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Sunday, April 17, 2016

For real numbers 0<x<π2, prove that cos2xcotx+sin2xtanx1. (Second Solution)

For real numbers 0<x<π2, prove that cos2xcotx+sin2xtanx1.

My solution:

First, note that we can rewrite cos2xcotx+sin2xtanx as cos3xsinx+sin3xcosx.

For the domain 0<x<π4, we have cos3x>sin3x,1sinx>1cosx, so by the rearrangement inequality we have:

cos2xcotx+sin2xtanx=cos3xsinx+sin3xcosxcos3xcosx+sin3xsinx=cos2x+sin2x=1

By the same token, for the domain π4x<π2, we have sin3x>cos3x,1cosx>1sinx, so by the rearrangement inequality we also have:

cos2xcotx+sin2xtanx=sin3xcosx+cos3xsinxsin3xsinx+cos3xcosx=sin2x+cos2x=1

Combining the two yields the result.

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