已知非零常数a、b满足asinπ5+bcosπ5acosπ5−bsinπ5=tan8π15，求ba．

Question

已知非零常数a、b满足(asin(π)/(5)+bcos(π)/(5))/(acos(π)/(5)−bsin(π)/(5))=tan(8 π)/(15)，求(b)/(a)．

Answer 1

由于(asin(π)/(5)+bcos(π)/(5))/(acos(π)/(5)−bsin(π)/(5))＝(sin(π)/(5)+(b)/(a)cos(π)/(5))/(cos(π)/(5)−(b)/(a)sin(π)/(5))，
则 (sin(π)/(5)+(b)/(a)cos(π)/(5))/(cos(π)/(5)−(b)/(a)sin(π)/(5))＝tan(8 π)/(15)．
整理，有(b)/(a)＝(sin(8 π)/(15)cos(π)/(5)−cos(8 π)/(15)sin(π)/(5))/(cos(8 π)/(15)cos(π)/(5)+sin(8 π)/(15)sin(π)/(5))＝(sin((8 π)/(15)−(π)/(5)))/(cos((8 π)/(15)−(π)/(5)))=tan(π)/(3)=√(3)．