{"id":2274,"date":"2025-12-24T17:19:15","date_gmt":"2025-12-24T08:19:15","guid":{"rendered":"https:\/\/math-travel.com\/?p=2274"},"modified":"2026-02-11T16:40:22","modified_gmt":"2026-02-11T07:40:22","slug":"waseinokousiki","status":"publish","type":"post","link":"https:\/\/math-travel.jp\/math-2\/waseinokousiki\/","title":{"rendered":"\u548c\u7a4d\u30fb\u7a4d\u548c\u306e\u516c\u5f0f\u306e\u899a\u3048\u65b9\uff1a\u4e38\u6697\u8a18\u4e0d\u8981\uff01\u52a0\u6cd5\u5b9a\u7406\u304b\u3089\u4e00\u77ac\u3067\u4f5c\u308b\u65b9\u6cd5\u3092\u4f1d\u6388"},"content":{"rendered":"\n
\u300c\u548c\u7a4d\u306e\u516c\u5f0f\u3063\u3066\u306a\u3093\u3060\u3063\u3051\u300d<\/span> \u548c\u7a4d\uff06\u7a4d\u548c\u306e\u516c\u5f0f\u304c\u899a\u3048\u3089\u308c\u307e\u305b\u3093\u2026<\/p>\n<\/span><\/span><\/span><\/div><\/div><\/div><\/div>\n\n\n \u4e09\u89d2\u95a2\u6570\u306e\u4e2d\u30671\u756a\u304f\u3089\u3044\u306b\u899a\u3048\u3065\u3089\u3044\u516c\u5f0f\u304c\u300c\u548c\u7a4d\u306e\u516c\u5f0f\u300d\u300c\u7a4d\u548c\u306e\u516c\u5f0f\u300d<\/span>\u3067\u3059\u3002<\/p>\n\n\n\n \u548c\u21d2\u7a4d\u306b\u5909\u63db\u3059\u308b\u306e\u304c\u548c\u7a4d\u306e\u516c\u5f0f\u3067\u3059<\/span>\u3002<\/p>\n\n\n\n \\begin{eqnarray} \u4e00\u65b9\u3067\u3001\u7a4d\u21d2\u548c\u306b\u5909\u63db\u3059\u308b\u306e\u304c\u7a4d\u548c\u306e\u516c\u5f0f\u3067\u3059\u3002<\/span><\/p>\n\n\n\n \\begin{eqnarray} \u548c\u7a4d\u306e\u516c\u5f0f\u3082\u7a4d\u548c\u306e\u516c\u5f0f\u3082\u975e\u5e38\u306b\u899a\u3048\u3065\u3089\u3044\u3067\u3059\u306d\u3002<\/p>\n\n\n\n \u672c\u8a18\u4e8b\u3067\u306f\u548c\u7a4d\u306e\u516c\u5f0f\u3068\u7a4d\u548c\u306e\u516c\u5f0f\u306b\u3064\u3044\u3066\u89e3\u8aac<\/span>\u3057\u307e\u3057\u305f\u3002<\/p>\n\n\n\n \u305d\u308c\u305e\u308c\u306e\u516c\u5f0f\u306e\u5c0e\u304d\u65b9\u3068\u899a\u3048\u65b9\u306b\u3064\u3044\u3066\u307e\u3068\u3081\u305f\u306e\u3067\u3001\u305c\u3072\u6700\u5f8c\u307e\u3067\u3054\u89a7\u304f\u3060\u3055\u3044\u3002<\/p>\n\n\n \u6c17\u306b\u306a\u308b\u898b\u51fa\u3057\u3092\u30af\u30ea\u30c3\u30af\u3057\u3066\u3001 \u307e\u305a\u306f\u7a4d\u548c\u306e\u516c\u5f0f<\/span>\u3092\u898b\u3066\u307f\u307e\u3057\u3087\u3046\u3002<\/p>\n\n\n\n \\begin{eqnarray} \u8907\u96d1\u306a\u5f62\u3092\u3057\u3066\u3044\u3066\u3001\u3053\u306e\u516c\u5f0f\u3092\u899a\u3048\u308b\u306e\u306f\u5927\u5909\u3067\u3059\u3088\u306d\u3002<\/p>\n\n\n\n \u306a\u305c\u7a4d\u548c\u306e\u516c\u5f0f\u304c\u3053\u3093\u306a\u5f62\u3092\u3057\u3066\u3044\u308b\u306e\u304b\u3001\u5c0e\u304d\u65b9\u3092\u7d39\u4ecb\u3057\u307e\u3059\u3002<\/p>\n\n\n\n \\(\\sin\\)\u306e\u52a0\u6cd5\u5b9a\u7406\u306e\u5f0f\u3092\u7528\u3044\u307e\u3059\u3002<\/p>\n\n\n\n \\[\\sin(\\alpha + \\beta)=\\sin \\alpha \\cos \\beta +\\cos \\alpha \\sin \\beta \\cdots \u2460\\] \u2460+\u2461\u304b\u3089<\/p>\n\n\n\n \\[\\sin(\\alpha + \\beta)+\\sin(\\alpha – \\beta)=2 \\sin \\alpha \\cos \\beta\\]<\/p>\n\n\n\n \u3057\u305f\u304c\u3063\u3066\u3001<\/p>\n\n\n\n \\[\\displaystyle \\sin \\alpha \\cos \\beta=\\frac{1}{2}\\{\\sin(\\alpha + \\beta)+\\sin(\\alpha – \\beta)\\}\\]<\/span><\/p>\n\n\n\n \u540c\u69d8\u306b\u2460-\u2461\u3092\u3059\u308b\u3068\u3001<\/p>\n\n\n\n \\[\\sin(\\alpha + \\beta)-\\sin(\\alpha – \\beta)=2 \\cos \\alpha \\sin \\beta\\]<\/p>\n\n\n\n \u3057\u305f\u304c\u3063\u3066\u3001<\/p>\n\n\n\n \\[\\displaystyle \\cos \\alpha \\sin \\beta=\\frac{1}{2}\\{\\sin(\\alpha + \\beta)-\\sin(\\alpha – \\beta)\\}\\]<\/span><\/p>\n\n\n\n \u6b21\u306b\\(\\sin \\alpha \\sin \\beta,\\cos \\alpha \\cos \\beta\\)\u306e\u5c0e\u304d\u65b9\u3092\u7d39\u4ecb\u3057\u307e\u3059\u3002<\/p>\n\n\n\n \\(\\cos\\)\u306e\u52a0\u6cd5\u5b9a\u7406\u3092\u601d\u3044\u51fa\u3057\u3066<\/p>\n\n\n\n \\[\\cos(\\alpha + \\beta)=\\cos \\alpha \\cos \\beta -\\sin \\alpha \\sin \\beta \\cdots \u2462\\] \u2462+\u2463\u3088\u308a<\/p>\n\n\n\n \\[\\cos(\\alpha + \\beta)+\\cos(\\alpha – \\beta)=2\\cos \\alpha \\cos \\beta\\]<\/p>\n\n\n\n \u3057\u305f\u304c\u3063\u3066\u3001<\/p>\n\n\n\n \\[\\displaystyle \\cos \\alpha \\cos \\beta=\\frac{1}{2}\\{\\cos(\\alpha + \\beta)+\\cos(\\alpha – \\beta)\\}\\]<\/span><\/p>\n\n\n\n \u540c\u69d8\u306b\u2462-\u2463\u3088\u308a\u3001<\/p>\n\n\n\n \\[\\cos(\\alpha + \\beta)-\\cos(\\alpha – \\beta)=-2\\sin \\alpha \\sin \\beta\\]<\/p>\n\n\n\n \u3057\u305f\u304c\u3063\u3066\u3001<\/p>\n\n\n\n \\[\\displaystyle \\sin \\alpha \\sin \\beta=-\\frac{1}{2}\\{\\cos(\\alpha + \\beta)-\\cos(\\alpha – \\beta)\\}\\]<\/span><\/p>\n\n\n\n \u3053\u306e\u3088\u3046\u306b\u7a4d\u548c\u306e\u516c\u5f0f\u306f\u52a0\u6cd5\u5b9a\u7406\u3092\u5229\u7528\u3057\u3066\u5c0e\u304f\u3053\u3068\u304c\u3067\u304d\u307e\u3059\u3002<\/p>\n\n\n\n \\begin{eqnarray} \u307c\u304f\u306f\u52a0\u6cd5\u5b9a\u7406\u304b\u3089\u5c0e\u3044\u3066\u3001\u7a4d\u548c\u306e\u516c\u5f0f\u3092\u4f7f\u3063\u3066\u3044\u308b\u3088<\/p>\n<\/span><\/span><\/span><\/div><\/div><\/div><\/div>\n\n\n \u6b21\u306b\u548c\u7a4d\u306e\u516c\u5f0f<\/span>\u306b\u3064\u3044\u3066\u89e3\u8aac\u3057\u307e\u3059\u3002<\/p>\n\n\n\n \u4e09\u89d2\u95a2\u6570\u306e\u548c\u306e\u5f62\u304b\u3089\u7a4d\u3078\u5909\u5f62\u3059\u308b\u516c\u5f0f\u3067\u3059\u3002<\/span><\/p>\n\n\n\n \\begin{eqnarray} \u53f3\u8fba\u306e\u89d2\u5ea6\\(\\displaystyle \\frac{A+B}{2},\\frac{A-B}{2}\\)\u306b\u6ce8\u610f\u3057\u307e\u3057\u3087\u3046\u3002<\/p>\n\n\n\n \u548c\u7a4d\u306e\u516c\u5f0f\u306f\u3001\u7a4d\u548c\u306e\u516c\u5f0f\u304b\u3089\u5909\u5f62\u3057\u3066\u5c0e\u304d\u307e\u3059\u3002<\/span><\/p>\n\n\n\n \\begin{eqnarray} \\(\\alpha+\\beta=A,\\alpha-\\beta=B\\)\u3068\u3059\u308b\u3068\u3001<\/p>\n\n\n\n \\[\\displaystyle \\alpha=\\frac{A+B}{2},\\beta=\\frac{A-B}{2}\\cdots\u2464\\]<\/p>\n\n\n\n \u2460,\u2464\u3088\u308a\u3001<\/p>\n\n\n\n \\[\\displaystyle \\sin \\frac{A+B}{2} \\cos \\frac{A-B}{2}=\\frac{1}{2}\\{\\sin A+\\sin B\\}\\]<\/p>\n\n\n\n \u3057\u305f\u304c\u3063\u3066\u3001<\/p>\n\n\n\n \\[\\displaystyle \\sin A+\\sin B=2\\sin \\frac{A+B}{2} \\cos \\frac{A-B}{2}\\]<\/span><\/p>\n\n\n\n \u540c\u69d8\u306b\u2461,\u2464\u304b\u3089<\/p>\n\n\n\n \\[\\displaystyle \\sin A-\\sin B=2\\cos \\frac{A+B}{2} \\sin \\frac{A-B}{2}\\]<\/span><\/p>\n\n\n\n \u2462,\u2464\u304b\u3089<\/p>\n\n\n\n \\[\\displaystyle \\cos A+\\cos B=2\\cos \\frac{A+B}{2} \\cos \\frac{A-B}{2}\\]<\/span><\/p>\n\n\n\n \u2463,\u2464\u304b\u3089<\/p>\n\n\n\n \\[\\displaystyle \\cos A-\\cos B=-2\\sin \\frac{A+B}{2} \\sin \\frac{A-B}{2}\\]<\/span><\/p>\n\n\n\n \u548c\u7a4d\u306e\u516c\u5f0f\u3092\u5c0e\u304f\u306b\u306f\u3001\u7a4d\u548c\u306e\u516c\u5f0f\u304c\u5fc5\u8981\u3067\u3059\u3002<\/p>\n\n\n\n \u305d\u3057\u3066\u7a4d\u548c\u306e\u516c\u5f0f\u3092\u5c0e\u304f\u305f\u3081\u306b\u52a0\u6cd5\u5b9a\u7406\u304c\u5fc5\u8981<\/span>\u306a\u306e\u3067\u3001\u52a0\u6cd5\u5b9a\u7406\u3092\u5fc5\u305a\u7406\u89e3\u3057\u3066\u304a\u304d\u307e\u3057\u3087\u3046\u3002<\/p>\n\n\n\n \u3053\u3053\u3067\u306f\u7a4d\u548c\u306e\u516c\u5f0f\u306e\u899a\u3048\u65b9\u3092\u8003\u3048\u307e\u3057\u3087\u3046\u3002<\/span><\/p>\n\n\n\n \\begin{eqnarray} \u7a4d\u548c\u306e\u516c\u5f0f\u3092\u899a\u3048\u3066\u3057\u307e\u3048\u3070\u3001\u548c\u7a4d\u306e\u516c\u5f0f\u3082\u5c0e\u304f\u3053\u3068\u304c\u3067\u304d\u307e\u3059\u3002<\/p>\n\n\n\n \u7a4d\u548c\u306e\u516c\u5f0f\u306f\u53f3\u8fba\u304c\u4ee5\u4e0b\u306e\u3088\u3046\u306a\u5f62\u3092\u3057\u3066\u3044\u307e\u3059\u3002<\/span><\/p>\n\n\n\n \u3064\u307e\u308a\u3001\u5404\u5f0f\u306e\u7b26\u53f7\u3068\\(\\sin,\\cos\\)\u3060\u3051\u3092\u899a\u3048\u308c\u3070\u5927\u4e08\u592b\u3067\u3059\u3002<\/span><\/p>\n\n\n \u5de6\u8fba\u306e\u4e09\u89d2\u95a2\u6570\u304c\u7570\u306a\u308b\u3068\u304d\u306f\u3001\u53f3\u8fba\u304c\\(\\sin\\)\u306e\u548c\u3068\u5dee\u306b\u306a\u308a\u307e\u3057\u305f\u3002<\/p>\n\n\n\n \u4e00\u65b9\u3067\u3001\u5de6\u8fba\u306e\u4e09\u89d2\u95a2\u6570\u304c\u540c\u3058\u3068\u304d\u3001\u53f3\u8fba\u304c\\(\\cos\\)\u306b\u306a\u308a\u307e\u3059\u3002<\/p>\n\n\n \\(\\sin \\times \\sin\\)\u306e\u3068\u304d\u306f\u3001\\(\\displaystyle -\\frac{1}{2}\\)\u306b\u306a\u308b\u306e\u3067\u6ce8\u610f\u3057\u307e\u3057\u3087\u3046\u3002<\/p>\n\n\n\n \u3053\u3053\u307e\u3067\u7a4d\u548c\u306e\u516c\u5f0f\u306e\u899a\u3048\u65b9\u3092\u7d39\u4ecb\u3057\u307e\u3057\u305f\u304c\u3001\u3053\u308c\u3067\u3082\u899a\u3048\u308b\u306e\u306f\u5927\u5909\u3067\u3059\u3002<\/span><\/p>\n\n\n\n \u7a4d\u548c\u306e\u516c\u5f0f\u304c\u3069\u3046\u3057\u3066\u3082\u899a\u3048\u3089\u308c\u306a\u3044\u4eba\u306f\u3001\u7a4d\u548c\u306e\u516c\u5f0f\u3092\u52a0\u6cd5\u5b9a\u7406\u304b\u3089\u5c0e\u3051\u308b\u3088\u3046\u306b\u3055\u3048\u3059\u308c\u3070\u554f\u984c\u306a\u3044\u3067\u3059\u3002<\/span><\/p>\n\n\n\n \u548c\u7a4d\u306e\u516c\u5f0f\uff06\u7a4d\u548c\u306e\u516c\u5f0f\u3092\u4f7f\u3063\u305f\u7df4\u7fd2\u554f\u984c\u306b\u6311\u6226\u3057\u307e\u3057\u3087\u3046\u3002<\/p>\n\n\n\n \u307e\u305a\u306f\u548c\u7a4d\u306e\u516c\u5f0f\u306b\u95a2\u3059\u308b\u554f\u984c\u304b\u3089\u3067\u3059\u3002<\/p>\n\n\n\n \u548c\u306e\u5f0f\u304b\u3089\u7a4d\u306e\u5f62\u3078\u5909\u5f62\u3057\u3066\u307f\u3088\u3046\u3002<\/p>\n\n\n\n \\[\\sin 5 \\theta +\\sin \\theta\\]<\/p>\n<\/div><\/div>\n\n\n\n \u3053\u308c\u306f\u548c\u7a4d\u306e\u516c\u5f0f\u3092\u7528\u3044\u307e\u3059\u3002<\/p>\n\n\n\n \\[\\displaystyle \\sin A+\\sin B=2 \\sin \\frac{A+B}{2} \\cos \\frac{A-B}{2}\\]<\/p>\n\n\n\n \\(A=5 \\theta,B=\\theta\\)\u3068\u3057\u3066\u3001<\/p>\n\n\n\n \\begin{eqnarray} \u6b21\u306b\u7a4d\u548c\u306e\u516c\u5f0f\u306b\u95a2\u3059\u308b\u554f\u984c\u3067\u3059\u3002<\/p>\n\n\n\n \u7a4d\u306e\u5f0f\u304b\u3089\u548c\u306e\u5f62\u3078\u5909\u5f62\u3057\u3066\u307f\u3088\u3046\u3002<\/p>\n\n\n\n \\[\\sin 3 \\theta \\cos \\theta\\]<\/p>\n<\/div><\/div>\n\n\n\n \u3053\u308c\u306f\u7a4d\u548c\u306e\u516c\u5f0f\u3092\u7528\u3044\u307e\u3059\u3002<\/p>\n\n\n\n \\[\\displaystyle \\sin \\alpha \\cos \\beta=\\frac{1}{2}\\{\\sin (\\alpha+\\beta)+\\sin (\\alpha-\\beta)\\}\\]<\/p>\n\n\n\n \\(\\alpha=3 \\theta,\\beta=\\theta\\)\u3068\u3057\u3066\u3001<\/p>\n\n\n\n \\begin{eqnarray} \u306a\u3093\u3068\u304b\u89e3\u304f\u3053\u3068\u304c\u3067\u304d\u307e\u3057\u305f\uff01<\/p>\n<\/span><\/span><\/span><\/div><\/div><\/div><\/div>\n\n \u516c\u5f0f\u3092\u4f7f\u3044\u3053\u306a\u305b\u308b\u3088\u3046\u306b\u7df4\u7fd2\u3057\u3066\u3044\u3053\u3046\uff01<\/p>\n<\/span><\/span><\/span><\/div><\/div><\/div><\/div>\n\n\n \u4eca\u56de\u306f\u548c\u7a4d\u306e\u516c\u5f0f\uff06\u7a4d\u548c\u306e\u516c\u5f0f\u306b\u3064\u3044\u3066\u307e\u3068\u3081\u307e\u3057\u305f\u3002<\/span><\/p>\n\n\n\n \\begin{eqnarray} \u4e00\u65b9\u3067\u3001\u7a4d\u21d2\u548c<\/span>\u306b\u5909\u63db\u3059\u308b\u306e\u304c\u7a4d\u548c\u306e\u516c\u5f0f<\/span>\u3067\u3059\u3002<\/p>\n\n\n\n \\begin{eqnarray} \u548c\u7a4d\u306e\u516c\u5f0f\u3082\u7a4d\u548c\u306e\u516c\u5f0f\u3082\u975e\u5e38\u306b\u899a\u3048\u3065\u3089\u3044\u3067\u3059\u306d\u3002<\/p>\n\n\n\n \u4eca\u56de\u306f\u548c\u7a4d\u306e\u516c\u5f0f\u3068\u7a4d\u548c\u306e\u516c\u5f0f\u306b\u3064\u3044\u3066\u307e\u3068\u3081\u307e\u3057\u305f\u304c\u3001\u4e09\u89d2\u95a2\u6570\u306b\u306f\u91cd\u8981\u306a\u516c\u5f0f\u304c\u305f\u304f\u3055\u3093\u3042\u308a\u307e\u3059\u3002<\/p>\n\n\n\n \u4e09\u89d2\u95a2\u6570\u306f\u516c\u5f0f\u3082\u591a\u304f\u3001\u5b9a\u671f\u30c6\u30b9\u30c8\u306e\u554f\u984c\u3082\u4f5c\u308a\u3084\u3059\u3044\u306e\u3067\u4e09\u89d2\u95a2\u6570\u306f\u8981\u6ce8\u610f\u306e\u5358\u5143<\/span>\u3067\u3059\u3002<\/p>\n\n\n\n \u307f\u3093\u306a\u306e\u52aa\u529b\u304c\u5831\u308f\u308c\u307e\u3059\u3088\u3046\u306b\uff01<\/p>\n","protected":false},"excerpt":{"rendered":" \u300c\u548c\u7a4d\u306e\u516c\u5f0f\u3063\u3066\u306a\u3093\u3060\u3063\u3051\u300d\u300c\u516c\u5f0f\u304c\u899a\u3048\u3089\u308c\u306a\u3044\u300d\u4eca\u56de\u306f\u548c\u7a4d\u306e\u516c\u5f0f\uff06\u7a4d\u548c\u306e\u516c\u5f0f\u306b\u3064\u3044\u3066\u307e\u3068\u3081\u307e\u3057\u305f\u3002 \u4e09\u89d2\u95a2\u6570\u306e\u4e2d\u30671\u756a\u304f\u3089\u3044\u306b\u899a\u3048\u3065\u3089\u3044\u516c\u5f0f\u304c\u300c\u548c\u7a4d\u306e\u516c\u5f0f\u300d\u300c\u7a4d\u548c\u306e\u516c\u5f0f\u300d\u3067\u3059\u3002 \u548c\u21d2\u7a4d\u306b\u5909\u63db\u3059\u308b\u306e\u304c\u548c\u7a4d\u306e\u516c\u5f0f\u3067\u3059\u3002 \u4e00 […]<\/p>\n","protected":false},"author":1,"featured_media":6948,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"swell_btn_cv_data":"","footnotes":""},"categories":[35,224],"tags":[36,14,11],"class_list":["post-2274","post","type-post","status-publish","format-standard","has-post-thumbnail","hentry","category-sincos","category-math-2","tag-36","tag-b","tag-11"],"yoast_head":"\n
\u300c\u516c\u5f0f\u304c\u899a\u3048\u3089\u308c\u306a\u3044\u300d<\/span>
\u4eca\u56de\u306f\u548c\u7a4d\u306e\u516c\u5f0f\uff06\u7a4d\u548c\u306e\u516c\u5f0f\u306b\u3064\u3044\u3066\u307e\u3068\u3081\u307e\u3057\u305f\u3002<\/p>\n\n\n
\u9ad8\u6821\u751f<\/span><\/div>
\\sin A+\\sin B &=& 2 \\sin \\frac{A+B}{2} \\cos \\frac{A-B}{2}\\\\
\\cos A+\\cos B &=& 2 \\cos \\frac{A+B}{2} \\cos \\frac{A-B}{2}\\\\
\\cos A-\\cos B &=& -2 \\sin \\frac{A+B}{2} \\sin \\frac{A-B}{2}
\\end{eqnarray}<\/p>\n<\/div><\/div>\n\n\n\n
\\sin \u03b1 \\cos \u03b2 &=& \\frac{1}{2}\\{\\sin (\u03b1+\u03b2)+\\sin (\u03b1-\u03b2)\\}\\\\
\\sin \u03b1 \\sin \u03b2 &=& \\frac{1}{2}\\{-\\cos (\u03b1+\u03b2)+\\cos (\u03b1-\u03b2)\\}\\\\
\\cos \u03b1 \\cos \u03b2 &=& \\frac{1}{2}\\{\\cos (\u03b1+\u03b2)+\\cos (\u03b1-\u03b2)\\}
\\end{eqnarray}<\/p>\n<\/div><\/div>\n\n\n\n
\u30b7\u30fc\u30bf<\/span><\/div>
\u305c\u3072\u6700\u5f8c\u307e\u3067\u3054\u89a7\u304f\u3060\u3055\u3044\u3002<\/p>\n<\/span><\/span><\/span><\/div><\/div><\/div><\/div>\n\n\n\u7a4d\u548c\u306e\u516c\u5f0f<\/h2>\n\n\n\n
\\sin \u03b1 \\cos \u03b2 &=& \\frac{1}{2}\\{\\sin (\u03b1+\u03b2)+\\sin (\u03b1-\u03b2)\\}\\\\
\\sin \u03b1 \\sin \u03b2 &=& \\frac{1}{2}\\{-\\cos (\u03b1+\u03b2)+\\cos (\u03b1-\u03b2)\\}\\\\
\\cos \u03b1 \\cos \u03b2 &=& \\frac{1}{2}\\{\\cos (\u03b1+\u03b2)+\\cos (\u03b1-\u03b2)\\}
\\end{eqnarray}<\/p>\n<\/div><\/div>\n\n\n\n\u7a4d\u548c\u306e\u516c\u5f0f\u300a\u5c0e\u304d\u65b9\u300b<\/h2>\n\n\n\n
\\[\\sin(\\alpha – \\beta)=\\sin \\alpha \\cos \\beta -\\cos \\alpha \\sin \\beta \\cdots \u2461\\]<\/p>\n\n\n\n
\\[\\cos(\\alpha – \\beta)=\\cos \\alpha \\cos \\beta +\\sin \\alpha \\sin \\beta \\cdots \u2463\\]<\/p>\n\n\n\n
\\sin \u03b1 \\cos \u03b2 &=& \\frac{1}{2}\\{\\sin (\u03b1+\u03b2)+\\sin (\u03b1-\u03b2)\\}\\\\
\\sin \u03b1 \\sin \u03b2 &=& \\frac{1}{2}\\{-\\cos (\u03b1+\u03b2)+\\cos (\u03b1-\u03b2)\\}\\\\
\\cos \u03b1 \\cos \u03b2 &=& \\frac{1}{2}\\{\\cos (\u03b1+\u03b2)+\\cos (\u03b1-\u03b2)\\}
\\end{eqnarray}<\/p>\n<\/div><\/div>\n\n\n
\u30b7\u30fc\u30bf<\/span><\/div>\u548c\u7a4d\u306e\u516c\u5f0f<\/h2>\n\n\n\n
\\sin A+\\sin B &=& 2 \\sin \\frac{A+B}{2} \\cos \\frac{A-B}{2}\\\\
\\cos A+\\cos B &=& 2 \\cos \\frac{A+B}{2} \\cos \\frac{A-B}{2}\\\\
\\cos A-\\cos B &=& -2 \\sin \\frac{A+B}{2} \\sin \\frac{A-B}{2}
\\end{eqnarray}<\/p>\n<\/div><\/div>\n\n\n\n\u548c\u7a4d\u306e\u516c\u5f0f\u300a\u5c0e\u304d\u65b9\u300b<\/h2>\n\n\n\n
\\sin \u03b1 \\cos \u03b2 &=& \\frac{1}{2}\\{\\sin (\u03b1+\u03b2)+\\sin (\u03b1-\u03b2)\\}\\\\
\\sin \u03b1 \\sin \u03b2 &=& \\frac{1}{2}\\{-\\cos (\u03b1+\u03b2)+\\cos (\u03b1-\u03b2)\\}\\\\
\\cos \u03b1 \\cos \u03b2 &=& \\frac{1}{2}\\{\\cos (\u03b1+\u03b2)+\\cos (\u03b1-\u03b2)\\}
\\end{eqnarray}<\/p>\n<\/div><\/div>\n\n\n\n\u548c\u7a4d\uff06\u7a4d\u548c\u306e\u516c\u5f0f\u3000\u899a\u3048\u65b9<\/h2>\n\n\n\n
\\sin \u03b1 \\cos \u03b2 &=& \\frac{1}{2}\\{\\sin (\u03b1+\u03b2)+\\sin (\u03b1-\u03b2)\\}\\\\
\\sin \u03b1 \\sin \u03b2 &=& \\frac{1}{2}\\{-\\cos (\u03b1+\u03b2)+\\cos (\u03b1-\u03b2)\\}\\\\
\\cos \u03b1 \\cos \u03b2 &=& \\frac{1}{2}\\{\\cos (\u03b1+\u03b2)+\\cos (\u03b1-\u03b2)\\}
\\end{eqnarray}<\/p>\n<\/div><\/div>\n\n\n\n
<\/figure>\n\n\n\n
<\/figure>\n<\/div>\n\n\n
<\/figure>\n<\/div>\n\n\n\u548c\u7a4d\uff06\u7a4d\u548c\u306e\u516c\u5f0f\u3000\u7df4\u7fd2\u554f\u984c<\/h2>\n\n\n
<\/figure>\n<\/div>\n\n\n
\\displaystyle \\sin 5 \\theta +\\sin \\theta&=&2 \\sin \\frac{5 \\theta+\\theta}{2} \\cos \\frac{5 \\theta-\\theta}{2}\\\\
&=&2 \\sin 3 \\theta \\cos 2 \\theta
\\end{eqnarray}<\/p>\n\n\n\n
\\displaystyle \\sin 3 \\theta \\cos \\theta&=&\\frac{1}{2}\\{\\sin (3 \\theta+\\theta)+\\sin (3 \\theta-\\theta)\\}\\\\
&=&\\frac{1}{2}\\left(\\sin 4 \\theta +\\sin 2 \\theta \\right)
\\end{eqnarray}<\/p>\n\n\n
\u9ad8\u6821\u751f<\/span><\/div>
\u30b7\u30fc\u30bf<\/span><\/div>\u548c\u7a4d\uff06\u7a4d\u548c\u306e\u516c\u5f0f\u3000\u307e\u3068\u3081<\/h2>\n\n\n\n
\\sin A+\\sin B &=& 2 \\sin \\frac{A+B}{2} \\cos \\frac{A-B}{2}\\\\
\\cos A+\\cos B &=& 2 \\cos \\frac{A+B}{2} \\cos \\frac{A-B}{2}\\\\
\\cos A-\\cos B &=& -2 \\sin \\frac{A+B}{2} \\sin \\frac{A-B}{2}
\\end{eqnarray}<\/p>\n<\/div><\/div>\n\n\n\n
\\sin \u03b1 \\cos \u03b2 &=& \\frac{1}{2}\\{\\sin (\u03b1+\u03b2)+\\sin (\u03b1-\u03b2)\\}\\\\
\\sin \u03b1 \\sin \u03b2 &=& \\frac{1}{2}\\{-\\cos (\u03b1+\u03b2)+\\cos (\u03b1-\u03b2)\\}\\\\
\\cos \u03b1 \\cos \u03b2 &=& \\frac{1}{2}\\{\\cos (\u03b1+\u03b2)+\\cos (\u03b1-\u03b2)\\}
\\end{eqnarray}<\/p>\n<\/div><\/div>\n\n\n\n