fx等于sin(2x-3/π)+cos(2x-6/π)+2cos²+1,求最小正周期

显示全部楼层 · 2013-6-30 17:11:03

f(x)=sin(2x-π/3)+cos(2x-π/6)+2cos2x+1
/顺带说一句，π/3、π/6你写反了。=sin(2x)cos(π/3)-cos(2x)sin(π/3)+cos(2x)cos(π/6)+sin(2x)sin(π/6)+cos(2x)+2=(1/2)sin(2x)-(√3/2)cos(2x)+(√3/2)cos(2x)+(1/2)sin(2x)+cos(2x)+2=sin(2x)+cos(2x)+2=√2sin(2x+π/4)+2最小正周期=2π/2=π...

千问 · 2013-6-30 17:11:03

答：f(x)=sin(2x-π/3)+cos(2x-π/6)+2cos2x+1=sin2xcosπ/3-cos2xsinπ/3+cos2xcosπ/6+sin2xsinπ/6+1+cos2x+1=sin2x+cos2x+2=√2*[(√2/2)sin2x+(√2/2)cos2x]+2=√2sin(2x+π/4)+2所以：...

千问 · 2013-6-30 17:11:03

.f(x)=sin(2x-π/3)+cos(2x-π/6)+2cos2x+1,
=sin2x*1/2-cos2x*根号3/2+cos2x*根号3/2+sin2x*1/2+1+cos2x+1
=sin2x+cos2x+2
=根号2sin(2x+π/4)+2T=2π/2=π...

千问 · 2013-6-30 17:11:03

f(x)=(sin2xcos3/π-cos2xsin3/π)+(cos2xcos6/π+sin2xsin6/π)+cos2x+2
=2sin2xsin6/π+cos2x+2 周期=2π/2=π...

千问 · 2013-6-30 17:11:03

你好这个题是个化简题你可以展开 ...