证明(3-4cos 2A+cos 4A) / (3+4cos 2A+cos 4A)=tan^4 A
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证明(3-4cos 2A+cos 4A) / (3+4cos 2A+cos 4A)=tan^4 A
(3-4cos 2A+cos 4A) / (3+4cos 2A+cos 4A)
=(3+2cos 2A^2-1-4cos 2A) / (3+2cos 2A^2-1+4cos 2A)
=(2+2cos 2A^2-4cos 2A) / (2+2cos 2A^2+4cos 2A)
= (1+cos 2A^2-2cos 2A) / (1+cos 2A^2+2cos 2A)
=( (cos 2A-1) / (cos 2A+1))^2
= ((2cos A^2-1-1) / (2cos A^2-1+1) )^2
=((cos A^2-1) / (cos A^2))^2
=((sinA^2) / (cosA^2))^2
=tan^4 A
=(3+2cos 2A^2-1-4cos 2A) / (3+2cos 2A^2-1+4cos 2A)
=(2+2cos 2A^2-4cos 2A) / (2+2cos 2A^2+4cos 2A)
= (1+cos 2A^2-2cos 2A) / (1+cos 2A^2+2cos 2A)
=( (cos 2A-1) / (cos 2A+1))^2
= ((2cos A^2-1-1) / (2cos A^2-1+1) )^2
=((cos A^2-1) / (cos A^2))^2
=((sinA^2) / (cosA^2))^2
=tan^4 A
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