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已知非零实数a,b满足关系式(asinπ/5+bcosπ/5)/(acosπ/5-bsinπ/5)=tan8π/15,则b/a的值是
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已知非零实数a,b满足关系式(asinπ/5+bcosπ/5)/(acosπ/5-bsinπ/5)=tan8π/15,则b/a的值是
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答案和解析
因a与b均为非零实数,故
(asinπ/5+bcosπ/5)/(acosπ/5-bsinπ/5) 两边均除以cosπ/5
=(atanπ/5+b)/(a-btanπ/5) 两边均除以a
=(b/a+tanπ/5)/(1-b/atanπ/5)
={tan[arctan(b/a)]+tanπ/5}/{1-tan[arctan(b/a)]*tanπ/5}
=tan[arctan(b/a)+tanπ/5]=tan(8π/15)
arctan(b/a)+π/5=8π/15+kπ
arctan(b/a)=π/3+kπ
tan[arctan(b/a)]=tan(π/3+kπ)
b/a=√3
(asinπ/5+bcosπ/5)/(acosπ/5-bsinπ/5) 两边均除以cosπ/5
=(atanπ/5+b)/(a-btanπ/5) 两边均除以a
=(b/a+tanπ/5)/(1-b/atanπ/5)
={tan[arctan(b/a)]+tanπ/5}/{1-tan[arctan(b/a)]*tanπ/5}
=tan[arctan(b/a)+tanπ/5]=tan(8π/15)
arctan(b/a)+π/5=8π/15+kπ
arctan(b/a)=π/3+kπ
tan[arctan(b/a)]=tan(π/3+kπ)
b/a=√3
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