已知(tan^2)a=2(tan^2)B+1,求证:(sin^2)B=2(sin^2)A-1
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已知(tan^2)a=2(tan^2)B+1,求证:(sin^2)B=2(sin^2)A-1
证明:
由(tan^2)A=2(tan^2)B+1得:
sin^2A/cos^2A=2sin^2B/cos^2B+1化简得:
cos^2Bsin^2A=2sin^2Bcos^2A+cos^2Acos^2B.(1)
又cos^2A=1-sin^2A cos^2B=1-sin^2B 将这两式代入(1)得:
(1-sin^2B)sin^2A=2sin^2B(1-sin^2A)+(1-sin^2A)(1-sin^2B)
即:sin^2A-sin^2Asin^2B=2sin^2B-2sin^2Asin^2B+1-sin^2B-sin^2A+sin^2Asin^2B
化简得:2sin^2A=sin^2B+1
即 sin^2B=2sin^2A-1
由(tan^2)A=2(tan^2)B+1得:
sin^2A/cos^2A=2sin^2B/cos^2B+1化简得:
cos^2Bsin^2A=2sin^2Bcos^2A+cos^2Acos^2B.(1)
又cos^2A=1-sin^2A cos^2B=1-sin^2B 将这两式代入(1)得:
(1-sin^2B)sin^2A=2sin^2B(1-sin^2A)+(1-sin^2A)(1-sin^2B)
即:sin^2A-sin^2Asin^2B=2sin^2B-2sin^2Asin^2B+1-sin^2B-sin^2A+sin^2Asin^2B
化简得:2sin^2A=sin^2B+1
即 sin^2B=2sin^2A-1
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