Question

一道三角函数化简题

请问等式化简的具体中间步骤是什么？

Answer 1

化简求值：sin(π/4-3x)cos(π/3-3x)-cos(π/6+3x)sin(π/4+3x) 解：原式=sin(π/4-3x)cos(π/3-3x)-cos[π/2-(π/3-3x)]sin(π/4+3x) =sin(π/4-3x)cos(π/3-3x)-sin(π/3-3x)sin(π/4+3x) =(1/2)[sin(π/4-π/3)+sin(π/4+π/3-6x)]-(1/2)[cos(π/3-π/4-6x)-cos(π/3+π/4)] =(1/2){[sin(π/4-π/3)-cos(π/3+π/4)]+[sin(7π/12-6x)-cos(π/12-6x)]} ******(其中sin(7π/12-6x)=sin(π/2+π/12-6x)=cos(π/12-6x))******* =(1/2){[sin(π/4-π/3)-cos(π/3+π/4)]+[cos(π/12-6x)-cos(π/12-6x)]} =(1/2){[sin(π/4-π/3)-cos(π/3+π/4)] =(1/2)[-sin(π/12)-cos(7π/12)]=(1/2)[sin(π/12)-cos(π/2+π/12)] =(1/2)[-sin(π/12)+sin(π/12)]=0