求证:4(cos^6 x +sin^6 x) = 1 + 3cos^2 2x
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求证:4(cos^6 x +sin^6 x) = 1 + 3cos^2 2x
4(cos^6 x +sin^6 x) = 1 + 3cos^2 2x
谢谢~~~
4(cos^6 x +sin^6 x) = 1 + 3cos^2 2x
谢谢~~~
左=4(cos^6 x +sin^6 x)
=4(cos^2x+sin^2x)(cos^4x-cos^2xsin^2x+sin^4x)
=4[(cos^2x+sin^2x)^2-3cos^2xsin^2x]
=4[1-3cos^2xsin^2x]
右=1+3cos^22x=1+3[cos^2x-sin^2x]^2=1+3(cos^4x+sin^4x-2cos^2xsin^2x)
=1+3[(cos^2x+sin^2x)^2-4cos^2xsin^2x]
=1+3[1-4cos^2xsin^2x]
=4[1-3cos^2xsin^2x]
所以:左=右
即:4(cos^6 x +sin^6 x) = 1 + 3cos^2 2x
=4(cos^2x+sin^2x)(cos^4x-cos^2xsin^2x+sin^4x)
=4[(cos^2x+sin^2x)^2-3cos^2xsin^2x]
=4[1-3cos^2xsin^2x]
右=1+3cos^22x=1+3[cos^2x-sin^2x]^2=1+3(cos^4x+sin^4x-2cos^2xsin^2x)
=1+3[(cos^2x+sin^2x)^2-4cos^2xsin^2x]
=1+3[1-4cos^2xsin^2x]
=4[1-3cos^2xsin^2x]
所以:左=右
即:4(cos^6 x +sin^6 x) = 1 + 3cos^2 2x
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