Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

$$\forall{x\in(0,\frac{\pi}{2})}\ \sin(x) > \frac{2}{\pi}x $$ I suppose that solving $ \sin x = \frac{2}{\pi}x $ is the top difficulty of this exercise, but I don't know how to think out such cases in which there is an argument on the right side of a trigonometric equation.

share|cite|improve this question
Try to draw graphs of lhs and rhs – Norbert Oct 14 '12 at 2:16
$\sin x$ is concave on this interval, so if you draw a line joining two points of the graph, the graph of this functions will lie above this line. What line do you get if you try the points $(0,\sin 0)$ and $(\pi/2, \sin \pi/2)$? – Martin Sleziak Oct 14 '12 at 7:51
up vote 3 down vote accepted

As one of the comments suggested, the easiest way is to draw a graph of sine and the line through $(0,0)$ and $(\frac{\pi}{2},1)$, and notice that one is above the other.

There's another way though; expanding on the hints above, consider the functions $f$ and $g$ defined by $$f(x) = \frac{\sin{x}}{x} \quad \text{and} \quad g(x) = x\cos{x} -\sin{x} $$ Then we have $$f'(x) = \frac{x\cos{x}-\sin{x}}{x^2} \quad \text{and} \quad g'(x) = -x\sin{x}$$ For $x \in [0,\frac{\pi}{2})$, we have $g'(x) \le 0$, so $g$ is decreasing. But we also have $g(0) = 0 $, so it follows that $g(x) \le 0$ on this interval. As a result, $f'(x) \le 0$ too, so $f$ is decreasing. As $x$ goes from (close to) $0$ to $\pi/2$, $f$ decreases from $1$ to $2/\pi$, and your result follows.

share|cite|improve this answer
How did you get the g(x) function? – 0x6B6F77616C74 Oct 15 '12 at 18:43
Perhaps it would have been better for me to have defined $g$ after finding $f'$. The motivation for choosing $g$ to be what I did is that then $f'(x) = \frac{g(x)}{x^2}$. Basically we want to conclude that $f$ is decreasing, but we don't see that immediately from its derivative, so we introduce $g$ to figure things out. – sourisse Oct 15 '12 at 19:18

Hint: Consider the monotone property of $f(x)=\frac{\sin(x)}{x}$ on interval $[0, \pi/2]$.

share|cite|improve this answer

Consider the properties of $\max$ and $\min$ of some function $f$,

$f = \dfrac{\sin x}{x}$ in this case.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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