已知tanθ+sinθ=a,tanθ-sinθ=b,求证:(a2-b2)2=16ab.
问题描述:
已知tanθ+sinθ=a,tanθ-sinθ=b,求证:(a2-b2)2=16ab.
答
证明:∵(a2-b2)2=[(a+b)(a-b)]2
=[(tanθ+sinθ+tanθ-sinθ)(tanθ+sinθ-tanθ+sinθ)]2
=16tan2θsin2θ.
又16ab=16(tan2θ-sin2θ)=16•
=16•tan2θsin2θ.
sin2θsin2θ
cos2θ
故有(a2-b2)2=16ab.