如题所述
tana=x tanb=y=-1/7
tan(a-b)=(x-y)/(1+xy)=1/2
代入y=-1/7
∴x=1/3
tan2a=2x/(1-x²)=3/4
tan(2a-b)=(tan2a-tanb)/(1+tan2atanb)=(3/4+1/7)/(1-3/4×1/7)=1
tan(2a-b)=1
tan(a-b)=(x-y)/(1+xy)=1/2
代入y=-1/7
∴x=1/3
tan2a=2x/(1-x²)=3/4
tan(2a-b)=(tan2a-tanb)/(1+tan2atanb)=(3/4+1/7)/(1-3/4×1/7)=1
tan(2a-b)=1
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第1个回答 2013-01-24
tan2(a-b)=2tan(a-b)/[1-tan²(a-b)]=2*1/2/(1-1/4)=4/3
tan(2a-b)=tan[2(a-b)+b]
=[tan2(a-b)+tanb]/[1-tan2(a-b)tanb]
=(4/3-1/7)[1-4/3*(-1/7)]
=27/25
tan(2a-b)=tan[2(a-b)+b]
=[tan2(a-b)+tanb]/[1-tan2(a-b)tanb]
=(4/3-1/7)[1-4/3*(-1/7)]
=27/25
