已知a=(3倍的根3cosx,根2cosx),b=(sinx,根2cosx)
问题描述:
已知a=(3倍的根3cosx,根2cosx),b=(sinx,根2cosx)
设函数f(x)=a*b+(b绝对值的平方)减去二分之五.(一)当x属于[兀/6,兀/2]时,求函数f(x)的值域.(二)若阿尔法属于[兀/6,兀/2]且f(阿尔法)=3,f(阿尔法减去兀/12)的值
答
f(x)=3√3sinxcosx+2cos²x+(sin²x+2cos²x)-(5/2)
=(3√3/2)sin2x+(3/2)(cos2x+1)-(3/2)
=3[(√3/2)sin2x+(1/2)cos2x]
=3sin(2x+π/6)
1、x∈[π/6,π/2]
则:2x+π/6∈[π/2,7π/6]
则:f(x)∈[-3/2,3]
2、f(a)=3sin(2a+π/6)=3,则:a=π/6,则:f(a-π/12)=f(π/12)=3sin(π/3)=3√3/2第二问错了2、f(a)=3sin(2a+π/6)=3,则:a=π/6,则:f(a-π/12)=f(π/6-π/12)=f(π/12)=3sin[2×(π/12)+π/6]=3sin(π/3)=3√3/2