[cos(ax+b)]'=-sin(ax+b)*a=acos(π/2+ax+b)
[cos(ax+b)]''=-acos(ax+b)*a=a^2cos(π/2*2+ax+b)
[cos(ax+b)]'''=a^2sin(ax+b)*a=a^3cos(π/2*3+ax+b)
.
所以：
[cos(ax+b)](n)=a^ncos(2nπ/2+ax+b).