Let \mathbf{0\leq \theta\leq \frac{\pi}{2}} . Prove that \mathbf{\sin \theta \geq \frac{2\theta}{\pi}}.


We consider the function \mathbf{ f(x) = \frac{\sin x }{x} } . The first derivative of this function is \mathbf{ f'(x) = \frac{x \cos x - \sin x} {x^2} } In the interval \mathbf{[0, \frac{\pi}{2}]} the numerator is always negative as x is less than tan x.

Hence f(x) is a monotonically decreasing function in the given interval. Hence f(x) attains least value at \mathbf{x = \frac{\pi}{2} } which equals \mathbf{ \frac{\sin\frac{\pi}{2}}{\frac{\pi}{2}} = \frac {2}{\pi}}

Therefore \mathbf{\frac{\sin \theta}{\theta} \ge \frac{2}{\pi}} in the given interval.