设f(x)=(sinax+cosax)²+2cos²ax，（a＞0）的最小正周期为π

显示全部楼层 · 2012-6-3 21:34:10

解：（1）f(x)=(sinax)^2+(cosax)^2+2sinaxcosax+2(cosax)^2=1+sin2ax+cos2ax-1=sin2ax+cos2ax=sin(2ax+π/4)所以T=2π/2a=π，∴a=1（2）由图像向右平移得到F（x）=sin(2(x-π/2)+π/4)=sin(2x-3π/4)g(x)=sin(4x-3π/4)-π/2+2kπ<4x-3π/4<π/2+2kπ
解得：π/16+kπ/2<x<5π/16+kπ/2 k属于自然数（3） f(x)=sin(2x+π/4) -π/2+2kπ<2x+π/4<π/2+2kπ
解得：-3π/8+kπ<x<π/8+kπ
k属于自...

千问 · 2012-6-3 21:34:10

解：（1）f(x)=(sinax)^2+(cosax)^2+2sinaxcosax+2(cosax)^2=1+sin2ax+cos2ax-1=sin2ax+cos2ax=sin(2ax+π/4)所以T=2π/2a=π，∴a=1（2）由图像向右平移得到F（x）=sin(2(x-π/2)+π/4)=sin(2x-3π/4)g(x)=sin(4x-...