こんにちは。いろいろ勉強して覚えたことです。忘備録として書いておきます。意外と覚えやすい証明方法ですので, ご参考にしてください。
斜辺ABの長さが1の直角三角形AEBがあり, それが図のように台形ABCDにちょうど収まっているとする。点Eは辺CD上にある。また, ,
,
とする。
このとき,
![Rendered by QuickLaTeX.com \mathrm{AE}=\sin\alpha](https://mathtext.info/blog/wp-content/ql-cache/quicklatex.com-7450c71a1ab3dd8306602f791ad4ef55_l3.png)
![Rendered by QuickLaTeX.com \mathrm{BE}=\cos\alpha](https://mathtext.info/blog/wp-content/ql-cache/quicklatex.com-a8552561f283de4989c2e9b775933a8d_l3.png)
![Rendered by QuickLaTeX.com \beta](https://mathtext.info/blog/wp-content/ql-cache/quicklatex.com-28f9b40a3308d16b41465da4899b77d9_l3.png)
![Rendered by QuickLaTeX.com \mathrm{BC}=\cos\alpha\cos\beta](https://mathtext.info/blog/wp-content/ql-cache/quicklatex.com-67487def9337056b7f74a2cc5b832a3a_l3.png)
![Rendered by QuickLaTeX.com \mathrm{EC}=\cos\alpha\sin\beta](https://mathtext.info/blog/wp-content/ql-cache/quicklatex.com-0f42932285a6a65be61cfbd706a33c5f_l3.png)
![Rendered by QuickLaTeX.com \kaku{AED}=\beta](https://mathtext.info/blog/wp-content/ql-cache/quicklatex.com-e8c5439348e0fcabee147eb2dce51407_l3.png)
![Rendered by QuickLaTeX.com \beta](https://mathtext.info/blog/wp-content/ql-cache/quicklatex.com-28f9b40a3308d16b41465da4899b77d9_l3.png)
![Rendered by QuickLaTeX.com \mathrm{ED}=\sin\alpha\cos\beta](https://mathtext.info/blog/wp-content/ql-cache/quicklatex.com-d6555294638d55dd9a3432e130f158ce_l3.png)
![Rendered by QuickLaTeX.com \mathrm{AD}=\sin\alpha\sin\beta](https://mathtext.info/blog/wp-content/ql-cache/quicklatex.com-c5a2bc0a4ee08e86eb0800ccf90fdcc6_l3.png)
ここで, 図のように, 点Aから下底へ垂線AFを引くと, 直角三角形ABFができる。このとき,
![Rendered by QuickLaTeX.com \kaku{ABF}=\alpha+\beta](https://mathtext.info/blog/wp-content/ql-cache/quicklatex.com-8b3eea13295006eed71ef18b26a8fa82_l3.png)
![Rendered by QuickLaTeX.com \mathrm{AB}=1](https://mathtext.info/blog/wp-content/ql-cache/quicklatex.com-c38c65387e8c0422d1c50d8488afed1e_l3.png)
![Rendered by QuickLaTeX.com \sin(\alpha+\beta)=\mathrm{AF}=\mathrm{ED}+\mathrm{EC}](https://mathtext.info/blog/wp-content/ql-cache/quicklatex.com-8c77930e426422468794e44d8226353f_l3.png)
![Rendered by QuickLaTeX.com \cos(\alpha+\beta)=\mathrm{BF}=\mathrm{BC}-\mathrm{AD}](https://mathtext.info/blog/wp-content/ql-cache/quicklatex.com-5254a4aa6ffc4e6ce6961169eeeaea9f_l3.png)
これより,
![Rendered by QuickLaTeX.com \sin(\alpha+\beta)=\sin\alpha\cos\beta+\cos\alpha\sin\beta\cdots\maru1](https://mathtext.info/blog/wp-content/ql-cache/quicklatex.com-2ef9dfc8e6fa16d31aea7cc33d710a6c_l3.png)
![Rendered by QuickLaTeX.com \cos(\alpha+\beta)=\cos\alpha\cos\beta-\sin\alpha\sin\beta\cdots\maru2](https://mathtext.info/blog/wp-content/ql-cache/quicklatex.com-c16fda1579cf15ffa935052231105bab_l3.png)
が得られる。
![Rendered by QuickLaTeX.com \maru1](https://mathtext.info/blog/wp-content/ql-cache/quicklatex.com-c8b8ec9c0d15342374d474f3407d687d_l3.png)
![Rendered by QuickLaTeX.com \maru2](https://mathtext.info/blog/wp-content/ql-cache/quicklatex.com-b55f3d993b1bbf2d3a2ca1e85ea19bd7_l3.png)
![Rendered by QuickLaTeX.com \beta](https://mathtext.info/blog/wp-content/ql-cache/quicklatex.com-28f9b40a3308d16b41465da4899b77d9_l3.png)
![Rendered by QuickLaTeX.com -\beta](https://mathtext.info/blog/wp-content/ql-cache/quicklatex.com-768cc1bfafa5f395cb28fba66efb67a5_l3.png)
![Rendered by QuickLaTeX.com \cos(-\beta)=\cos\beta](https://mathtext.info/blog/wp-content/ql-cache/quicklatex.com-449529b85d6c9e166860cb486a69fb6b_l3.png)
![Rendered by QuickLaTeX.com \sin(-\beta)=-\sin\beta](https://mathtext.info/blog/wp-content/ql-cache/quicklatex.com-9d62a6aca0ae16b9a1bdaf61314ea1ca_l3.png)
![Rendered by QuickLaTeX.com \sin(\alpha-\beta)=\sin\alpha\cos\beta-\cos\alpha\sin\beta\cdots\maru3](https://mathtext.info/blog/wp-content/ql-cache/quicklatex.com-00bc7dac63dd64dd3bd4a2d81840b0d2_l3.png)
![Rendered by QuickLaTeX.com \cos(\alpha-\beta)=\cos\alpha\cos\beta+\sin\alpha\sin\beta\cdots\maru4](https://mathtext.info/blog/wp-content/ql-cache/quicklatex.com-ec63c2e0ce62b3165fc96b8be116477e_l3.png)
が得られる。
最後に,
![Rendered by QuickLaTeX.com \tan](https://mathtext.info/blog/wp-content/ql-cache/quicklatex.com-012bc3b87fa42feb109dba905c4ab6ac_l3.png)
![Rendered by QuickLaTeX.com \maru1](https://mathtext.info/blog/wp-content/ql-cache/quicklatex.com-c8b8ec9c0d15342374d474f3407d687d_l3.png)
![Rendered by QuickLaTeX.com \maru4](https://mathtext.info/blog/wp-content/ql-cache/quicklatex.com-759f766af6f2a34c5000079a99ce3363_l3.png)
![Rendered by QuickLaTeX.com \tan\left(\alpha\pm\beta\right)=\dfrac{\sin\left(\alpha\pm\beta\right)}{\cos\left(\alpha\pm\beta\right)}\cdots\textcircled{\scriptsize 5}](https://mathtext.info/blog/wp-content/ql-cache/quicklatex.com-c71efb4d0a62e8dc31893c5c6fb027d8_l3.png)
これまでの証明を利用すると,
![Rendered by QuickLaTeX.com \textcircled{\scriptsize 5}](https://mathtext.info/blog/wp-content/ql-cache/quicklatex.com-df889b7f3a63ddd9c8f7453dad48313f_l3.png)
![Rendered by QuickLaTeX.com \tan\left(\alpha\pm\beta\right)=\dfrac{\sin\alpha\cos\beta\pm\cos\alpha\sin\beta}{\cos\alpha\cos\beta\mp\sin\alpha\sin\beta}](https://mathtext.info/blog/wp-content/ql-cache/quicklatex.com-1d28aca8d999085fff78b01da8b626e2_l3.png)
右辺の分子分母を
![Rendered by QuickLaTeX.com \cos\alpha\cos\beta](https://mathtext.info/blog/wp-content/ql-cache/quicklatex.com-4990797f754d524bceca71f748078984_l3.png)
![Rendered by QuickLaTeX.com \begin{array}{lll}\tan\left(\alpha\pm\beta\right)&=&\dfrac{\dfrac{\sin\alpha\cancel{\cos\beta}}{\cos\alpha\cancel{\cos\beta}}\pm\dfrac{\cancel{\cos\alpha}\sin\beta}{\cancel{\cos\alpha}\cos\beta}}{1\mp\dfrac{\sin\alpha\sin\beta}{\cos\alpha\cos\beta}}\\&=&\dfrac{\tan\alpha\pm\tan\beta}{1\mp\tan\alpha\tan\beta}\end{array}](https://mathtext.info/blog/wp-content/ql-cache/quicklatex.com-0e6a7ba822d9f2dad4b03cc70de3a621_l3.png)
以上より,
![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/wp-content/ql-cache/quicklatex.com-b80a17189d26f2aa3d89a54364de2730_l3.png)
このような感じで加法定理が証明できる。
案外覚えやすいので, 覚えられるなら覚えておいた方がいいと思う。
加法定理