大师作文网计算cos36-cos72如何解要过程帮帮忙
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计算cos36-cos72如何解要过程帮帮忙
1）先解出sin18,设sin18=x.
因为cos36=sin54,注意36=18*2,54=18*3,由2 倍角和3倍角公式,得
1-2x^2=3x-4x^3,即4x^3-2x^2-3x+1=（x-1)(4x^2+2x-1)=0,
由于sin18≠1,解4x^2+2x-1=0得x=（√5-1）/4=sin18
2)x=cos36-cos72=1-2(sin18)^2-sin18=0.5
另解
cos36-cos72
=cos(54-18)-cos(54+18)
=cos54cos18+sin54sin18-[cos54cos18-sin54sin18]
=2sin54sin18
=-2sin54sin(-18)
=2sin54sin18
=2cos36sin18
=2cos36sin18cos18/cos18
=cos36sin36/cos18
=sin72/(2cos18)
=sin72/(2sin72)=1/2
因为cos36=sin54,注意36=18*2,54=18*3,由2 倍角和3倍角公式,得
1-2x^2=3x-4x^3,即4x^3-2x^2-3x+1=（x-1)(4x^2+2x-1)=0,
由于sin18≠1,解4x^2+2x-1=0得x=（√5-1）/4=sin18
2)x=cos36-cos72=1-2(sin18)^2-sin18=0.5
另解
cos36-cos72
=cos(54-18)-cos(54+18)
=cos54cos18+sin54sin18-[cos54cos18-sin54sin18]
=2sin54sin18
=-2sin54sin(-18)
=2sin54sin18
=2cos36sin18
=2cos36sin18cos18/cos18
=cos36sin36/cos18
=sin72/(2cos18)
=sin72/(2sin72)=1/2