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# sin^2x – cos^2x – 1 assumes the least value for set of given values of x given bya. x= npie +(-1)^n+1(pie/6) x=npie+(-1)^n(pie/6)x+npie+9-1)^n(pie/3)x=npie-(-1)^n(pie/6)

Arun Kumar IIT Delhi
7 years ago
I'm removing -1 because it doesn't matter
$y=sin^2x-cos^2x \\=>\frac{\mathrm{dy} }{\mathrm{d} x}=2sin(2x) \\=>\frac{\partial^2 y}{\partial x^2}=4cos(2x) \\=>x=\pi/4$
but here this is not the answer we have to chose from the options which gives the least value
$y=sin^2x-cos^2x \\=>y=2sin^2x-1$
A,B,D
Arun Kumar
IIT Delhi