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

**Discussion:**

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.

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