问题
解答题
证明三角恒等式2sin4x+
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答案
证明：左边=2sin4x+
（2sinxcosx）2+5cos4x-cos（2x+x）cosx3 4
=2sin4x+3sin2xcos2x+5cos4x-（cos2xcosx-sin2xsinx）cosx
=2sin4x+3sin2xcos2x+5cos4x-[（2cos2x-1）cosx-2sin2xcosx]cosx
=2sin4x+3sin2xcos2x+5cos4x-[2cos3x-cosx-2（1-cos2x）cosx]cosx
=2sin4x+3sin2xcos2x+5cos4x-（4cos3x-3cosx）cosx
=2sin4x+3sin2xcos2x+cos4x+3cos2x
=（2sin2x+cos2x）（sin2x+cos2x）+3cos2x
=2sin2x+cos2x+3cos2x
=2+2cos2x=2（1+cos2x）=右边