已知cos^4a/cos^2β+sin^4a/sin2^β=1 求证cos^4β/cos^2a+sin^4β/sin^2a=1
要证cos^4β/cos^2a+sin^4β/sin^2a=1
只需证cos^4βsin2α+sin^4βcos2α=cos2αsin2α
(1-sin2β)cos2βsin2α+(1-cos2β)sin2βcos2α=cos2αsin2α
cos2βsin2α+sin2βcos2α-sin2βcos2βsin2α-cos2βsin2βcos2α=cos2αsin2α
cos2βsin2α+sin2βcos2α-sin2βcos2β(sin2α+cos2α)=cos2αsin2α
cos2βsin2α+sin2βcos2α-sin2βcos2β=cos2αsin2α
cos2βsin2α+sin2βcos2α-sin2βcos2β-cos2αsin2α=0
cos2β(sin2α-sin2β)-cos2α(sin2α-sin2β)=0
(cos2β-cos2α)(sin2α-sin2β)=0
即证cos2β=cos2α,sin2α=sin2β然后···?