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Question: Given x > sinx for x > 0, prove πx - 2x^2 > sin2x for 0 < x < π/2

This is a problem about inequality that involves trig, I tried to divide the RHSs and LHSs by x > sinx to get π /2 - x > cosx, and stuck on it.

I am not sure the way I am trying to work out is correct or not. If correct, then how to eliminate the cosx? Is it related to the π/2?

Winnie Chan
  • 11

1 Answers1

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Work from the answer,

πx-2x^2 > sin 2x

π/2*x-x^2 > sinxcosx

π/2 - x > cosx

π/2 > cox+sinx

Since 1<cosx+sinx< squared root of 2, (as above) then work back to reach the statement.

Winnie Chan
  • 11