Question

已知tan²a=2tan²b+1,求证sin²b=2sin²a-1

Answer 1

证明：
tan²a=2tan²b+1

∴ sin²a/cos²a=2sin²b/cos²b+1
∴ sin²a/cos²a+1=2（sin²b/cos²b+1）
∴ (sin²a+cos²a)/cos²a=2(sin²b+cos²b)/cos²b
∴ 1/cos²a=2/cos²b
即 2cos²a=cos²b
即 2(1-sin²a)=1-sin²b
∴ sin²b=2sin²a-1

Answer 2

解：tan^2A=2tan^B+1
(sinA/cosA)^2=2(sinB/cosB)^2+1
sin^2A/cos^2A=2sin^2B/cos^2B+1
sin^2A*cos^2B=2sin^2Bcos^2A+cos^2Acos^2B
sin^2A(1-sin^2B)=2sin^2B(1-sin^2A)+(1-sin^2A)(1-sin^2B)
sin^2A-sin^2Asin^2B=2sin^2B-2sin^2Asin^2B+1-sin^2A-sin^2B+sin^2Asn^2B
sin^2B=2sin^2A-1

Answer 3

sin²a/(1-sin²a)=2sin²b/(1-sin²b)+1
sin²a-sin²asin²b=2sin²b-2sin²asin²b+1-sin²b-sin²a+sin²asin²b
sin²b=2sin²a-1