已知函数f(x)=sin^2x+2√3sinx*cosx+sin(x+π/4)*sin(x-π/4)
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已知函数f(x)=sin^2x+2√3sinx*cosx+sin(x+π/4)*sin(x-π/4)
(1)求f(x)的最小正周期及值域.(2)若x=x0 (0≤x0≤π/2)为f(x)的一个零点,求f(2x0)的值.
(1)求f(x)的最小正周期及值域.(2)若x=x0 (0≤x0≤π/2)为f(x)的一个零点,求f(2x0)的值.
f(x)=-(cos²x-sin²x)+2√3cos[π/2-(x+π/4)]cos(x-π/4)-√3=-cos2x+2√3cos(-x+π=-cos2x+2√3cos²(x-π/4)-√3=-cos2x+2√3[1+cos2(x-π/4)]/2-√3=-cos2x+√3[1+cos(2x-π/2)]-√3=-cos2x+sin2x=√2(√2/2*sin2x-√2/2cos2x)=√2(sin2xcosπ/4-cos2xsinπ/4)=√2sin(2x-π/4)所以T=2π/2=π递减则2kπ+π/2<2x-π/4<2kπ+3π/22kπ+3π/4<2x<2kπ+7π/4kπ+3π/8<x<kπ+7π/8减区间(kπ+3π/84062kπ+7π/8)2、-π/12<=x<=25π/36-5π/12<=36所以2x-π/4=-5π/12最小ox=-π/122x-π/4=π/2最大406x=3π/8sin(-5π/12)=-sin(5π/12)=-sin(π/4+π/6)=-sinπ/4cosπ/6-cosπ/4sinπ/6=-(√6+√2)/4sinπ/2=1x=-π/12,y最小=-(√3+1)/2x=3π/8y最大=√2
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