求f(x)=2cos^2x+sin2x的最值小
问题描述:
求f(x)=2cos^2x+sin2x的最值小
答
f(x)=2cos^2x+sin2x
=1+cos2x+sin2x
=sin2x+cos2x+1
=√2*(√2/2*sin2x+√2/2*cos2x)+1
=√2*(sin2xcosπ/4+cos2xsinπ/4)+1
=√2*sin(2x+π/4)+1
-1