Question

求y=cos^2x+sin2x最小值

Answer 1

y=sin2x+(1+cos2x)/2
=sin²x+1/2*cos2x+1/2
=√(1+1/4)si(2x+z)+1/2
所以最小是是-√5/2+1/2

Answer 2

y=cos²x+sin2x
cos2x=2cos²x-1
cos²x=(cos2x+1)/2

y=cos²x+sin2x
=(cos2x+1)/2+sin2x
=(cos2x+2sin2x+1)/2
=(cos2x+2sin2x)/2+1/2
=(√5)[(1/√5)cos2x+(2/√5)sin2x]/2+1/2
=(√5)sin(2x+θ)/2+1/2 cosθ=2/√5
sin(2x+θ)≥-1

y=(√5)sin(2x+θ)/2+1/2≥(1-(√5))/2

Answer 3

y=1-sin^2x+sin2x
令t=sin2x
y=-t^2+t+1
配方即可。同时要注意sin2x 的取值范围。