已知函数f(x)=(根号3sinωx+cosωx)cosωx

显示全部楼层 · 2012-4-7 23:19:35

解：f(x)=√3sinωxcosωx+cos2ωx=√3/2sin2ωx+1/2（cos2ωx-1）=√3/2sin2ωx+1/2cos2ωx-1/2=sin2ωxcosπ/6+cos2ωxsinπ/6-1/2=sin（2ωx+π/6）-1/2（1）由正弦函数的单调性，当-π/2+2kπ≤2ωx+π/6≤π/2+2kπ时，f(x)单调递增所以，f(x)的单调递增区间为[-π/3+kπ，π/6+kπ][k属于Z]（2）由（2a-c）cosB=bcosC，两边同除b得：（2a/b-c/b）cosB=cosC由正弦定理得：（2sinA/sinB-sinC/sinB）cosB=cosC,化简得：2sinAcosB=c...