求值遇y=sin^2x+cosx+2 (pi/3

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求值遇y=sin^2x+cosx+2 (pi/3
求值遇y=sin^2x+cosx+2 (pi/3

求值遇y=sin^2x+cosx+2 (pi/3
y=1-(cosx)^2+cosx+2
令a=cosx
则y=-a^2+a+3
π/3

y=sin^2x+cosx+2
=3-cos^2x+cosx
=-(cosx-1/2)^2+13/4
因为pi/3<=x<=pi,(cosx-1/2)^2>=0
故当cosx=-1时,y有最小值=-9/4+13/4=1
故当cosx-1/2=0时,y有最大值=13/4
所以y=sin^2x+cosx+2 的值域为[1,13/4]

y=sin^2x+cosx+2=-cos^2x+cosx+3
令t=cosx,t值域为[-1,1/2]
y=-t^2+t+3,定义域为t的值域[-1,1/2]
由此算得y的值域