谁会？？
高一数学解答题:用角A,B,C的正切表示tan(A+B+C)
答案:2 悬赏:30 手机版
解决时间 2021-05-05 17:17
- 提问者网友：感性作祟
- 2021-05-05 04:54
最佳答案
- 五星知识达人网友：山河有幸埋战骨
- 2021-05-05 06:32
tg(A+B+C)
=tg[A+(B+C)]
=[tgA+tg(B+C)]/[1-tgA*tg(B+C)]
=tgA+[(tgB+tgC)/(1-tgBtgC)]/[1-tgA*(tgB+tgC)/(1-tgBtgC)]
=[tgA(1-tgBtgC)+tgB+tgC]/[1-tgBtgC-tgA*(tgB+tgC)]
=(tgA+tgB+tgC-tgAtgBtgC)/(1-tgAtgB-tgBtgC-tgAtgC)
=tg[A+(B+C)]
=[tgA+tg(B+C)]/[1-tgA*tg(B+C)]
=tgA+[(tgB+tgC)/(1-tgBtgC)]/[1-tgA*(tgB+tgC)/(1-tgBtgC)]
=[tgA(1-tgBtgC)+tgB+tgC]/[1-tgBtgC-tgA*(tgB+tgC)]
=(tgA+tgB+tgC-tgAtgBtgC)/(1-tgAtgB-tgBtgC-tgAtgC)
全部回答
- 1楼网友：想偏头吻你
- 2021-05-05 07:29
tan(A+B+C)
=tan((A+B)+C)
=(tanA+tanB+tanC)/(1-(tanA+tanB)*tanC)
而tan(A+B)=(tanA+tanB)/(1-tanA*tanB) 将其带入上式即可
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