cos(2k-1)π/12=cos(kπ/6-π/12)=cos(kπ/6)cos(π/12)+sin(kπ/6)sin(π/12)cos(2k+1)π/12=cos(kπ/6+π/12)=cos(kπ/6)cos(π/12)-sin(kπ/6)sin(π/12)上述两式相减,得2*sin(kπ/6)sin(π/12),因...

　　不是，有什么方法可以一眼看出左边可以化为右边吗

　　下面的算不算规律？cos(a-b)-cos(a+b)=[cos(a)cos(b)+sin(a)sin(b)]-[cos(a)cos(b)-sin(a)sin(b)]=2*sin(a)*sin(b)原式中a=kπ/6b=π/12

　　好像不算。。

　　下面的呢？和差化积,积化和差公式是怎样推导出来的?sin(a+b)=sinacosb+cosasinbsin(a-b)=sinacosb-cosasinb两式相加得：sinacosb=1/2[sin(a+b)+sin(a-b)]...(1)两式相减得：cosasinb=1/2[sin(a+b)-sin(a-b)]...(2)cos(a+b)=cosacosb-sinasinbcos(a-b)=cosacosb+sinasinb两式相加得：cosacosb=1/2[cos(a+b)+cos(a-b)]...(3)两式相减得：sinasinb=-1/2[cos(a+b)-cos(a-b)]...(4)用(a+b)/2、(a-b)/2分别代替上面四式中的a,b就可得到和差化积的四个式子。如：(1)式可变为：sina+sinb=2sin[(a+b)/2]*cos[(a-b)/2]其它依次类推即可。