已知cos^4β/cos^2a+sin^4β/sin^2a=1,求证cos^4a/cos^2β+sin^4a/sin2^
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已知cos^4β/cos^2a+sin^4β/sin^2a=1,求证cos^4a/cos^2β+sin^4a/sin2^β=1
跪谢,我一定会追加50分的!
跪谢,我一定会追加50分的!
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
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
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