求f(x)=cos2x-sin2x的最大值和最小值，x属于[0，π/2]

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解：f(x)=cos2x-sin2x=√2sin(2x-π/4)
∵x属于[0，π/2]
∴枝春哗2x属于[0，π]
∴2x-π/4属于[-π/4，3π/4]
∴f(x)的最大值和森孝最小值分别为√2，猛行-1

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值域是[-1，√2]

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f(x)=-sin2x+cos2x
f(x)=-(sin2x-cos2x)
f(x)=-2^1/唤兄2sin(2x-pai/4)
0<=x<=pai/2
0<=2x<=pai
-pai/4<=2x-pai/4<=3pai/4
2x-pai/4=pai/2,sin(2x-pai/4)=1
fmin=-2^1/2
2x-pai/4=-pai/4
sin(2x-pai/4)=sin(-pai/4)=-2^1/皮链首2/2
fmax=-2^1/2*(-2^1/2/燃数2)=1

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