分析:证明的思路是化简左边式子,方法是利用2倍角公式和同角三角函数的基本关系,得到式子与右边相等即可.
解答:证明:左边=2sin
4x+
(2sinxcosx)
2+5cos
4x-cos(2x+x)cosx
=2sin
4x+3sin
2xcos
2x+5cos
4x-(cos2xcosx-sin2xsinx)cosx
=2sin
4x+3sin
2xcos
2x+5cos
4x-[(2cos
2x-1)cosx-2sin
2xcosx]cosx
=2sin
4x+3sin
2xcos
2x+5cos
4x-[2cos
3x-cosx-2(1-cos
2x)cosx]cosx
=2sin
4x+3sin
2xcos
2x+5cos
4x-(4cos
3x-3cosx)cosx
=2sin
4x+3sin
2xcos
2x+cos
4x+3cos
2x
=(2sin
2x+cos
2x)(sin
2x+cos
2x)+3cos
2x
=2sin
2x+cos
2x+3cos
2x
=2+2cos
2x=2(1+cos
2x)=右边
点评:考查学生理解三角函数恒等式的证明思路,运用和差倍分的三角函数及同角三角函数的基本关系的能力.