问题标题：
【已知:5cos(a-b/2)+7cos(b/2)=0求:tan(a/2)*tan((a-b)/2)已知5cos(a-b/2)+7cos(b/2)=0,求tan(a/2)*tan[(a-b)/2].由5cos(a-b/2)+7cos(b/2)=0得5[cosacos(b/2)+sinasin(b/2)]+7cos(b/2)=05{[2cos^2(a/2)-1]cos(b/2)+sinasin(b/2)}+7cos(b/2)=010[cos^2(a/2)cos(b】
问题描述：
已知:5cos(a-b/2)+7cos(b/2)=0
求:tan(a/2)*tan((a-b)/2)
已知5cos(a-b/2)+7cos(b/2)=0,求tan(a/2)*tan[(a-b)/2].
由5cos(a-b/2)+7cos(b/2)=0
得5[cosacos(b/2)+sinasin(b/2)]+7cos(b/2)=0
5{[2cos^2(a/2)-1]cos(b/2)+sinasin(b/2)}+7cos(b/2)=0
10[cos^2(a/2)cos(b/2)+10sin(a/2)cos(a/2)sin(b/2)+2cos(b/2)=0
7cos(b/2)——>2cos(b/2)这为什么会从7跑到2的?
用2cos(b/2)除两边得:
5cos^2(a/2)+5sin(a/2)cos(a/2)tan(b/2)+1=0
∴tan(b/2)=-[1+5cos^2(a/2)]/5sin(a/2)cos(a/2)
=-[sin^2(a/2)+6cos^2(a/2)]/5sin(a/2)cos(a/2)
=-[tan^2(a/2)+6]/5tan(a/2).
于是tan(a/2)tan[(a-b)/2]=tan(a/2)[tan(a/2)-tan(b/2)]/[1+tan(a/2)tan(b/2],将tan(b/2)的表达式代入,化简得:
tan(a/2)tan[(a-b)/2]=-6.
白丹回答：
[2cos^2(a/2)-1]cos(b/2)这里有个-1*cos(b/2)
括号外乘以5变成了-5cos(b/2)+7cos(b/2)=2cos(b/2)
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