选修2-2 阅读与思考 平面与空间中的余弦定理 课件15张PPT
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| 名称 | 选修2-2 阅读与思考 平面与空间中的余弦定理 课件15张PPT |  | |
| 格式 | zip | ||
| 文件大小 | 569.2KB | ||
| 资源类型 | 教案 | ||
| 版本资源 | 人教新课标A版 | ||
| 科目 | 数学 | ||
| 更新时间 | 2019-08-15 11:21:59 | ||
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课件15张PPT。平面与空间中的余弦定理复习:合情推理 推理演绎推理归纳推理类比推理“我珍视类比胜过任何别的东西，它是我最可信赖的老师，它能揭示自然界的秘密。”约翰尼斯·开普勒德国杰出的天文学家，物理学家，数学家类比平面内直角三角形的勾股定理，给出空间中四面体性质的猜想．c2=a2+b2余弦定理：余弦定理：类比？四面体P－ABC设二面角P—AB—C,P—BC—A,
P—AC—B,C—PA—B,A—PB—C,
B—PC—A分别为α1，α2，α3，β1，β2，β3猜想：S2△ABC = S2△PAB+S2 △PBC+S2 △PAC—2S △PABS △PBC ·cosβ2
—2S △PBCS △PAC ·cosβ3—2S△PACS△PAB ·cosβ1
在△ABC,在a上作高，将边a写作：a=c cosB+b cosC ①c=a cosB+b cosA ③a2=ac cosB+ab cosC ④ b2=ab cosC+bc cosA ⑤ c2=ac cosB+bc cosA ⑥⑤+ ⑥得: b2+c2=ab cosC+ac cosB+2bc cosA a2=b2+c2 －2bc cosA ac cosB+ab cosC= b2+c2－2bc cosA b=a cosC+c cosA ② S2△ABC = S2△PAB+S2 △PBC+S2 △PAC—2S △PABS △PBC ·cosβ2
—2S △PBCS △PAC ·cosβ3—2S△PACS△PAB ·cosβ1
类比得空间余弦定理的证明过程S△ABC = S△PAB cosα1 + S △PBC ·cosα2 + S△PAC ·cosα3
S△PAB = S△ABC cosα1 + S △PBC ·cosβ2 + S△PAC ·cosβ1
S△PBC = S△ABC cosα2 + S △PAB ·cosβ2 + S△PAC ·cosβ3
S△PAC = S△ABC cosα3 + S △PBC ·cosβ3 + S△PAB ·cosβ1
证明：S2△ABC =S△ABCS△PABcosα1 +S△ABCS△PBC cosα2 +S△ABCS△PACcosα3
S2△PAB =S△ABCS△PAB cosα1+S△PBC S△PAB cosβ2 +S△PAB S△PACcosβ1
S2△PBC =S△ABCS△PBC cosα2 +S△PBC S △PABcosβ2 +S△PBC S△PAC cosβ3
S2△PAC =S△PACS△ABCcosα3 +S△PACS△PBCcosβ3 +S△PACS△PAB cosβ1
S△ABC = S△PAB cosα1 + S △PBC ·cosα2 + S△PAC ·cosα3
S△PAB = S△ABC cosα1 + S △PBC ·cosβ2 + S△PAC ·cosβ1
S△PBC = S△ABC cosα2 + S △PAB ·cosβ2 + S△PAC ·cosβ3
S△PAC = S△ABC cosα3 + S △PBC ·cosβ3 + S△PAB ·cosβ1
S2△ABC=S△ABCS△PABcosα1+S△ABCS△PBC cosα2 +S△ABC S△PACcosα3
S2△PAB =S△ABCS△PABcosα1 +S△PBCS△PABcosβ2 +S△PABS△PACcosβ1
S2△PBC =S△ABCS△PBCcosα2 +S△PBC S△PAB cosβ2 +S△PBCS△PACcosβ3
S2△PAC =S△PACS△ABCcosα3 +S△PACS△PBCcosβ3 +S△PACS△PABcosβ1
S2△PAB +S2△PBC +S2△PAC=S△ABCS△PAB cosα1 +S△ABC S△PBC cosα2
+S△PACS△ABCcosα3＋2S△PAB S△PACcosβ1＋2S△PBC S△PAB cosβ2 ＋2S△PBCS△PACcosβ3 S△ABCS△PAB cosα1 +S△ABCS△PBCcosα2 +S△PAC S△ABCcosα3
=S2△PAB +S2△PBC +S2△PAC－2 S△PABS△PAC cosβ1
－2S △PBC S△PAB cosβ2 － 2S△PBC S△PAC cosβ3S2△ABC =S△ABCS△PABcosα1 +S△ABCS△PBCcosα2 +S△ABCS△PAC cosα3
S2△PAB =S△ABCS△PABcosα1 +S △PBCS△PAB cosβ2 +S△PABS△PACcosβ1
S2△PBC =S△ABCS△PBCcosα2 +S△PBC S△PABcosβ2 +S△PBC S△PACcosβ3
S2△PAC =S△PACS△ABCcosα3 +S△PACS△PBCcosβ3 +S△PACS△PABcosβ1
S△ABCS△PAB cosα1 +S△ABCS△PBCcosα2 +S△PAC S△ABCcosα3
=S2△PAB +S2△PBC +S2△PAC－2S△PAB S△PACcosβ1
－2S△PBCS△PAB cosβ2 － 2S△PBC S△PAC cosβ3S2△ABC = S2△PAB+S2 △PBC+S2 △PAC—2S△PACS△PAB ·cosβ1
－2S △PABS △PBC ·cosβ2 —2S △PBCS △PAC ·cosβ3四面体P--ABC类比：S2△ABC = S2△PAB+S2 △PBC+S2 △PAC
— 2S△PACS△PAB ·cosβ1
—2S △PABS △PBC ·cosβ2
—2S △PBCS △PAC ·cosβ3
P—AC—B,C—PA—B,A—PB—C,
B—PC—A分别为α1，α2，α3，β1，β2，β3猜想：S2△ABC = S2△PAB+S2 △PBC+S2 △PAC—2S △PABS △PBC ·cosβ2
—2S △PBCS △PAC ·cosβ3—2S△PACS△PAB ·cosβ1
在△ABC,在a上作高，将边a写作：a=c cosB+b cosC ①c=a cosB+b cosA ③a2=ac cosB+ab cosC ④ b2=ab cosC+bc cosA ⑤ c2=ac cosB+bc cosA ⑥⑤+ ⑥得: b2+c2=ab cosC+ac cosB+2bc cosA a2=b2+c2 －2bc cosA ac cosB+ab cosC= b2+c2－2bc cosA b=a cosC+c cosA ② S2△ABC = S2△PAB+S2 △PBC+S2 △PAC—2S △PABS △PBC ·cosβ2
—2S △PBCS △PAC ·cosβ3—2S△PACS△PAB ·cosβ1
类比得空间余弦定理的证明过程S△ABC = S△PAB cosα1 + S △PBC ·cosα2 + S△PAC ·cosα3
S△PAB = S△ABC cosα1 + S △PBC ·cosβ2 + S△PAC ·cosβ1
S△PBC = S△ABC cosα2 + S △PAB ·cosβ2 + S△PAC ·cosβ3
S△PAC = S△ABC cosα3 + S △PBC ·cosβ3 + S△PAB ·cosβ1
证明：S2△ABC =S△ABCS△PABcosα1 +S△ABCS△PBC cosα2 +S△ABCS△PACcosα3
S2△PAB =S△ABCS△PAB cosα1+S△PBC S△PAB cosβ2 +S△PAB S△PACcosβ1
S2△PBC =S△ABCS△PBC cosα2 +S△PBC S △PABcosβ2 +S△PBC S△PAC cosβ3
S2△PAC =S△PACS△ABCcosα3 +S△PACS△PBCcosβ3 +S△PACS△PAB cosβ1
S△ABC = S△PAB cosα1 + S △PBC ·cosα2 + S△PAC ·cosα3
S△PAB = S△ABC cosα1 + S △PBC ·cosβ2 + S△PAC ·cosβ1
S△PBC = S△ABC cosα2 + S △PAB ·cosβ2 + S△PAC ·cosβ3
S△PAC = S△ABC cosα3 + S △PBC ·cosβ3 + S△PAB ·cosβ1
S2△ABC=S△ABCS△PABcosα1+S△ABCS△PBC cosα2 +S△ABC S△PACcosα3
S2△PAB =S△ABCS△PABcosα1 +S△PBCS△PABcosβ2 +S△PABS△PACcosβ1
S2△PBC =S△ABCS△PBCcosα2 +S△PBC S△PAB cosβ2 +S△PBCS△PACcosβ3
S2△PAC =S△PACS△ABCcosα3 +S△PACS△PBCcosβ3 +S△PACS△PABcosβ1
S2△PAB +S2△PBC +S2△PAC=S△ABCS△PAB cosα1 +S△ABC S△PBC cosα2
+S△PACS△ABCcosα3＋2S△PAB S△PACcosβ1＋2S△PBC S△PAB cosβ2 ＋2S△PBCS△PACcosβ3 S△ABCS△PAB cosα1 +S△ABCS△PBCcosα2 +S△PAC S△ABCcosα3
=S2△PAB +S2△PBC +S2△PAC－2 S△PABS△PAC cosβ1
－2S △PBC S△PAB cosβ2 － 2S△PBC S△PAC cosβ3S2△ABC =S△ABCS△PABcosα1 +S△ABCS△PBCcosα2 +S△ABCS△PAC cosα3
S2△PAB =S△ABCS△PABcosα1 +S △PBCS△PAB cosβ2 +S△PABS△PACcosβ1
S2△PBC =S△ABCS△PBCcosα2 +S△PBC S△PABcosβ2 +S△PBC S△PACcosβ3
S2△PAC =S△PACS△ABCcosα3 +S△PACS△PBCcosβ3 +S△PACS△PABcosβ1
S△ABCS△PAB cosα1 +S△ABCS△PBCcosα2 +S△PAC S△ABCcosα3
=S2△PAB +S2△PBC +S2△PAC－2S△PAB S△PACcosβ1
－2S△PBCS△PAB cosβ2 － 2S△PBC S△PAC cosβ3S2△ABC = S2△PAB+S2 △PBC+S2 △PAC—2S△PACS△PAB ·cosβ1
－2S △PABS △PBC ·cosβ2 —2S △PBCS △PAC ·cosβ3四面体P--ABC类比：S2△ABC = S2△PAB+S2 △PBC+S2 △PAC
— 2S△PACS△PAB ·cosβ1
—2S △PABS △PBC ·cosβ2
—2S △PBCS △PAC ·cosβ3