# let 0<= x <=90 degees.  prove that sin(x) >= (2x)/pi.

jitender lakhanpal
62 Points
12 years ago

Dear aritra,

between 0 to 90 value of 2x/∏ will be 0 to 1 so the and sin is increasing in the interval between 0 to ∏/2 from 0 to 1

sinx - 2x/∏ is increasing we can check that from differentiation its value also lie between 0 to 1

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jagdish singh singh
173 Points
12 years ago

$\hspace{-16}Here \mathbf{0 \leq x \leq \frac{\pi}{2}} and \mathbf{f(x)=\sin x}\\\\\\ Let \mathbf{x=x_{0}} , Where \mathbf{x=x_{0}} and \mathbf{0 \leq x_{0} \leq \frac{\pi}{2}}\\\\\\ Now Here Slope of Line \mathbf{OA} is \mathbf{m_{OA}=\frac{\sin x_{0}}{x_{0}}}\\\\\\ and Slope of Line \mathbf{OB} is \mathbf{m_{OB}=\frac{1}{\frac{\pi}{2}}=\frac{2}{\pi}}\\\\\\ Now Here \mathbf{m_{OA}\geq m_{OB}}\\\\\\ So \mathbf{\frac{\sin x_{0}}{x_{0}}\geq \frac{2}{\pi}}\\\\\\ So \boxed{\boxed{\mathbf{\sin x_{0}\geq \frac{2}{\pi}.x_{0}}}}$