Question

Cosπx/2=2/π这个x怎么求？

Answer 1

Cosπx/2=2/π这个x求：

cos(π/2)cosπx/2+sinπ/2sinπx/2

=0+sinπx/2

=sinπx/2

h→0

f'(x)=lim[f(x+h)-f(x)]/h

=lim[cos(x+h)-cosx]/h

cos(x+h)=cosxcosh-sinxsinh

lim[cosx(cosh-1)+sinxsinh]/h

=limsinxsinh/h

=sinxlimsinh/h

=sinx*sin'h

=sinxcosh

=sinx

例

求函数y=|sinx|+|cosx|的最小正周期。

解

∵ =|sinx|+|cosx|

=|－sinx|+|cosx|

=|cos(x+π/2)|+|sin(x+π/2)|

=|sin(x+π/2)|+|cos(x+π/2)|

=f(x+π/2)