颐和园谐趣园平面分析:高一 三角函数题

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若 ab不等于0

[asin(16∏5)+bcos(16∏5)]/[acos(16∏5)-bin(16∏5)]=tan(8∏/5)
则 b/a =?

令tan(8∏/5)=x
sin(16∏/5)=2x/(1+x^2)
cos(16∏5)=(1-x^2)/(1+x^2)
[asin(16∏5)+bcos(16∏5)]/[acos(16∏5)-bin(16∏5)]
=[2ax+b(1-x^2)]/[a(1-x^2)-2bx]=x
2ax+b(1-x^2)=ax(1-x^2)-2bx^2
ax[x^2+1]=-b[1+x^2]
b/a=-x=-tan(8∏/5)