cos^4a/(cos^2b)+sin^4a/(sin^2a)=1 求证cos^4b/(cos^2a)+sin^4a/(sin^2b)=1
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cos^4a/(cos^2b)+sin^4a/(sin^2a)=1求证cos^4b/(cos^2a)+sin^4a/(sin^2b)=1cos^4a/(cos^2b)+sin^4a/(sin^2a)=
cos^4a/(cos^2b)+sin^4a/(sin^2a)=1 求证cos^4b/(cos^2a)+sin^4a/(sin^2b)=1
cos^4a/(cos^2b)+sin^4a/(sin^2a)=1 求证cos^4b/(cos^2a)+sin^4a/(sin^2b)=1
cos^4a/(cos^2b)+sin^4a/(sin^2a)=1 求证cos^4b/(cos^2a)+sin^4a/(sin^2b)=1
证明：
输入过于麻烦,用换元法吧
设A=sin²A,B=sin²B
∵ sin^4a/sin^2b+cos^4a/cos^2b=1
即A²/B+(1-A)²/(1-B)=1
∴ A²(1-B)+(1-A)²B=B(1-B)
∴ A²-A²B+B-2AB+A²B=B-B²
∴ A²-2AB=-B²
∴ A²-2AB+B²=0
∴ (A-B)²=0
∴ A=B
∴ sin^4b/sin^2a+cos^4b/cos^2a
=A²/B+(1-B)²/(1-A)
=A²/A+(1-A)²/(1-A)
=A+1-A
=1
∴ 等式成立.
啊
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