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An inequality related to (sin x)/x function (I.S.I. B.Math 2007 problem 7 solution)

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.

May 6, 2014