解:以下用#代替派爱,用E代替属于,
(1)f(x)=2sin(x+#/6)-2cosx=2sinx*V3/2+2cosx*1/2-2cosx=v3sinx-cosx
=2sin(x-#/6)
2k#-#/2<=x-#/6<=2k#+#/2 2k#-#/3<=x<=2k#+2#/3(kEZ)为单调递增区间
2k#+#/2<=x-#/6<=2k#+3#/2 2k#+2#/3<=x<=2k#+5#/6(kEZ)为单调递减区间
(2)2sin(x-#/6)=6/5 sin(x-#/6)=3/5
Cos(2x-#/3)=cos 2(x-#/6)=1-2*(3/5)^2=7/25
解:由题得,f(x)=2sin(x+π/6)-2cosx【2sin(x+π/6)=√3sinx+cosx,这个你应该能推算吧】
=√3sinx+cosx-2cosx
=√3sinx-cosx
=2sin(x-π/6)
因为f(x)=6/5,sin(x-π/6)=3/5
【下面用到这个公式估计你也学了吧,[sin(x/2)]^2=(1-cosx)/2】
[sin(x-π/6)]^2=1-cos(2x-π/3)=9/25,得
cos(2x-π/3)=16/25