6
$\begingroup$

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 ?

$\endgroup$
3
  • $\begingroup$ $$ \frac{\sin x} x = \cos c = \frac 1 {\sqrt{1+\tan^2 c}} $$ but we cannot go on to say$\displaystyle {} > \frac 1 {\sqrt{1+x^2}}$ unless we have $\tan c < x. \qquad$ $\endgroup$ Oct 18, 2020 at 17:02
  • $\begingroup$ @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. $\endgroup$
    – Sibi
    Oct 18, 2020 at 17:10
  • 1
    $\begingroup$ $$\begin{align} & \frac{\sin x} x = \cos c \\ {} \\ > {} & \cos x \text{ since the cosine function decreases on this interval} \\ {} \\ = {} & \sqrt{1-\sin^2 x} \\ {} \\ > {} & \sqrt{1-x^2} \text{ as shown in part (a) of the same exercise} \\{} \\ \not> {} & \frac 1 {\sqrt{1+x^2}}\quad \text{So this is not there yet.} \end{align}$$ $\endgroup$ Oct 18, 2020 at 17:30

2 Answers 2

1
$\begingroup$

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.

$\endgroup$
3
  • $\begingroup$ Can you expand on how squaring has given you $\sin^2(x)(1+x^2)>x$ ? $\endgroup$
    – Sibi
    Oct 18, 2020 at 17:38
  • $\begingroup$ @Sibi, I think it's supposed to be an $x^2$ on the right hand side of the $\gt$. $\endgroup$ Oct 18, 2020 at 23:14
  • $\begingroup$ @BarryCipra Thanks! If that's the case, then do you where did the $(1 - sin^2(x))$ go from the RHS ? $\endgroup$
    – Sibi
    Oct 19, 2020 at 4:17
0
$\begingroup$

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

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.