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

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 ?

• $$\frac{\sin x} x = \cos c = \frac 1 {\sqrt{1+\tan^2 c}}$$ but we cannot go on to say$\displaystyle {} > \frac 1 {\sqrt{1+x^2}}$ unless we have $\tan c < x. \qquad$ Oct 18, 2020 at 17:02
• @MichaelHardy I don't think we have an assumption of $\tan c < x$. But I have attached the snapshot of the question from the book here: imgur.com/a/UHtGZGb The question number is 12(b) and I hope I haven't missed anything.
– Sibi
Oct 18, 2020 at 17:10
• \begin{align} & \frac{\sin x} x = \cos c \\ {} \\ > {} & \cos x \text{ since the cosine function decreases on this interval} \\ {} \\ = {} & \sqrt{1-\sin^2 x} \\ {} \\ > {} & \sqrt{1-x^2} \text{ as shown in part (a) of the same exercise} \\{} \\ \not> {} & \frac 1 {\sqrt{1+x^2}}\quad \text{So this is not there yet.} \end{align} Oct 18, 2020 at 17:30

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.
• Can you expand on how squaring has given you $\sin^2(x)(1+x^2)>x$ ?
• @Sibi, I think it's supposed to be an $x^2$ on the right hand side of the $\gt$. Oct 18, 2020 at 23:14
• @BarryCipra Thanks! If that's the case, then do you where did the $(1 - sin^2(x))$ go from the RHS ?