cos(a) < 2pi+[pi/2 - sin(b)] ........... [ for all val of a and b ]
sin(cos a) < cos(sin b )
a < b
cos(a) < 2pi+[pi/2 - sin(b)] ........... [ for all val of a and b ]
sin(cos a) < cos(sin b )
a < b
Why is it that cos(a) < 2pi+[pi/2 - sin(b)] ........... [ for all val of a and b ]?
Where does pi/2 - sinb come from?
RHS is greater than 2pi (as pi/2-sinx >0 for all x )
just apply sin on both sides of first statement you will see imp of pi/2-sinb,,,
very neat trick indeed..
When i did the question i used graphs..