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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

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