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) 這為什么會(huì)從7跑到2的?
用2cos(b/2)除兩邊得:
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],將tan(b/2)的表達(dá)式代入,化簡得:
tan(a/2)tan[(a-b)/2]=-6.