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An inequality related to (sin x)/x function | ISI BMath 2007

This is a problem number 7 from ISI B.Math 2007 based on an inequality related to (sin x)/x function. Try out this problem.

Problem: An inequality related to (sin x)/x function

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.

Some Useful Links:

Our ISI CMI Program

How to use invariance in Combinatorics – ISI Entrance Problem – Video

This is a problem number 7 from ISI B.Math 2007 based on an inequality related to (sin x)/x function. Try out this problem.

Problem: An inequality related to (sin x)/x function

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.

Some Useful Links:

Our ISI CMI Program

How to use invariance in Combinatorics – ISI Entrance Problem – Video

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