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已知tan(a+6π/5)=m(m≠1),求[sin(11π/5+a)+3cos(a-9π/5)]/[sin(14π/5-a)+cos(a+16π/5)]的值.
题目详情
已知tan(a+6π/5)=m(m≠1),求[sin(11π/5+a)+3cos(a-9π/5)]/[sin(14π/5-a)+cos(a+16π/5)]的值.
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答案和解析
tan(a+6π/5)=m
则tan(a+π/5)=m
[sin(π/5+a)+3cos(a+π/5)]/[sin(-π/5-a)-cos(a+π/5)]
=[sin(π/5+a)+3cos(a+π/5)]/[-sin(π/5+a)-cos(a+π/5)]
上下除以cos(a+π/5)
且sin(a+π/5)/cos(a+π/5)=tan(a+6π/5)
所以原式=[tan(a+6π/5)+3]/[-tan(a+6π/5)-1]
=-(m+3)/(m+1)
则tan(a+π/5)=m
[sin(π/5+a)+3cos(a+π/5)]/[sin(-π/5-a)-cos(a+π/5)]
=[sin(π/5+a)+3cos(a+π/5)]/[-sin(π/5+a)-cos(a+π/5)]
上下除以cos(a+π/5)
且sin(a+π/5)/cos(a+π/5)=tan(a+6π/5)
所以原式=[tan(a+6π/5)+3]/[-tan(a+6π/5)-1]
=-(m+3)/(m+1)
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