已知函数f(x)=[2√3cos(x/2)+2sin(x/2)]cos(x/2)

显示全部楼层 · 2011-10-27 09:31:22

解：f(x)=[2√3cos(x/2)+2sin(x/2)]cos(x/2)=2√3[cos(x/2)]^2+sinx=√3{2[cos(x/2)]^2-1}+√3+sinx=√3cosx+sinx+√3=2(sinx*1/2+√3/2*cosx)+√3=2sin(x+π/3)+√3(1)求f(17π/12)=2sin(7π/4)+√3=√3-√2(2)f(C)=√3+1得2sin(C+π/3)+√3=√3+1故C=π/2，sinC=1,A+B=π/2,则sinB=cosA且b^2=ac由正弦定理得sinB^2=sinAsinC=sinA代入的cosA^2=sinA1-sinA^2=sinAsinA^...

千问 · 2011-10-27 09:31:22

1。先把[2√3cos(x/2)+2sin(x/2)]cos(x/2)转化为4[sin(π/3)cos(x/2)+cos(π/3)sin(x/2)]cos(x/2)下面就利用两角和公式和积化和差公式就能算了。（提示）1/24+17/24=3/4 1/24-17/24=-2/32。将上式化为f(x)=√3(1+cosx)+sinx=2sin(x+π/3)+...