【三角恒等变换】cos(33°-x)sin(63°-x)-cos(x+57°)sin(27°+x)化简并求值
问题描述：
【三角恒等变换】cos(33°-x)sin(63°-x)-cos(x+57°)sin(27°+x)
化简并求值
答
原式=sin[90-(33-x)]cos[90-(63-x)]-cos(x+57)sin(27+x)
=sin(x+57)cos(27+x)-cos(x+57)sin(27+x)
=sin[(x+57)-(27+x)]
=sin30
=1/2
答
cos(33°-x)sin(63°-x)-cos(x+57°)sin(27°+x)=cos[90°-(x+57°)]sin[90°-(x+27°]-cos(x+57°)sin(x+27°)=sin(x+57°)cos(x+27°)-cos(x+57°)sin(x+27°)=sin[(x+57°)-(x+27°)]=sin30°=1/2