid posting complete solutions, so that others can work on it.

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TopNewestOn \([0,\pi/2]\), consider the two functions : \(f(x)=\sin x-x\), and : \(g(x)=\tan x -x\).

\(f'(x)=\cos x -1\leq 0\), which means that \(f(x)\leq f(0)=0\).

And : \(g'(x)=\sec^2 x -1\leq 0\), which means that \(g(x)\geq g(0)=0\).

And since these two function are not constant on any open interval we get sharp inequalities on the open interval \((0,\pi/2)\), which get us to what we want. – Haroun Meghaichi · 3 years ago

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– Calvin Lin Staff · 2 years, 12 months ago

Do you know a non-calculus approach to this problem?Log in to reply

– Akash Shah · 2 years, 11 months ago

By looking at the graphs of sinx , tanx and x we can say it is true.Log in to reply

– Haroun Meghaichi · 2 years, 11 months ago

I can't draw a circle here, the whole idea is for a point \(M\) on the first quadrant (and not on \((1,0)\) ) of the unit circle, the arc from \((1,0)\) to \(M\) is longer than the distance between \(M\) and the x-axis which gives us \(\sin x< x\). A similar approach would be with the other side of the inequality.Log in to reply

– Sambit Senapati · 2 years, 12 months ago

We can use Maclaurin expansion of sinx.Log in to reply