Question

已知5cos(a-b/2)+7caos(b/2)=0 求(tan a/2)*tan((a-b)/2)

Answer 1

5cos(a-b/2)+7cos(b/2)=5cosacos(b/2)+5sinasin(b/2)+7cos(b/2)
=10cos(a/2)cos[(a-b)/2]+2cos(b/2)=0
cos(a/2)cos[(a-b)/2]=(-1/5)*2cos(b/2) (1)
又5cos(a-b/2)+7cos(b/2)=5cosacos(b/2)+5sinasin(b/2)+7cos(b/2)
=-10sin(a/2)sin[(a-b)/2]+12cos(b/2)=0
sin(a/2)sin[(a-b)/2]=(6/5)*cos(b/2) (2)
(2)/(1) 得(tan a/2)*tan((a-b)/2) =(6/5)*cos(b/2)÷[(-1/5)*2cos(b/2)]
=-6