Studies and fun

TODAY'S PROBLEM

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.

Studies and fun

Sunday, 27 July 2014

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