欧阳金先
幼苗
共回答了16个问题采纳率:93.8% 举报
由已知f(x) = 6cos2x – √3sin2x = 6*(1 + cos2x)/2 –√3sin2x = 3 + 3cos2x – √3sin2x = √[32 + (-√3)2]sin[2x + π + arctan(3/-√3)] + 3 = 2√3sin(2x + 2π/3) + 3,即f(x) = 2√3sin(2x + 2π/3) + 3 ;
因为x∈R,所以(2x + 2π/3)∈R,所以sin(2x + 2π/3)∈[-1,1],所以2√3sin(2x + 2π/3)∈[-2√3,2√3],所以[2√3sin(2x + 2π/3) + 3]∈[3 – 2√3,3 + 2√3],所以f(x)max = 3 + 2√3 .
而且2π/2= π,因此函数f(x)的最小正周期为π ;
(2)锐角a满足f(a) = 2 – √3,即2√3sin(2a + 2π/3) + 3 = 2 – √3,因此2√3sin(2a + 2π/3) = -1 – √3,所以sin(2a + 2π/3) = (-1 – √3)/(2√3) = -(3 + √3)/6,原题求tan(4a/5),由于不是特殊角,只能求出近似值,不能求精确值.a近似等于56.03°,tan(4a/5)近似等于0.99.(检查一下你打的题目有没有问题)
1年前
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