高中数学

显示全部楼层 · 2011-9-19 15:15:18

若sin(π/6-α）=1/3，则cos（2π/3+2α）等于：A。-7/9 B。-1/3 C。1/3D。7/9解：sin(π/6-α)=sin[π/2-(π/3+α)]=cos(π/3+α)=1/3cos(2π/3+2α)=cos[2(π/3+α)]=2cos2(π/3+α)-1=2(1/3)2-1=2/9-1=-7/9.故选A。...

千问 · 2011-9-19 15:15:18

因为sin(派/6-α）=1/3，而cos（2派/3+2α）=2cos^2(α+π/3)-1=2sin^2[π/2-(α+π/3)]-1=2sin^2(π/6-α)-1=2*(1/3)^2-1=2/9-1=-7/9所以选A...

千问 · 2011-9-19 15:15:18

答案选A呀！！sin（π/6-a）=cos[π/2-(π/6-a)]=cos(π/3+a)=1/3,则cos（2π/3+2a）=2[cos（π/3+a）]^2-1=-7/9平方不好弄就加了中括号！！其实这类的问题只是对基本公式的灵活运用,并不难!祝学业有成！！...

千问 · 2011-9-19 15:15:18

cos(2π/3+2α) =-cos[π-(2π/3+2α)] =-cos(π/3-2α) =-cos[2(π/6-α)] =-{1-2[sin(π/6-α)]^2} =-(1-2/9) =-7/9...