快!快!快! 高2数学!

显示全部楼层 · 2009-2-11 21:54:00

曲线f(x)=cos(2x+φ)(-π<=φ<=0)与直线y=x-m在x=π/6处相切,求
⑴φ的值.
⑵曲线f(x)在x∈[0,π/2]时与x所围成的图形的面积

千问 · 2009-2-11 21:54:00

曲线f(x)=cos(2x+φ)(-π<=φ<=0)与直线y=x-m在x=π/6处相切，则f'(π/6)=1，sin(π/3+φ)=-1/2，φ=-π/2 。f(x)=cos(2x-π/2)，曲线f(x)在 x∈[0,π/2] 时（f(x)非负）与x所围成的图形的面积 [0,π/2]∫cos(2x-π/2)dx=[0,π/2]∫d{[sin(2x-π/2)]/2}=1 。...