求答案,三角形ABC中,a+b=10c=8 求:tant(A/2)*tant(B/2)
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解决时间 2021-04-04 23:39
- 提问者网友：謫仙
- 2021-04-04 02:52
求答案,三角形ABC中,a+b=10c=8 求:tant(A/2)*tant(B/2)
最佳答案
- 五星知识达人网友：神也偏爱
- 2021-04-04 03:02
(a+b)/c=10,(sin(A)+sin(B))/sin(A+B)=10,即
2sin((A+B)/2)cos((A-B)/2)=20sin((A+B)/2)cos((A+B)/2),cos((A-B)/2)=10cos((A+B)/2),
展开：cos(A/2)cos(B/2)+sin(A/2)sin(B/2)=10*(cos(A/2)cos(B/2)-sin(A/2)sin(B/2))
9cos(A/2)cos(B/2)=11sin(A/2)sin(B/2)
得tan(A/2)*tan(B/2)=9/11
2sin((A+B)/2)cos((A-B)/2)=20sin((A+B)/2)cos((A+B)/2),cos((A-B)/2)=10cos((A+B)/2),
展开：cos(A/2)cos(B/2)+sin(A/2)sin(B/2)=10*(cos(A/2)cos(B/2)-sin(A/2)sin(B/2))
9cos(A/2)cos(B/2)=11sin(A/2)sin(B/2)
得tan(A/2)*tan(B/2)=9/11
全部回答
- 1楼网友：罪歌
- 2021-04-04 03:59
应该是9/11
- 2楼网友：人间朝暮
- 2021-04-04 03:52
(a+b)(a+b)/c=10,(sin(A)+sin(B))/sin(A+B)=10,
2sin((A+B)/2)cos((A-B)/2)=20sin((A+B)/2)cos((A+B)/2),cos((A-B)/2)=10cos((A+B)/2),
/c=10,(sin(A)+sin(B))/sin(A+B)=10
cos(A/2)cos(B/2)+sin(A/2)sin(B/2)=10*(cos(A/2)cos(B/2)-sin(A/2)sin(B/2))
9cos(A/2)cos(B/2)=11sin(A/2)sin(B/2)
得tan(A/2)*tan(B/2)=9/11
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