# $$\sin x < x < \tan x$$

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.

Note by Calvin Lin
4 years, 3 months ago

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On $$[0,\pi/2]$$, 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.

- 4 years, 3 months ago

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

Staff - 4 years, 2 months ago

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

- 4 years, 2 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.

- 4 years, 2 months ago

We can use Maclaurin expansion of sinx.

- 4 years, 2 months ago