已知非零常数 a,b满足(acosπ/5-bsinπ/5)/(asinπ/5+bcosπ/5)=1/(tan8π/15),求b/a.
人气:316 ℃ 时间:2020-06-13 10:33:33
解答
(acosπ/5-bsinπ/5)/(asinπ/5 bcosπ/5)=1/(tan8π/15),左边上下除以a.设b/a=k
cos8π/15sinπ/5+kcos8π/15cosπ/5=sin8π/15cosπ/5-ksinπ/5cos8π/15
kcos(8π/15-π/5)=sin(8π/15-π/5)
k=-tan(π/3)=-√3
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