こんにちは。今回は単位円を使った加法定理の証明を行います。
加法定理
![Rendered by QuickLaTeX.com \textcircled{\scriptsize 1}](https://mathtext.info/blog/wordpress/wp-content/ql-cache/quicklatex.com-c6c409597cd1edd40f9cf5a0814f6312_l3.png)
![Rendered by QuickLaTeX.com \sin\left(\alpha\pm\beta\right)=\sin\alpha\cos\beta\pm\cos\alpha\sin\beta](https://mathtext.info/blog/wordpress/wp-content/ql-cache/quicklatex.com-6c8c034d0d7da31f7327aea7f5a78211_l3.png)
![Rendered by QuickLaTeX.com \textcircled{\scriptsize 2}](https://mathtext.info/blog/wordpress/wp-content/ql-cache/quicklatex.com-0caddd1d027af04be04074eaacc3ce9d_l3.png)
![Rendered by QuickLaTeX.com \cos\left(\alpha\pm\beta\right)=\cos\alpha\cos\beta\mp\sin\alpha\sin\beta](https://mathtext.info/blog/wordpress/wp-content/ql-cache/quicklatex.com-4b63b7056368065656791032d3490669_l3.png)
![Rendered by QuickLaTeX.com \textcircled{\scriptsize 3}](https://mathtext.info/blog/wordpress/wp-content/ql-cache/quicklatex.com-efa2fc7c19846faf88f9f496179d3fb1_l3.png)
![Rendered by QuickLaTeX.com \tan\left(\alpha\pm\beta\right)=\dfrac{\tan\alpha\pm\tan\beta}{1\mp\tan\alpha\tan\beta}](https://mathtext.info/blog/wordpress/wp-content/ql-cache/quicklatex.com-b80a17189d26f2aa3d89a54364de2730_l3.png)
cosの加法定理の証明
下図の単位円で, 点A( 1, 0 )を回転させた点Pの座標は,
となる。
このとき,
となります。
次にこの△AOPを点Oを回転の中心として,
![Rendered by QuickLaTeX.com -\alpha](https://mathtext.info/blog/wordpress/wp-content/ql-cache/quicklatex.com-d47cc3699340d6d2bc520bc7cadda975_l3.png)
![Rendered by QuickLaTeX.com '](https://mathtext.info/blog/wordpress/wp-content/ql-cache/quicklatex.com-e1c30b534ff750f73acf6a200f33fd32_l3.png)
![Rendered by QuickLaTeX.com '](https://mathtext.info/blog/wordpress/wp-content/ql-cache/quicklatex.com-e1c30b534ff750f73acf6a200f33fd32_l3.png)
![Rendered by QuickLaTeX.com '](https://mathtext.info/blog/wordpress/wp-content/ql-cache/quicklatex.com-e1c30b534ff750f73acf6a200f33fd32_l3.png)
![Rendered by QuickLaTeX.com (\cos\left(-\alpha\right), \sin\left(-\alpha\right))](https://mathtext.info/blog/wordpress/wp-content/ql-cache/quicklatex.com-c742e7119fdfba3d8fcc36950d406db0_l3.png)
![Rendered by QuickLaTeX.com '](https://mathtext.info/blog/wordpress/wp-content/ql-cache/quicklatex.com-e1c30b534ff750f73acf6a200f33fd32_l3.png)
![Rendered by QuickLaTeX.com \cos\alpha, -\sin\alpha](https://mathtext.info/blog/wordpress/wp-content/ql-cache/quicklatex.com-8e7d4fa4126b02c7273d4494ad09fd73_l3.png)
![Rendered by QuickLaTeX.com '](https://mathtext.info/blog/wordpress/wp-content/ql-cache/quicklatex.com-e1c30b534ff750f73acf6a200f33fd32_l3.png)
![Rendered by QuickLaTeX.com '](https://mathtext.info/blog/wordpress/wp-content/ql-cache/quicklatex.com-e1c30b534ff750f73acf6a200f33fd32_l3.png)
![Rendered by QuickLaTeX.com \cos\beta, -\sin\beta](https://mathtext.info/blog/wordpress/wp-content/ql-cache/quicklatex.com-2e89a9193beeb3a6d0e2c9b811533073_l3.png)
このとき,
![Rendered by QuickLaTeX.com \begin{array}{lll}\text{A}'\text{P}'^2&=&(\cos\beta-\cos\alpha)^2+(\sin\beta+\sin\alpha)^2\\&=&\cos^2\beta-2\cos\alpha\cos\beta+\cos^2\alpha+\sin^2\beta+2\sin\alpha\sin\beta+\sin^2\alpha\\&=&-2\cos\alpha\cos\beta+2\sin\alpha\sin\beta+2\cdots\textcircled{\scriptsize 2}\end{array}](https://mathtext.info/blog/wordpress/wp-content/ql-cache/quicklatex.com-5e796102a0c192fbbf21e4f8bba1e8de_l3.png)
となる。
△AOP
![Rendered by QuickLaTeX.com \equiv](https://mathtext.info/blog/wordpress/wp-content/ql-cache/quicklatex.com-366b6092f94455eb2f8df6d17fbeaf46_l3.png)
![Rendered by QuickLaTeX.com '](https://mathtext.info/blog/wordpress/wp-content/ql-cache/quicklatex.com-e1c30b534ff750f73acf6a200f33fd32_l3.png)
![Rendered by QuickLaTeX.com '](https://mathtext.info/blog/wordpress/wp-content/ql-cache/quicklatex.com-e1c30b534ff750f73acf6a200f33fd32_l3.png)
![Rendered by QuickLaTeX.com \textcircled{\scriptsize 1}=\textcircled{\scriptsize 2}](https://mathtext.info/blog/wordpress/wp-content/ql-cache/quicklatex.com-72a90ab9a5da0f87c26949429d6abfba_l3.png)
![Rendered by QuickLaTeX.com \begin{array}{rcl}-2\cos\left(\alpha+\beta\right)+2&=&-2\cos\alpha\cos\beta+2\sin\alpha\sin\beta+2\\-2\cos\left(\alpha+\beta\right)&=&-2\cos\alpha\cos\beta+2\sin\alpha\sin\beta\\\cos\left(\alpha+\beta\right)&=&\cos\alpha\cos\beta-\sin\alpha\sin\beta\end{array}](https://mathtext.info/blog/wordpress/wp-content/ql-cache/quicklatex.com-b04153fdebb33cb60d435e9f01ae09f8_l3.png)
よって,
![Rendered by QuickLaTeX.com \cos\left(\alpha+\beta\right)=\cos\alpha\cos\beta-\sin\alpha\sin\beta](https://mathtext.info/blog/wordpress/wp-content/ql-cache/quicklatex.com-813230b3b53fdd5f04b68848c0e0d702_l3.png)
が成立する。
![Rendered by QuickLaTeX.com \beta](https://mathtext.info/blog/wordpress/wp-content/ql-cache/quicklatex.com-28f9b40a3308d16b41465da4899b77d9_l3.png)
![Rendered by QuickLaTeX.com -\beta](https://mathtext.info/blog/wordpress/wp-content/ql-cache/quicklatex.com-768cc1bfafa5f395cb28fba66efb67a5_l3.png)
![Rendered by QuickLaTeX.com \cos\left(\alpha-\beta\right)=\cos\alpha\cos\left(-\beta\right)-\sin\alpha\sin\left(-\beta\right)](https://mathtext.info/blog/wordpress/wp-content/ql-cache/quicklatex.com-8c400a208a2864324362436b11bc93e9_l3.png)
![Rendered by QuickLaTeX.com \cos\left(\alpha-\beta\right)=\cos\alpha\cos\beta+\sin\alpha\sin\beta](https://mathtext.info/blog/wordpress/wp-content/ql-cache/quicklatex.com-cdfd40d935f1be10b01fd046ad006473_l3.png)
が得られる。
以上より,
![Rendered by QuickLaTeX.com \cos\left(\alpha\pm\beta\right)=\cos\alpha\cos\beta\mp\sin\alpha\sin\beta](https://mathtext.info/blog/wordpress/wp-content/ql-cache/quicklatex.com-4b63b7056368065656791032d3490669_l3.png)
sinの加法定理の証明
で
を
とすると,
よって,
が得られ, この式で, を
で置き換えると,
が得られる。
以上より,
tanの加法定理の証明
の証明
これまでの証明を利用すると, は次のように書き換えることができる。
右辺の分子分母をで割ると,
以上より,