α为锐角,a,b为常数,sinα=asinβ,且tanα=btanβ,求cosα为什么不是cosa=√(1-a^2)/(1-b^2).

sinα=asinβ,且tanα=btanβ
相除cosβ=bcosα/a,sinβ=sinα/a
（cosβ）∧2+（sinβ）∧2=（bcosα/a）∧2+（sinα/a））∧2=1
b∧2（cosα）∧2+（sinα）∧2=a∧2
（cosα）∧2=(a∧2-1)/(b∧2-1)
cosα=√〔(a∧2-1)/(b∧2-1)〕

(sinβ)^2=(sinα)^2/a^2
所以(cosβ)^2=1-(sinβ)^2=[a^2-(sinα)^2]/a^2
两式相除得
(tanβ)^2=(sinα)^2/[a^2-(sinα)^2]
(tanα)^2=b^2*(tanβ)^2=b^2(sinα)^2/[a^2-(sinα)^2]
两边除以(sina)^2
1/(cosα)^2=b^2/[a^2-(sinα)^2]
对角相乘得
b^2(cosα)^2=a^2-1+(cosα)^2
移项并化简
(cosα)^2=(a^2-1)/(b^2-1)
两边再开方即可

tanα=btanβ(α为锐角) =>cosa=sina/(btanβ).sina=asinβ.cosa=acosβ/b.(sinβ)^2=(sinα)^2/a^2,(cosβ)^2=1-(sinβ)^2=[a^2-(sinα)^2]/a^2. sin^2(a)+cos^2(a)=1.=>1/(cosα)^2=b^2/[a^2-(sinα)^2]. b^2(cos...