こんにちは。相城です。今回は余弦定理ができるまで。ということでお話ししていきます。
余弦定理
△ABCのBC
![Rendered by QuickLaTeX.com =a](https://mathtext.info/blog/wordpress/wp-content/ql-cache/quicklatex.com-ef5e023296fbcd93fca7ae7d8fb99e99_l3.png)
![Rendered by QuickLaTeX.com =b](https://mathtext.info/blog/wordpress/wp-content/ql-cache/quicklatex.com-81fba85cf686691bdff648859f638669_l3.png)
![Rendered by QuickLaTeX.com =c](https://mathtext.info/blog/wordpress/wp-content/ql-cache/quicklatex.com-9dbab1b5e2afc23adec9c82b4edb6b32_l3.png)
![Rendered by QuickLaTeX.com a^2=b^2+c^2-2bc\cos\text{A}](https://mathtext.info/blog/wordpress/wp-content/ql-cache/quicklatex.com-261256ab6c2cada14c4e76b6ffaa08fa_l3.png)
![Rendered by QuickLaTeX.com b^2=a^2+c^2-2ac\cos\text{B}](https://mathtext.info/blog/wordpress/wp-content/ql-cache/quicklatex.com-dede5444873b6f4a15e65651962b218e_l3.png)
![Rendered by QuickLaTeX.com c^2=a^2+b^2-2ab\cos\text{C}](https://mathtext.info/blog/wordpress/wp-content/ql-cache/quicklatex.com-9d9d5ea58bc6bacb9bc56e48d20dc21a_l3.png)
が成り立つ。これを余弦定理という。
どうやって余弦定理ができるの?
余弦定理は中学3年生で習った三平方の定理の拡張版だと思ってください。
三平方の定理はのときだけでしたが, 余弦定理では
は
まで拡張できます。
次の余弦定理①を下の図を使って証明しましょう。ただし, は鋭角とします。
![](https://www.mathtext.info/blog/wordpress/wp-content/uploads/2020/04/yogen1.png)
△ABCで,
![Rendered by QuickLaTeX.com \angle{\text{BAC}}=\theta](https://mathtext.info/blog/wordpress/wp-content/ql-cache/quicklatex.com-4e25c72989494c8f1d44aaf27ff9e71b_l3.png)
辺ABの交点をDとする。このとき, 右の図のように, AD
![Rendered by QuickLaTeX.com =b\cos\theta](https://mathtext.info/blog/wordpress/wp-content/ql-cache/quicklatex.com-95fd2b344dee51dbef951b8b3ad68da7_l3.png)
![Rendered by QuickLaTeX.com =b\sin\theta](https://mathtext.info/blog/wordpress/wp-content/ql-cache/quicklatex.com-3b59a5285b29f3eac5db3303ced97f68_l3.png)
![Rendered by QuickLaTeX.com =c-b\cos\theta](https://mathtext.info/blog/wordpress/wp-content/ql-cache/quicklatex.com-f64a0b5e31920b560ac65f3e23aa7ab4_l3.png)
![Rendered by QuickLaTeX.com a^2=(c-b\cos\theta)^2+b^2\sin^2\theta](https://mathtext.info/blog/wordpress/wp-content/ql-cache/quicklatex.com-35f59898346f78a1efed976b28c06c33_l3.png)
![Rendered by QuickLaTeX.com =c^2-2bc\cos\theta+b^2\cos^2\theta+b^2\sin^2\theta](https://mathtext.info/blog/wordpress/wp-content/ql-cache/quicklatex.com-1e02ea8c5e73a32afc3e4ec2ed020103_l3.png)
![Rendered by QuickLaTeX.com =c^2-2bc\cos\theta+b^2(\cos^2\theta+\sin^2\theta)](https://mathtext.info/blog/wordpress/wp-content/ql-cache/quicklatex.com-5c05ea146f9ff40bd23080959e2070c0_l3.png)
![Rendered by QuickLaTeX.com =b^2+c^2-2bc\cos\theta](https://mathtext.info/blog/wordpress/wp-content/ql-cache/quicklatex.com-75793f70e3fb4be1c80967046431796b_l3.png)
![Rendered by QuickLaTeX.com \theta](https://mathtext.info/blog/wordpress/wp-content/ql-cache/quicklatex.com-df36b52cea0081617d2fc178107fe54d_l3.png)
が鈍角の場合
![](https://www.mathtext.info/blog/wordpress/wp-content/uploads/2020/04/yogen2-1.png)
頂点BからCAの延長線上に垂線を下ろして, 交点をDとする。このとき, 図のように, より, AD
, BD
となります。
ここで,
である。これを適用し, △BCDで三平方の定理を用いると,
![Rendered by QuickLaTeX.com a^2=(b-c\cos\theta)^2+c^2\sin^2\theta](https://mathtext.info/blog/wordpress/wp-content/ql-cache/quicklatex.com-6096bdd1497c59f11b941b81124d65a7_l3.png)
![Rendered by QuickLaTeX.com =b^2-2bc\cos\theta+c^2\cos^2\theta+c^2\sin^2\theta](https://mathtext.info/blog/wordpress/wp-content/ql-cache/quicklatex.com-45e78e3f921957fda81113d2984f7ffa_l3.png)
![Rendered by QuickLaTeX.com =b^2-2bc\cos\theta+c^2(\cos^2\theta+\sin^2\theta)](https://mathtext.info/blog/wordpress/wp-content/ql-cache/quicklatex.com-f6a76c9cb1f6155a631f0b4d45fb8630_l3.png)
![Rendered by QuickLaTeX.com =b^2+c^2-2bc\cos\theta](https://mathtext.info/blog/wordpress/wp-content/ql-cache/quicklatex.com-75793f70e3fb4be1c80967046431796b_l3.png)