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求函数f(x)=1/2sin^2x+cos^2x+根号3/4sin2x的最大值
人气:154 ℃ 时间:2019-08-20 01:27:18
解答
f(x)=sin^2x/2+cos^2x+√3/4sin2x
=1/2+cos^2x/2+√3/4sin2x
=1/2+(cos2x+1)/4+√3/4sin2x
=1/2+1/4*cos2x+1/4+√3/4sin2x
=1/4*cos2x+√3/4sin2x+3/4
=1/2*(1/2*cos2x+√3/2sin2x)+3/4
=1/2*(sinπ/6*cos2x+cosπ/6sin2x)+3/4
=1/2*sin(π/6+2x)+3/4
-1<=sin(π/6+2x)<=1
-1/2<=1/2sin(π/6+2x)<=1/2
1/4<=1/2sin(π/6+2x)+3/4<=5/4
所以函数f(x)=sin^2x/2+cos^2x+√3/4sin2x的最大值为:5/4
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