Question

1.已知5cos(A-B/2)+7cosB/2=0,求tanA/2*tan(A-B)/2

Answer 1

1.

5cos(A-B/2)+7cosB/2=0

5cos[A/2+(A-B)/2]+7cos[A/2-(A-B)/2]=0

5*{cos(A/2)*cos[(A-B)/2]-sin(A/2)*sin[(A-B)/2]}+7*{cos(A/2)*cos[(A-B)/2]+sin(A/2)*sin[(A-B)/2]}=0

同除以cos(A/2)*cos[(A-B)/2]

5*{1-tan(A/2)*tan[(A-B)/2]}+7*{1+tan(A/2)*tan[(A-B)/2]}=0

tan(A/2)*tan[(A-B)/2]=6

2.

设3a-2b为向量c

c-b=3a-3b=3(a-b)

c-a=3a-2b-a=2(a-b)

c-b、c-a与a-b的方向相同

终点在一条直线上