和弦外音_zz
幼苗
共回答了19个问题采纳率:78.9% 举报
解法一:
(1-√3tan75°)/(√3+tan75°)
=(1-tan60°tan75°) /(tan60°+tan75°)
=1/ tan(60°+75°)
=1/tan135°
= -1
解法二:
tan75°=tan(45°+30°)
=(tan45°+tan30°) /(1-tan45°tan30°)
=(1+√3/3) /(1-√3/3)
=(3+√3) /(3-√3)
=(√3+1)/(√3-1)
=(√3+1)*(√3+1) /(3-1)
=(3+2√3+1) /2
=2+√3
所以:
(1-√3tan75°)/(√3+tan75°)
= [ 1-√3*(2+√3) ] / (√3+2+√3)
=(1-2√3-3) /(2+2√3)
=(-2√3-2) / (2√3+2)
= -1
1年前
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零下半度
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tanx=1/3,tan(x-y)=-2,且π/2<y<x,求y
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和弦外音_zz
答:
tanx=1/3,tan(x-y)=-2
所以:
tan(x-y)=(tanx-tany)/(1-tanxtany)=-2
所以:
(1/3-tany)/(1-tany /3)=-2
1/3-tany=-2+(2/3)tany
tany=7/3