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已知sinx+sinx2次方=1,求cos2次方x+cos6次方x的值
人气:117 ℃ 时间:2020-05-08 10:35:48
解答
sinx+sin^2x=1,sinx=(-1+√5)/2
sinx=cos^2x
cos^2x+cos^6x
=cos^2x(cos^4x+1)
=sinx(sin^2x+1)
=sinx(1-cos^2x+1)
=sinx(2-cos^2x)
=sinx(2-sinx)
=2sinx-sin^2x
=sinx+1
=1+(-1+√5)/2
=(1+√5)/2
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