求解速度的高一数学函数问题！

显示全部楼层 · 2013-6-19 06:45:31

(1) x = π/6, f(x) = (1/2)sin(π/3)sinφ + [cos2(π/6)]cosφ - (1/2)cosφ= (√3/4)sinφ + (3/4)cosφ - (1/2)cosφ= (1/2)[(√3/2)sinφ + (1/2)cosφ]= (1/2)[sinφcos(π/6) + cosφsin(π/6)]= (1/2)sin(φ + π/6) = 1/2sin(φ + π/6) = 1φ + π/6 = π/2φ = π/3f(x) = (√3/4)sin2x + (1/2)cos2x - 1/4(2)横坐标为原来的一半，x变为x/2:g(x)...

千问 · 2013-6-19 06:45:31

cos^2x = 1/2 (1+cos2x)f(x) = 1/2 sin2x sinPhi + 1/2 cosPhi + 1/2 cos2xcosPhi - 1/2 cosPhi = 1/2 cos(2x-Phi)f(pi/6) = 1/2, cos(2pi/3 - phi) = 1, 2pi/3 - phi = 0, phi = 2pi/3 ...