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已知sinα=asinβ,tanα=btanβ,α为锐角,求证:(cosα)^2=(a^2-1)/(b^2-1).

分类: 作业答案 • 2021-12-18 17:05:49
问题描述：

已知sinα=asinβ,tanα=btanβ,α为锐角,求证:(cosα)^2=(a^2-1)/(b^2-1).

答

(sinβ)^2=(sinα)^2/a^2,(cosβ)^2=1-(sinβ)^2=[a^2-(sinα)^2]/a^2.
(tanβ)^2=(sinα)^2/[a^2-(sinα)^2].
(tanα)^2=b^2*(tanβ)^2=b^2(sinα)^2/[a^2-(sinα)^2].
1/(cosα)^2=b^2/[a^2-(sinα)^2].
b^2(cosα)^2=a^2-1+(cosα)^2.
(cosα)^2=(a^2-1)/(b^2-1).

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