2cos(A/2)-1-cosA/2cos(A/2)+1+cosA=1-cos(A/2)/1+cos(A/2) 请问如何
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2cos(A/2)-1-cosA/2cos(A/2)+1+cosA=1-cos(A/2)/1+cos(A/2) 请问如何证明,
等式左边={2cos(A/2)-1-cosA}/{2cos(A/2)+1+cosA}
={2cos(A/2)-1-(2(cos(A/2))^2-1}/{2cos(A/2)+1+2(cos(A/2))^2-1}
={2cos(A/2)-2(cos(A/2))^2}/{2cos(A/2)+2(cos(A/2)^2)
=2cos(A/2)(1-cos(aA/2)/(2cos(A/2)(1+cos(A/2))
={1-cos(A/2)}/{1+cos(A/2)}
=等式右边
={2cos(A/2)-1-(2(cos(A/2))^2-1}/{2cos(A/2)+1+2(cos(A/2))^2-1}
={2cos(A/2)-2(cos(A/2))^2}/{2cos(A/2)+2(cos(A/2)^2)
=2cos(A/2)(1-cos(aA/2)/(2cos(A/2)(1+cos(A/2))
={1-cos(A/2)}/{1+cos(A/2)}
=等式右边
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