已知非零函数a.b满足：(asin(π/5)+bcos(π/5))/(acos(π/5)-bsin(π/5))=tan8π/5,求b/a的值

如题所述

推荐答案 2011-05-21

(asin(π/5)+bcos(π/5))/(acos(π/5)-bsin(π/5))=tan8π/5=tan3π/5=sin3π/5/cos3π/5
asinπ/5cos3π/5+bcosπ/5cos3π/5=acosπ/5sin3π/5-bsinπ/5sin3π/5
bcosπ/5cos3π/5+bsinπ/5sin3π/5=acosπ/5sin3π/5-asinπ/5cos3π/5
b(cosπ/5cos3π/5+sinπ/5sin3π/5)=a(cosπ/5sin3π/5-sinπ/5cos3π/5)
bcos2π/5=asin2π/5
b/a=tan2π/5

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第1个回答 2011-05-21

设tanβ=b/a
(asin(π/5)+bcos(π/5))/(acos(π/5)-bsin(π/5))=sin(π/5+β)/cos(π/5+β)=tan(π/5+β)=tan8π/5
所以β=7π/5，b/a=tan（7π/5）

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