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=cos2αsin2αcos2βsin2α+sin2βcos2α-sin2

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=cos2αsin2αcos2βsin2α+sin2βcos2α-sin2
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···?

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=cos2αsin2αcos2βsin2α+sin2βcos2α-sin2
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证明：
cos^4β/cos^2a+sin^4β/sin^2a=1
cos^4βsin²α+sin^4βcos²α=cos²αsin²α
(1-sin²β)cos²βsin²α+(1-cos²β)sin²βcos²α=cos²αsin²α
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证明：
cos^4β/cos^2a+sin^4β/sin^2a=1
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
所以cos²β=cos²α，sin²α=sin²β
所以
cos^4a/cos^2β+sin^4a/sin2^β
=(cos²β)²/cos²β+(sin²β)²/sin²β
=cos²β+sin²β
=1
设(sinα)^2＝n、(sinβ)^2＝m
则(cosα)^2＝1－n、(cosβ)^2＝1－m
由已知得：[(1－m)^2/(1－n)]＋(m^2/n)＝1
化简得：(m－n)^2＝0
所以 m＝n
所以(cos^4a/cos^2β)＋(sin^4a/sin2^β)
＝[(1－n)^2/(1－m)]＋(n^2/m)
＝[(1－m)^2/(1－n)]＋(m^2/n)
＝1

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