5cos（a－（b/2)） 7cos(b/2)=0 则 tg（a/2)*tg（(a－b）/2）＝？？？？
tg(a/2)*tg[(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tgx=sinx/cosx，然后将分子/分母积化和差）
...全部
tg(a/2)*tg[(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tgx=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)
∴tg(a/2)*tg[(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。
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