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Sunday 27 July 2014

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.
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