三角函数化简题目，谢了！~

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

解：∵根据诱导公式，有
cos(π/6 + 3x) = sin[π/2 - (π/6 + 3x)] = sin(π/3 - 3x)
sin(π/4 + 3x) = cos[π/2 - (π/4 + 3x)] = cos(π/4 - 3x)

∴原式 ＝ sin(π/4 - 3x)*cos(π/3 - 3x) － sin(π/4 + 3x)*cos(π/6 + 3x)
＝ sin(π/4 - 3x)*cos(π/3 - 3x) － cos(π/4 - 3x)*sin(π/3 - 3x)
＝ sin【(π/4 - 3x)－(π/3 - 3x)】
＝ sin（- π/12）
＝ - sin(π/12)

过程有点复杂，结果是 【根号下2 - 根号下6】/4
你先把减号两边化成CC-SS的形式{利用 sin(90-x)=cosx 等公式｝

公式记住了 往上面套就行了，学很多年了，都忘了