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)上述兩式相減,得 2*sin(kπ/6)sin(π/12),因...不是，有什么方法可以一眼看出左邊可以化為右邊嗎下面的算不算規(guī)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/6 b=π/12好像不算。。下面的呢？和差化積,積化和差公式是怎樣推導(dǎo)出來的?sin(a+b)=sinacosb+cosasinb sin(a-b)=sinacosb-cosasinb 兩式相加得：sinacosb=1/2[sin(a+b)+sin(a-b)]...(1) 兩式相減得：cosasinb=1/2[sin(a+b)-sin(a-b)]...(2) cos(a+b)=cosacosb-sinasinb cos(a-b)=cosacosb+sinasinb 兩式相加得： cosacosb=1/2[cos(a+b)+cos(a-b)]...(3) 兩式相減得：sinasinb=-1/2[cos(a+b)-cos(a-b)]...(4) 用(a+b)/2、(a-b)/2分別代替上面四式中的a,b 就可得到和差化積的四個(gè)式子。 如：(1)式可變?yōu)椋?sina+sinb=2sin[(a+b)/2]*cos[(a-b)/2] 其它依次類推即可。