Question

y=(sinx^4+cosx^4+sinx^2*cosx^2)/2-sin2x的最值和最小正周期

Answer 1

y = [(sinx)^4 + (cosx)^4 + (sinx)^2(cosx)^2]/2 - sin(2x)

= [(sinx)^4 + (cosx)^4 + 2(sinx)^2(cosx)^2 - (sinx)^2(cosx)^2]/2 - sin(2x)

= [1 - (sinx)^2(cosx)^2]/2 - sin(2x)

= 1/2 - [sin(2x)]^2/8 - sin(2x)

= {4 - 8sin(2x) - [sin(2x)]^2}/8

= {20 - 16 - 8sin(2x) - [sin(2x)]^2}/8

= {20 - [4 + sin(2x)]^2}/8

-5/8 = {20 - [4 + 1]^2}/8