已知sinα=msinβ,ncosα=mcosβ,且α、β为锐角,求证cosα=根号下(m方-1)除以(n方-1)
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![已知sinα=msinβ,ncosα=mcosβ,且α、β为锐角,求证cosα=根号下(m方-1)除以(n方-1)](/uploads/image/z/11471487-15-7.jpg?t=%E5%B7%B2%E7%9F%A5sin%CE%B1%EF%BC%9Dmsin%CE%B2%2Cncos%CE%B1%3Dmcos%CE%B2%2C%E4%B8%94%CE%B1%E3%80%81%CE%B2%E4%B8%BA%E9%94%90%E8%A7%92%2C%E6%B1%82%E8%AF%81cos%CE%B1%3D%E6%A0%B9%E5%8F%B7%E4%B8%8B%EF%BC%88m%E6%96%B9-1%EF%BC%89%E9%99%A4%E4%BB%A5%EF%BC%88n%E6%96%B9-1%EF%BC%89)
已知sinα=msinβ,ncosα=mcosβ,且α、β为锐角,求证cosα=根号下(m方-1)除以(n方-1)
已知sinα=msinβ,ncosα=mcosβ,且α、β为锐角,求证cosα=根号下(m方-1)除以(n方-1)
已知sinα=msinβ,ncosα=mcosβ,且α、β为锐角,求证cosα=根号下(m方-1)除以(n方-1)
α、β为锐角,sinα=msinβ,ncosα=mcosβ,
所以m>0,n>0
cosα
=(m/n)*cosβ
=(m/n)*根号"1-sinβ*sinβ"
=(m/n)*根号"1-1/m^2*sinα*sinα"
=根号"1-sinα*sinα"
所以sinα*sinα=(n^2-m^2)/(n^2-1)
cosα=根号下(m方-1)除以(n方-1)
sinα=msinβ,ncosα=mcosβ
sin^2α=m^2*sin^2β......(1)
n^2*cos^2α=m^2*cos^2β......(2)
(1)+(2):
sin^2α+n^2*cos^2α=m^2*sin^2β+m^2*cos^2β=m^2
1-cos^2α+n^2*cos^2α=m^2
(n^1-1)*cos^2α=m^2-1
cos^2α=(m^2-1)/(n^2-1)
α、β为锐角
cosα=√[(m^2-1)/(n^2-1)]