已知向量a=(3sinx,根号3),b=(cosx,cos^2 x-1/2),函数f(x)=向量a向量b (1)求函数f
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已知向量a=(3sinx,根号3),b=(cosx,cos^2 x-1/2),函数f(x)=向量a向量b (1)求函数f(x)的周期
(2)写出函数f(x)的递减区间(3)求f(x)在[0,2分之π]上最值并求出相应的x的值.
优质解答:
f(x)=3sinxcosx+√3cos²x-√3/2
= 3/2 sin2x+√3/2cos2x
= √3(√3/2 sin2x+1/2cos2x)
=√3 sin(2x+π/6)
(1) 最小正周期 T=2π/2=π;
(2)递减区间满足2kπ+π/2≦2x+π/6≦2kπ+3π/2
即减区间为[kπ+π/6,kπ+2π/3]
(3) x∈[0,π/2], 2x+π/6∈[π/6,7π/6], sin(2x+π/6)∈[-1/2,1],
最小值为 -√3/2,这时 x=π/2;
最大值为 √3,这时 2x+π/6=π/2 即 x=π/3
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