求y=cosx+sin^2x+1的最值

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求y=cosx+sin^2x+1的最值求y=cosx+sin^2x+1的最值求y=cosx+sin^2x+1的最值y=cosx+sin^2x+1=cosx+1-cos^2x+1=-cos^2x+cos

求y=cosx+sin^2x+1的最值
求y=cosx+sin^2x+1的最值

求y=cosx+sin^2x+1的最值
y=cosx+sin^2x+1
=cosx+1-cos^2x+1
=-cos^2x+cosx+2
=-(cos^2x-cosx+1/4)+9/4
=-(cosx-1/2)^2+9/4
因此
当cosx=1/2时有最大值9/4
当cosx=-1时有最小值0

y=cosx+sin²x+1
=cosx+1-cos²x+1
=-cos²x+cosx+2
=-(cosx-1/2)²+9/4
当cos=1/2时,y有最大值=9/4
当cosx=-1时,y有最小值=0