Question

高一数学三角函数题，要快，高分~~！！！

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！！！）

Answer 1

tan(a/2)*tan[(a－b）/2]
＝{sin(a/2)/cos(a/2)}×{sin[(a－b）/2]/cos[(a－b）/2]}
＝{sin(a/2)×sin[(a－b）/2]}/{cos(a/2)×cos[(a－b）/2]}
＝(-1/2){cos[(a/2)+(a－b）/2]-cos[(a/2)-(a－b）/2]} / (1/2){cos[(a/2)+(a－b）/2]+cos[(a/2)-(a－b）/2]}
＝-{cos[a-(b/2)]-cos(b/2)} / {cos[a-(b/2)]+cos(b/2)}

(提示：根据tanx=sinx/cosx，然后将分子/分母积化和差）

∵5cos[a－(b/2)]+7cos(b/2)＝0
∴cos[a－(b/2)]＝(-7/5)cos(b/2)
∴tan(a/2)*tan[(a－b）/2]
＝-{cos[a-(b/2)]-cos(b/2)} / {cos[a-(b/2)]+cos(b/2)}
＝-{(-7/5)cos(b/2)-cos(b/2)} / {(-7/5)cos(b/2)+cos

(b/2)}
＝-[(-12/5)cos(b/2)]/[(-2/5)cos(b/2)]
＝(12/5)/(-2/5)
＝-6

Answer 2

5cos(A-B/2)+7cos(B/2)=0,求tan(A/2)tan(A/2-B/2)

解:条件化为
5cos[(A/2-B/2)+ (A/2)]
=- 7cos(-B/2)
=- 7cos[(A/2-B/2)-(A/2)]

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

12 cos(A/2-B/2)cos(A/2)
=-2 sin(A/2-B/2)sin(A/2)

sin(A/2-B/2)sin(A/2)/cos(A/2-B/2)cos(A/2)=-12/2

故tan(A/2)tan(A/2-B/2)= -6