Question

已知函数f（x）=1/2cos^2-√3sinxcosx-1/2sin^2x+1 x∈R

（1）求f（x）的最小正周期以及在区间【0，π/2】上的最大值和最小值
（2）若f（x1）=9/5，x1∈[-π/6,π/6]，求cos2x的值

Answer 1

(1)f(x)=(1/2)(cos^2x-sin^2x)-(√3/2)(2sinxcosx)+1
=1/2cos2x-(√3/2)*sin2x+1
=cos(2x+π/3)+1
最小正周期为π，区间上最大值3/2，最小值1/2。
(2)cos(2x1+π/3)+1=9/5 x1∈[-π/6,π/6]
则 (2x1+π/3)∈[0,2π/3]
所以： cos(2x1+π/3)=4/5 sin(2x1+π/3)=3/5
故： cos2x1=cos(2x1+π/3-π/3)=1/2*4/5+√3/2*3/5=(4+3√3)/10

Answer 2

f(x)=(1/2)(cosx)^2-3sinxcosx-(1/2)(sinx)^2 +1
==(1/2)[(cosx)^2-(sinx)^2]-(1/2)2sinxcosx -sin2x
=(1/2)(cos2x+sin2x)
=(√2/2)[cos2xcos(pi/4)+sin2xsin(pi/4)]
=(√2/2)cos(2x-pi/4)
1)周期T=2pi/2=pi
2)对称轴2x-pi/4=kpi--->x=kpi/2+pi/8
3)递减区间2kpi=<2x-pi/4=<2kpi+pi--->kpi+pi/8=<x=<kpi+5pi/8
递增区间2kpi+pi=<2x+pi/4=<2kpi--->kpi+5pi/8=<x=<kpi+9pi/8
(2)cos(2x1+π/3)+1=9/5 x1∈[-π/6,π/6]
so (2x1+π/3)∈[0,2π/3]
so cos(2x1+π/3)=4/5 sin(2x1+π/3)=3/5
so cos2x1=cos(2x1+π/3-π/3)=1/2*4/5+√3/2*3/5=(4+3√3)/10
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