Prove $0 < x < \pi /2 \implies \sin x > x/\sqrt{x^2+1}$ using Mean Value Theorem

Question

I'm solving the following problem:

Show that if $0 < x < \pi /2$ then $\sin x > \dfrac{x}{\sqrt{x^2+1}}$.

One of the hints given is to apply mean value theorem for $\sin (x)$ on the interval $[0,x]$

This is my attempt so far:

Let $f(x) = \sin(x)$

Since all trigonometric functions are continuous and $\sin (x)$ is differentiable, mean value theorem can be applied.

$$\frac{\sin x - \sin 0}{x - 0} = \cos c$$

$$\frac{\sin x}{x} = \cos c$$

We know that $0 < c < x$

So, $0 < c < x \leq \pi / 2$

So, $0 \leq \cos c < 1$

Also, $\cos c > \cos x$

$$\cos c > \sqrt{1 - \sin^2 x}$$

$$\frac{\sin x}{x} > \sqrt{1 - \sin^2 x}$$

$$\sin x > x\sqrt{1 - \sin^2 x}$$

Now after this I'm stuck. I'm not sure how to bring $\sqrt{x^2 + 1}$ into the proof! I did think over it and was able to find some relations involving it like:

$$\sqrt{x^2 + 1} > 1$$

But I think I'm going the wrong path. How should I complete the proof ?

Answer 1

You are on the right path: square both sides of $\sin(x)>x (1-\sin^2(x))^{1/2}$ and get $\sin^2(x)>x^2(1-\sin^2(x))=x^2-x^2\sin^2(x)$, then add $x^2\sin^2(x)$ to both sides to obtain $\sin^2(x)(1+x^2)>x^2$. Now divide both sides by $(1+x^2)$ and take the square root.

Answer 2

you've almost done it , just square both the sides and you'll find the factor you've been looking for