求证(cos^2 x-sin^2 x)(cos^4 x+sin^4 x)+1/4 sin 2x sin 4x=cos 2
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求证(cos^2 x-sin^2 x)(cos^4 x+sin^4 x)+1/4 sin 2x sin 4x=cos 2x
证明:∵cos²x-sin²x=cos2x
cos⁴x+sin⁴x=1-2cos²xsin²x=1-(1-cos4x)/4=3/4+(cos4x)/4
∴(cos²x-sin²x)(cos⁴x+sin⁴x)=cos2x(3/4+(cos4x)/4)=3cos(2x)/4+cos(2x)cos(4x)/4
∴(cos²x-sin²x)(cos⁴x+sin⁴x)+(1/4)sin 2x sin 4x
=3cos(2x)/4+(cos2xcos4x+sin2xsin4x)/4
=3cos(2x)/4+cos(4x-2x)/4
=3cos(2x)/4+cos(2x)/4
=cos2x
∴(cos²x-sin²x)(cos⁴x+sin⁴x)+(1/4)sin 2x sin 4x=cos2x
cos⁴x+sin⁴x=1-2cos²xsin²x=1-(1-cos4x)/4=3/4+(cos4x)/4
∴(cos²x-sin²x)(cos⁴x+sin⁴x)=cos2x(3/4+(cos4x)/4)=3cos(2x)/4+cos(2x)cos(4x)/4
∴(cos²x-sin²x)(cos⁴x+sin⁴x)+(1/4)sin 2x sin 4x
=3cos(2x)/4+(cos2xcos4x+sin2xsin4x)/4
=3cos(2x)/4+cos(4x-2x)/4
=3cos(2x)/4+cos(2x)/4
=cos2x
∴(cos²x-sin²x)(cos⁴x+sin⁴x)+(1/4)sin 2x sin 4x=cos2x
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