Prove the inequality 2x/π < sin x < x for 0 < x < π/2

Prove the inequality

$\frac{2}{\pi}{x} < \sin x < x \qquad \text{for} \quad 0 < x < \frac{\pi}{2}.$

Proof. Define a function $f(x) = x - \sin x$ . Then, since $\cos x < 1$ for $0 < x < \frac{\pi}{2}$ we have

$f'(x) = 1 - \cos x > 0 \qquad \text{for } 0 < x < \frac{\pi}{2}.$

Since $f(0) = 0$ and $f$ is increasing on $0 < x < \frac{\pi}{2}$ (since its derivative is positive) we have

$f(x) > 0 \quad \implies \quad x - \sin x > 0 \quad \implies \quad x > \sin x.$

For the inequality on the left (which is much more subtle), we want to show

$\frac{2}{\pi}x < \sin x \qquad \text{which implies} \qquad frac{2}{\pi} < \frac{\sin x}{x}.$

So, we consider the function

$f(x) = \frac{\sin x}{x} \quad \implies \quad f'(x) = \frac{x \cos x - \sin x}{x^2}.$

We know

$\lim_{x \to 0} f(x) = \lim_{x \to 0} \frac{\sin x}{x} = 1$

and

$f \left( \frac{\pi}{2} \right) = \frac{2}{\pi}.$

Now, if we can show that $f(x)$ is decreasing on the whole interval $\left( 0, \frac{\pi}{2} \right)$ then we will be done (since this would mean $f(x) > \frac{2}{\pi}$ on the whole interval since if it were less than $\frac{2}{\pi}$ somewhere then it would have to increase to get back to $\frac{2}{\pi}$ on the right end of the interval).

To show $f(x)$ is decreasing on the whole interval we will show that its derivative is negative. To that end, define a function

$g(x) = x \cos x - \sin x$

(This is the numerator in the expression we got for $f'(x)$ . Since we know the denominator of that expression is always positive, we are going to show this $g(x)$ is always negative to conclude $f'(x)$ is always negative.) Now, $g(0) = 0$ and

$g'(x) = \cos x - x \sin x - \cos x = -x \sin x < 0$

for $0 < x < \frac{\pi}{2}$ . Thus, $g'(x)$ is negative, and so $g(x)$ is decreasing. Since $g(0) = 0$ we then have $g(x)$ is negative on the whole interval. Therefore, $f'(x) \leq 0$ on the interval. Hence, $f(x)$ is decreasing. Hence, we indeed have

$\frac{\sin x}{x} > \frac{2}{\pi} \quad \implies \quad \sin x > \frac{2}{\pi}x$

for all $x \in \left( 0, \frac{\pi}{2} \right). \qquad \blacksquare$

Stumbling Robot

Prove the inequality 2x/π < sin x < x for 0 < x < π/2

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