Question

已知tanA+sinA=a,tanA-sinA=b,求证(a2-b2)2=16ab.

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Answer 1

(a^2-b^2)^2=[(a+b)(a-b)]^2=16tan^2(A)sin^2(A)
16ab=16[tan^2(A)-sin^2(A)]
要证(a2-b2)2=16ab成立
需证tan^2(A)sin^2(A)=tan^2(A)-sin^2(A)
即证tan^2(A)sin^2(A)-tan^2(A)=-sin^2(A)
即证-tan^2(A)[1-sin^2(A)]=-sin^2(A)
即证-sin^2(A)=-sin^2(A)
显然-sin^2(A)=-sin^2(A)成立
∴(a2-b2)2=16ab成立.

其中-tan^2(A)[1-sin^2(A)]=[-sin^2(A)/cos^2(A)]*cos^2(A)=-sin^2(A)

Answer 2

tanA+sinA=a tanA-sinA=b 解得a+b=2tanA a-b=2sinA
证式左边：(a^2-b^2)^2=(a-b)^2*(a+b)^2=4sin^2A*4tan^2A
=16sin^2A*tan^2A
证式右边: 16ab=16(tanA+sinA)*(tanA-sinA)
=16*tan^2A*sin^2A
左边=右边 得证

因为(a2-b2)=4tanθsinθ
所以(a2-b2)2=16(tanθsinθ)^2
而16ab=16(tanθ)^2-(sinθ)^2
=16(sinθ)^2/(cosθ)^2-(sinθ)^2
=16(sinθ)^2[1/(cosθ)^2-1]
=16(sinθ)^2[1-(cosθ)^2]/(cosθ)^2
=16(sinθ)^2(sinθ)^2/(cosθ)^2
=16(tanθsinθ)^2
所以(a2-b2)2=16ab