f(x)=cosx^2-sinx^2+(2根号3)sinxcosx+1
= cos2x + 根号3sin2x + 1
= 2( 根号3/2 sin2x+ 1/2 cos2x ) + 1
= 2( sin2xcosπ/6 + cos2xsinπ/6 ) + 1
= 2 sin(2x+π/6) + 1
最小正周期 = 2π/2 = π
最小值 = 2*(-1)+1 = -1
f(a)=2
2 sin(2a+π/6) + 1 = 2
sin(2a+π/6) = 1/2
a属于【π/4,π/2】
2a∈【π/2,π】
2a+π/6∈【2π/3,7π/6】
2a+π/6 = 5π/6
a = π/3
已知函数f(x)=cosx^2-sinx^2+(2根号3)sinxcosx+1
已知函数f(x)=cosx^2-sinx^2+(2根号3)sinxcosx+1
1.求f(x)的最小正周期及最小值
2.若f(a)=2,且a属于【π/4,π/2】,求a的值
1.求f(x)的最小正周期及最小值
2.若f(a)=2,且a属于【π/4,π/2】,求a的值
数学人气:747 ℃时间:2019-12-09 08:21:40
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