Question

已知tanα+sinα=a,tanα-sinα=b,求证(a^2-b^2)^2=16ab

这道题应该用分析法或者是综合法证明这道题。

Answer 1

左=(a^2-b^2)^2
=(a+b)^2·(a-b)^2
=[(tanα+sinα)+(tanα-sinα)]^2·[(tanα+sinα)-(tanα-sinα)]^2
=[2tanα]^2·[2sinα]^2
=16[tanα·sinα]^2
=16(tanα)^2·(sinα)^2;
右=16ab
=16(tanα+sinα)·(tanα-sinα)
=16[(tanα)^2-(sinα)^2]
=16[(sinα)^2/(cosα)^2-(sinα)^2]
=16(sinα)^2[(secα)^2-1]
=16(sinα)^2·(tanα)^2
∴左=右

Answer 2

a^2-b^2=(tana+sina)-(tana-sina)^2=4tana x sina.
(a^2-b^2)^2=16tan^2a x sin^2a=16 x(sin^4a/cos^2a)
又16ab=16(tan^2a-sin^2a)=16 x[ (sin^2a-cos^2a x sin^2a)/cos^2a]=16 x (sin^4a/cos^2a).所以(a^2-b^2)^2=16ab