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

来源：学生作业帮助网编辑：作业帮时间：2024/04/30 01:04:56

$已知:5cos(a-b/2)+7cos(b/2)=0 求:tan(a/2)*tan((a-b)/2)已知 5cos(a-b/2)+7cos(b/2)=0,求tan(a/2)*tan[(a-b)/2].由5cos(a-b/2)+7cos(b/2)=0得5[cosacos(b/2)+sinasin(b/2)]+7cos(b/2)=05{[2cos^2(a/2)-1]cos(b/2)+sinasin(b/2)}+7cos(b/2)=010[cos^2(a/2)cos(b$

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
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.

5{[2cos^2(a/2)-1]cos(b/2)+sinasin(b/2)}+7cos(b/2)=0
省略了一部
10cos^2(a/2）cos(b/2)-5cos(b/2)+sinasin(b/2)+7cos(b/2)=0

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 已知cos(a+b)=4/5,cos(a-b)=-4/5,3π/2 .已知Cos(a+b)=4/5,cos(a-b)= -4/5,3/2π Cos(a+b)*cos(a-b)=1/5 求cos ^2-sin^2 已知cos(a+b)cos(a-b)=1/3,则cos^2a-sin^2b的值是? 已知5cos(a-b/2)+7cos(b/2)=0,求tan(a/2)*tan((a-b)/2)的值.（要有具体过程!）已知tanA tanB为3x^2+5x-7=0的两根（sin(A+B)）/(COS(A-B))cos^2(A+B) 已知cos（a+b）=4/5,cos（a-b）=-4/5,2/3派已知cos（a+b）=2／3,cos(a-b)=1／5,则tanatanb=? 已知2cos^ 4a+5cos^ 2a-7=Asin^ 4a+Bsin^ 2a+C是恒等式,求A,B,C的值谢谢了, 已知2cos^4a+5cos^2a-7=Asin^4a+Bsin^2a+C是恒等式.求A,B,C 已知a=(2cosα,2sinα),b=(5cosβ,5sinβ),若a·b=-5,则cos= 已知cos(a+b)cos(a-b)=1/4,则cos^2a+cos^2b=是（cosa)^2 已知A+B=2/3π,则cos^2A+cos^2B的最大值已知cos(a+b)cos(a-b)=1/3 ,求(cosa)^2-(cosb)^2 已知cos(a-b)=-12/13 cos(a+b)=12/13且π/2 高一数学题已知cos(A+B)cos(A)+sin(A+B)sin(A)=3/5,求cos(高一数学题已知cos(A+B)cos(A)+sin(A+B)sin(A)=3/5,求cos(B-3π/2)的值已知4sin asin b=根号2,4cos a cos b=根号6,则cos 2a+cos 2b的值是?