tan(8π/15)=tan(π/3+π/5)=(tanπ/3+tanπ/5)/(1-tanπ/3*tanπ/5)
(asinπ/5+bcosπ/5)/(acosπ/5-bsinπ/5)=(atanπ/5+b)/(a-btanπ/5)
=(b/a+tanπ/5)/(1-(b/a)tanπ/5)
所以b/a=tanπ/3=根号下3
tan(8π/15)=tan(π/3+π/5)=(tanπ/3+tanπ/5)/(1-tanπ/3*tanπ/5)
(asinπ/5+bcosπ/5)/(acosπ/5-bsinπ/5)=(atanπ/5+b)/(a-btanπ/5)
=(b/a+tanπ/5)/(1-(b/a)tanπ/5)
所以b/a=tanπ/3=根号下3