已知asin(θ+α)=bsin(θ+β),求证

问题描述:

已知asin(θ+α)=bsin(θ+β),求证
tanθ=(bsinβ-asinα)/(acosα-bcosβ)
asin是a乘以sin,同理bsin acos bcos

asin(θ+α)=bsin(θ+β)a(sinθcosα+cosθsinα)=b(sinθcosβ+cosθsinβ)asinθcosα+acosθsinα=bsinθcosβ+bcosθsinβ移项asinθcosα-bsinθcosβ=bcosθsinβ-acosθsinαsinθ(acosα-bcosβ)=cosθ(bsi...