Trigonometry> trigo inequality...

let 0<= x <=90 degees.

prove that sin(x) >= (2x)/pi.

ARITRA SINHA , 13 Years ago

Grade 11

2 Answers

jitender lakhanpal

Last Activity: 13 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

Last Activity: 13 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}}}}$$