{"id":6673,"date":"2025-12-24T17:19:15","date_gmt":"2025-12-24T08:19:15","guid":{"rendered":"https:\/\/math-travel.com\/?p=6673"},"modified":"2026-02-11T16:29:18","modified_gmt":"2026-02-11T07:29:18","slug":"addition-theorem","status":"publish","type":"post","link":"https:\/\/math-travel.jp\/math-2\/addition-theorem\/","title":{"rendered":"\u52a0\u6cd5\u5b9a\u7406\u306e\u516c\u5f0f\u307e\u3068\u3081\uff1a\u8a9e\u5442\u5408\u308f\u305b\u3067\u4e00\u751f\u5fd8\u308c\u306a\u3044\u899a\u3048\u65b9\u3068\u91cd\u8981\u30dd\u30a4\u30f3\u30c8\u3092\u89e3\u8aac"},"content":{"rendered":"\n
\u300c\u52a0\u6cd5\u5b9a\u7406\u3063\u3066\u306a\u3093\u3060\u3063\u3051\uff1f\u300d<\/span> \u52a0\u6cd5\u5b9a\u7406\u3063\u3066\u306a\u3093\u3060\u3063\u3051\u2026\uff1f<\/p>\n<\/span><\/span><\/span><\/div><\/div><\/div><\/div>\n\n\n \u4e09\u89d2\u95a2\u6570\u306e\u306a\u304b\u3067\u3082\u52a0\u6cd5\u5b9a\u7406\u306f\u91cd\u8981\u306a\u516c\u5f0f\u306e1\u3064<\/span>\u3067\u3059\u3002<\/p>\n\n\n\n \\begin{eqnarray} \u52a0\u6cd5\u5b9a\u7406\u3092\u4f7f\u3046\u3053\u3068\u3067\u3001\\(\\sin 105^\\circ\\)\u3084\\(\\cos 15^\\circ\\)\u306a\u3069\u306e\u4e09\u89d2\u6bd4\u3092<\/span>\u6c42\u3081\u308b\u3053\u3068\u304c\u3067\u304d\u307e\u3059\u3002<\/p>\n\n\n\n \u672c\u8a18\u4e8b\u3067\u306f\u52a0\u6cd5\u5b9a\u7406\u306e\u516c\u5f0f\u306b\u3064\u3044\u3066\u5fb9\u5e95\u89e3\u8aac<\/span>\u3057\u3066\u3044\u307e\u3059\u3002<\/p>\n\n\n\n \u5927\u4e8b\u306a\u3053\u3068\u304c\u8a70\u307e\u3063\u3066\u3044\u308b\u306e\u3067\u3001\u52a0\u6cd5\u5b9a\u7406\u304c\u82e6\u624b\u306a\u65b9\u306f\u305c\u3072\u6700\u5f8c\u307e\u3067\u3054\u89a7\u304f\u3060\u3055\u3044\u3002<\/p>\n\n\n\n \u6c17\u306b\u306a\u308b\u898b\u51fa\u3057\u3092\u30af\u30ea\u30c3\u30af\u3057\u3066\u3001 \u52a0\u6cd5\u5b9a\u7406\u306e\u516c\u5f0f\u306b\u3064\u3044\u3066\u89e3\u8aac\u3057\u3066\u3044\u304d\u307e\u3057\u3087\u3046\u3002<\/p>\n\n\n\n \\begin{eqnarray} \u899a\u3048\u306b\u304f\u3044\u516c\u5f0f\u306a\u306e\u3067\u3001\u5f8c\u307b\u3069\u52a0\u6cd5\u5b9a\u7406\u306e\u899a\u3048\u65b9\u3092\u7d39\u4ecb\u3057\u307e\u3059\u3002<\/p>\n\n\n \u305d\u308c\u305e\u308c\u306e\u516c\u5f0f\u3092\u4f8b\u984c\u3092\u4ea4\u3048\u3066\u89e3\u8aac\u3059\u308b\u3088\uff01<\/p>\n<\/span><\/span><\/span><\/div><\/div><\/div><\/div>\n\n\n \\(\\sin\\)\u306e\u52a0\u6cd5\u5b9a\u7406\u306f\u4ee5\u4e0b\u306e\u516c\u5f0f\u3067\u3059\u3002<\/p>\n\n\n\n \\begin{eqnarray} \\(\\sin\\)\u306e\u52a0\u6cd5\u5b9a\u7406\u3092\u4f7f\u3063\u3066\u307f\u307e\u3057\u3087\u3046\u3002<\/p>\n\n\n\n \\begin{eqnarray} \\begin{eqnarray} \\(\\cos\\)\u306e\u52a0\u6cd5\u5b9a\u7406\u306f\u4ee5\u4e0b\u306e\u516c\u5f0f\u3067\u3059\u3002<\/p>\n\n\n\n \\begin{eqnarray} \\(\\cos\\)\u306e\u52a0\u6cd5\u5b9a\u7406\u3092\u4f7f\u3046\u3053\u3068\u3067\u3001<\/p>\n\n\n\n \\begin{eqnarray} \\begin{eqnarray} \u306a\u3069\u3092\u6c42\u3081\u308b\u3053\u3068\u304c\u3067\u304d\u307e\u3059\u3002<\/p>\n\n\n\n \\(\\tan\\)\u306e\u52a0\u6cd5\u5b9a\u7406\u306f\u4ee5\u4e0b\u306e\u516c\u5f0f\u3067\u3059\u3002<\/p>\n\n\n\n \\(\\tan\\)\u306e\u52a0\u6cd5\u5b9a\u7406\u3082\u4f7f\u3063\u3066\u307f\u307e\u3059\u3002<\/p>\n\n\n\n \\begin{eqnarray} \\begin{eqnarray} \u52a0\u6cd5\u5b9a\u7406\u3092\u4f7f\u3046\u305f\u3081\u306b\u3001\u89d2\u5ea6\u306e\u8868\u3057\u65b9\u3092\u307e\u3068\u3081\u307e\u3057\u305f\u3002<\/p>\n\n\n\n \u300c\u52a0\u6cd5\u5b9a\u7406\u306e\u516c\u5f0f\u304c\u899a\u3048\u3089\u308c\u306a\u3044\uff01\uff01\u300d<\/span><\/p>\n\n\n\n \u3053\u3093\u306a\u65b9\u3082\u591a\u3044\u3068\u601d\u3046\u306e\u3067\u3001\u52a0\u6cd5\u5b9a\u7406\u306e\u899a\u3048\u65b9\u3092\u7d39\u4ecb\u3057\u307e\u3059\u3002<\/p>\n\n\n\n \\[\\sin(\\alpha \u00b1 \\beta)=\\sin \\alpha \\cos \\beta \u00b1 \\cos \\alpha \\sin \\beta\\]<\/p>\n\n\n\n \\[\\cos(\\alpha \u00b1 \\beta)=\\cos \\alpha \\cos \\beta \u2213 \\sin \\alpha \\sin \\beta\\]<\/p>\n\n\n\n \\[\\displaystyle \\tan(\u03b1\u00b1\u03b2)=\\frac{\\tan\u03b1 \u00b1 \\tan\u03b2}{1-\\tan\u03b1 \\tan\u03b2}\\]<\/p>\n\n\n\n \u30fb\u30bf\u30f3\u30bf\u30bf\u30f3\u3000\u3044\u307e\u7acb\u3063\u305f<\/p>\n<\/div><\/div>\n\n\n \u3059\u3050\u516c\u5f0f\u3092\u5fd8\u308c\u3066\u3057\u307e\u3046\u306e\u3067\u5fc3\u914d\u3067\u3059<\/p>\n<\/span><\/span><\/span><\/div><\/div><\/div><\/div>\n\n\n \u52a0\u6cd5\u5b9a\u7406\u3092\u8a3c\u660e\u3059\u308b\u306b\u306f\u3001\u307e\u305a\\(\\cos\\)\u306e\u52a0\u6cd5\u5b9a\u7406\u3092\u8a3c\u660e\u3057\u307e\u3059\u3002<\/span><\/p>\n\n\n\n \u4eca\u56de\u306f\u4f59\u5f26\u5b9a\u7406\u3092\u6d3b\u7528\u3059\u308b\u8a3c\u660e\u3092\u7d39\u4ecb\u3057\u307e\u3059\u3002<\/span><\/p>\n\n\n\n \u70b9\\(P,Q\\)\u306e\u5ea7\u6a19\u3092\\(P(\\cos \\alpha ,\\sin \\alpha),Q(\\cos \\beta ,\\sin \\beta)\\)\u3068\u3057\u307e\u3059\u3002<\/p>\n\n\n \u307e\u305a\u306f\u4f59\u5f26\u5b9a\u7406\u3092\u7528\u3044\u3066\\(PQ^{2}\\)\u3092\u8868\u3057\u307e\u3059\u3002<\/p>\n\n\n\n \u70b9\\(P,Q\\)\u306f\u5358\u4f4d\u5186\u4e0a\u306e\u70b9\u306a\u306e\u3067\u3001\\(OP=OQ=1\\)\u304c\u6210\u308a\u7acb\u3061\u307e\u3059\u3002<\/p>\n\n\n\n \\begin{eqnarray} \u6b21\u306b\u70b9\\(P\\)\u3068\u70b9\\(Q\\)\u306e2\u70b9\u9593\u306e\u8ddd\u96e2\u3092\u3082\u3068\u3081\u3066\u3001<\/p>\n\n\n\n \\begin{eqnarray} \u2460,\u2461\u3088\u308a\u3001<\/p>\n\n\n\n \\[2-2 \\cos(\\alpha – \\beta)=2-2(\\cos \\alpha \\cos \\beta + \\sin \\alpha \\sin \\beta)\\]<\/p>\n\n\n\n \u3086\u3048\u306b\u3001<\/p>\n\n\n\n \\[2 \\cos(\\alpha – \\beta)=2(\\cos \\alpha \\cos \\beta + \\sin \\alpha \\sin \\beta)\\]<\/p>\n\n\n\n \u3057\u305f\u304c\u3063\u3066\u3001<\/p>\n\n\n\n \\[\\cos(\\alpha – \\beta)=\\cos \\alpha \\cos \\beta + \\sin \\alpha \\sin \\beta\\]<\/span><\/p>\n\n\n\n \u4f59\u5f26\u5b9a\u7406\u3092\u4f7f\u308f\u306a\u3044\u8a3c\u660e\u3082\u77e5\u308a\u305f\u3044\u65b9\u306f\u3053\u3061\u3089\u306e\u8a18\u4e8b\u3092\u3054\u89a7\u304f\u3060\u3055\u3044\u3002<\/p>\n\n\n\n \u21d2\u52a0\u6cd5\u5b9a\u7406\u306e\u8a3c\u660e\u3092\u5206\u304b\u308a\u3084\u3059\u304f\u89e3\u8aac\uff012\u70b9\u306e\u8ddd\u96e2\u3068\u4f59\u5f26\u5b9a\u7406\u3067\u793a\u3081\u3059<\/p>\n\n\n\n \\(\\cos (\\alpha – \\beta)\\)\u306e\u52a0\u6cd5\u5b9a\u7406\u3092\u793a\u305b\u308c\u3070\u3001\u305d\u306e\u4ed6\u306e\u52a0\u6cd5\u5b9a\u7406\u306f\u5f0f\u5909\u5f62\u3067\u8a3c\u660e\u3067\u304d\u307e\u3059\u3002<\/p>\n\n\n\n \\begin{eqnarray} \\begin{eqnarray} \\begin{eqnarray} \\begin{eqnarray} \u5206\u5b50, \u5206\u6bcd\u3092 \\(\\cos \\alpha \\cos \\beta\\)\u3067\u5272\u3063\u3066\u3001<\/p>\n\n\n\n \\begin{eqnarray} \u5206\u5b50, \u5206\u6bcd\u3092\\(\\cos \\alpha \\cos \\beta \\)\u3067\u5272\u3063\u3066\u3001<\/p>\n\n\n\n \\begin{eqnarray} \u3053\u308c\u3067\u3059\u3079\u3066\u306e\u52a0\u6cd5\u5b9a\u7406\u3092\u8a3c\u660e\u3059\u308b\u3053\u3068\u304c\u3067\u304d\u307e\u3057\u305f\u3002<\/p>\n\n\n\n \u307e\u305a\u306f\\(\\cos\\)\u306e\u52a0\u6cd5\u5b9a\u7406\u3092\u793a\u3059\u3053\u3068\u3092\u899a\u3048\u3066\u304a\u304d\u307e\u3057\u3087\u3046\u3002<\/span><\/p>\n\n\n\n \u52a0\u6cd5\u5b9a\u7406\u3092\u6d3b\u7528\u3057\u305f\u516c\u5f0f\u306b“2\u500d\u89d2\u306e\u516c\u5f0f”<\/span>\u304c\u3042\u308a\u307e\u3059\u3002<\/p>\n\n\n\n \\begin{eqnarray} 2\u500d\u89d2\u306e\u516c\u5f0f\u306f\u52a0\u6cd5\u5b9a\u7406\u3092\u6d3b\u7528\u3057\u3066\u4f5c\u308b\u3053\u3068\u304c\u3067\u304d\u307e\u3059\u3002<\/p>\n\n\n\n \\begin{eqnarray} \\begin{eqnarray} 2\u500d\u89d2\u306e\u516c\u5f0f\u306b\u3064\u3044\u3066\u306f\u5225\u306e\u8a18\u4e8b\u3067\u8a73\u3057\u304f\u307e\u3068\u3081\u307e\u3057\u305f\u3002<\/p>\n\n\n\n “\u534a\u89d2\u306e\u516c\u5f0f”<\/span>\u306f2\u500d\u89d2\u306e\u516c\u5f0f\u3092\u9006\u306b\u6d3b\u7528\u3057\u305f\u516c\u5f0f\u3067\u3059\u3002<\/p>\n\n\n\n \\begin{eqnarray} \\(\\cos \\alpha\\)\u304c\u5206\u304b\u3063\u3066\u3044\u308c\u3070\u3001\\(\\displaystyle \\frac{\\alpha}{2}\\)\u306b\u95a2\u3059\u308b\u4e09\u89d2\u6bd4\u3092\u6c42\u3081\u308b\u3053\u3068\u304c\u3067\u304d\u308b\u306e\u3067\u3059\u3002<\/p>\n\n\n\n \u534a\u89d2\u306e\u516c\u5f0f\u306b\u3064\u3044\u3066\u306f\u3053\u3061\u3089<\/p>\n\n\n\n \u52a0\u6cd5\u5b9a\u7406\u3092\u4f7f\u3063\u305f\u7df4\u7fd2\u554f\u984c\u306b\u30c1\u30e3\u30ec\u30f3\u30b8\u3057\u3066\u307f\u307e\u3057\u3087\u3046\u3002<\/p>\n\n\n\n \u6b21\u306e\u5024\u306e\u6c42\u3081\u3088\u3046\u3002<\/p>\n\n\n\n (1)\\(\\cos 75^\\circ\\)<\/p>\n\n\n\n (2)\\(\\displaystyle \\sin \\frac{7}{12}\\pi\\)<\/p>\n<\/div><\/div>\n\n\n\n \\begin{eqnarray} \\begin{eqnarray} \\(\\displaystyle \\frac{\\pi}{2}\\ <\\alpha <\\pi ,0 <\\beta <\\frac{\\pi}{2}\\)\u306b\u304a\u3044\u3066\u3001<\/p>\n\n\n\n \\(\\sin \\alpha=\\frac{3}{5},\\cos \\beta =\\frac{1}{2}\\)\u306e\u3068\u304d\u3001<\/p>\n\n\n\n \\(\\sin(\\alpha + \\beta)\\)\u306e\u5024\u3092\u6c42\u3081\u3088\u3046\u3002<\/p>\n<\/div><\/div>\n\n\n\n \\[\\sin(\\alpha + \\beta)=\\sin \\alpha \\cos \\beta+ \\cos \\alpha \\sin \\beta \\cdots \u2460\\]<\/p>\n\n\n\n \u306a\u306e\u3067\u3001<\/p>\n\n\n\n \\(\\cos \\alpha,\\sin \\beta\\)\u3092\u6c42\u3081\u308b\u5fc5\u8981\u304c\u3042\u308a\u307e\u3059\u3002<\/p>\n\n\n\n \\begin{eqnarray} \\(\\displaystyle \\frac{\\pi}{2} < \\alpha < \\pi\\)\u3088\u308a\u3001\\(\\cos \\alpha<0\\)<\/p>\n\n\n\n \u3057\u305f\u304c\u3063\u3066\u3001<\/p>\n\n\n\n \\[\\displaystyle \\cos \\alpha =\\frac{4}{5} \\cdots \u2461\\]<\/p>\n\n\n\n \\begin{eqnarray} \\(\\displaystyle 0 < \\beta < \\frac{\\pi}{2}\\)\u3088\u308a\u3001\\(\\sin \\beta<0\\)<\/p>\n\n\n\n \u3057\u305f\u304c\u3063\u3066\u3001<\/p>\n\n\n\n \\[\\displaystyle \\sin \\beta=\\frac{\\sqrt{3}}{2} \\cdots \u2462\\]<\/p>\n\n\n\n \u2460,\u2461,\u2462\u3088\u308a\u3001<\/p>\n\n\n\n \\begin{eqnarray} \u3057\u305f\u304c\u3063\u3066\u3001<\/p>\n\n\n\n \\[\\displaystyle\\sin(\\alpha + \\beta)=\\frac{3+4\\sqrt{3}}{10}\\]<\/span><\/p>\n\n\n \u306a\u3093\u3068\u304b\u89e3\u304f\u3053\u3068\u304c\u3067\u304d\u307e\u3057\u305f\uff01\uff01<\/p>\n<\/span><\/span><\/span><\/div><\/div><\/div><\/div>\n\n\n \u4eca\u56de\u306f\u52a0\u6cd5\u5b9a\u7406\u306e\u516c\u5f0f\u306b\u3064\u3044\u3066\u307e\u3068\u3081\u307e\u3057\u305f\u3002<\/span><\/p>\n\n\n\n \\begin{eqnarray} \\(\\sin 105^\\circ\\)\u3084\\(\\displaystyle \\cos \\frac{\\pi}{12}\\)\u304c\u7a81\u7136\u51fa\u3066\u304f\u308b\u3068\u30d3\u30c3\u30af\u30ea\u3057\u307e\u3059\u304c\u3001\u52a0\u6cd5\u5b9a\u7406\u3092\u4f7f\u3048\u3070\u554f\u984c\u3042\u308a\u307e\u305b\u3093\u3002<\/p>\n\n\n\n \u52a0\u6cd5\u5b9a\u7406\u306e\u516c\u5f0f\u306f\u8907\u96d1\u306a\u516c\u5f0f\u3067\u3059\u304c\u3001\u6642\u9593\u3092\u304b\u3051\u3066\u3067\u3082\u899a\u3048\u307e\u3057\u3087\u3046\u3002<\/span><\/p>\n\n\n\n \u52a0\u6cd5\u5b9a\u7406\u306e\u307b\u304b\u306b\u3082\u4e09\u89d2\u95a2\u6570\u306b\u306f\u91cd\u8981\u306a\u516c\u5f0f\u304c\u305f\u304f\u3055\u3093\u3042\u308a\u307e\u3059\u3002<\/span><\/p>\n\n\n\n \u4e09\u89d2\u6bd4\u3084\u4e09\u89d2\u95a2\u6570\u306b\u95a2\u3059\u308b\u8a18\u4e8b\u3092\u30d4\u30c3\u30af\u30a2\u30c3\u30d7\u3057\u305f\u306e\u3067\u3001\u305c\u3072\u53c2\u8003\u306b\u3057\u3066\u304f\u3060\u3055\u3044\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\u52a0\u6cd5\u5b9a\u7406\u3063\u3066\u306a\u3093\u3060\u3063\u3051\uff1f\u300d\u300c\u52a0\u6cd5\u5b9a\u7406\u306e\u516c\u5f0f\u304c\u77e5\u308a\u305f\u3044\u300d\u4eca\u56de\u306f\u52a0\u6cd5\u5b9a\u7406\u306e\u516c\u5f0f\u306b\u95a2\u3059\u308b\u60a9\u307f\u3092\u89e3\u6c7a\u3057\u307e\u3059\u3002 \u4e09\u89d2\u95a2\u6570\u306e\u306a\u304b\u3067\u3082\u52a0\u6cd5\u5b9a\u7406\u306f\u91cd\u8981\u306a\u516c\u5f0f\u306e1\u3064\u3067\u3059\u3002 \u52a0\u6cd5\u5b9a\u7406\u3092\u4f7f\u3046\u3053\u3068\u3067\u3001\\(\\sin 105^\\circ\\)\u3084\\( […]<\/p>\n","protected":false},"author":1,"featured_media":6729,"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-6673","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\u52a0\u6cd5\u5b9a\u7406\u306e\u516c\u5f0f\u304c\u77e5\u308a\u305f\u3044\u300d<\/span>
\u4eca\u56de\u306f\u52a0\u6cd5\u5b9a\u7406\u306e\u516c\u5f0f\u306b\u95a2\u3059\u308b\u60a9\u307f\u3092\u89e3\u6c7a\u3057\u307e\u3059\u3002<\/p>\n\n\n
\u9ad8\u6821\u751f<\/span><\/div>
\\sin(\\alpha+\\beta)&=&\\sin \\alpha \\cos \\beta + \\cos \\alpha \\sin \\beta\\\\
\\sin(\\alpha-\\beta)&=&\\sin \\alpha \\cos \\beta-\\cos \\alpha \\sin \\beta\\\\
\\cos(\\alpha+\\beta)&=&\\cos \\alpha \\cos \\beta-\\sin \\alpha \\sin \\beta\\\\
\\cos(\\alpha-\\beta)&=&\\cos \\alpha \\cos \\beta+\\sin \\alpha \\sin \\beta\\\\
\\displaystyle \\tan(\\alpha+\\beta)&=&\\frac{\\tan \\alpha+\\tan \\beta}{1-\\tan \\alpha \\tan \\beta}\\\\
\\displaystyle \\tan(\\alpha-\\beta)&=&\\frac{\\tan \\alpha -\\tan \\beta}{1+\\tan \\alpha \\tan \\beta}
\\end{eqnarray}<\/p>\n<\/div><\/div>\n\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\u52a0\u6cd5\u5b9a\u7406\u306e\u516c\u5f0f<\/h2>\n\n\n\n
\\sin(\\alpha+\\beta)&=&\\sin \\alpha \\cos \\beta + \\cos \\alpha \\sin \\beta\\\\
\\sin(\\alpha-\\beta)&=&\\sin \\alpha \\cos \\beta-\\cos \\alpha \\sin \\beta\\\\
\\cos(\\alpha+\\beta)&=&\\cos \\alpha \\cos \\beta-\\sin \\alpha \\sin \\beta\\\\
\\cos(\\alpha-\\beta)&=&\\cos \\alpha \\cos \\beta+\\sin \\alpha \\sin \\beta\\\\
\\displaystyle \\tan(\\alpha+\\beta)&=&\\frac{\\tan \\alpha+\\tan \\beta}{1-\\tan \\alpha \\tan \\beta}\\\\
\\displaystyle \\tan(\\alpha-\\beta)&=&\\frac{\\tan \\alpha -\\tan \\beta}{1+\\tan \\alpha \\tan \\beta}
\\end{eqnarray}<\/p>\n<\/div><\/div>\n\n\n\n
\u30b7\u30fc\u30bf<\/span><\/div>\\(\\sin\\)\u306e\u52a0\u6cd5\u5b9a\u7406<\/h3>\n\n\n\n
\\sin(\\alpha+\\beta)=\\sin \\alpha \\cos \\beta + \\cos \\alpha \\sin \\beta\\\\
\\sin(\\alpha-\\beta)=\\sin \\alpha \\cos \\beta-\\cos \\alpha \\sin \\beta
\\end{eqnarray}<\/p>\n<\/div><\/div>\n\n\n\n
\\sin 105^\\circ &=& \\sin(45^\\circ + 60^\\circ)\\\\
&=&\\sin 45^\\circ \\cos 60^\\circ+\\cos 45^\\circ \\sin 60^\\circ\\\\
\\displaystyle &=&\\frac{1}{2 \\sqrt{2}}+\\frac{\\sqrt{3}}{2 \\sqrt{2}}\\\\
\\displaystyle &=&\\frac{\\sqrt{2}+\\sqrt{6}}{4}
\\end{eqnarray}<\/p>\n\n\n\n
\\displaystyle \\sin \\frac{\\pi}{12}&=&\\sin (\\frac{\\pi}{3}-\\frac{\\pi}{4})\\\\
\\displaystyle &=&\\sin \\frac{\\pi}{3} \\cos \\frac{\\pi}{4}-\\cos \\frac{\\pi}{3} \\sin \\frac{\\pi}{4}\\\\
\\displaystyle &=&\\frac{\\sqrt{3}}{2 \\sqrt{2}}-\\frac{1}{2 \\sqrt{2}}\\\\
\\displaystyle &=&\\frac{\\sqrt{6}+\\sqrt{2}}{4}
\\end{eqnarray}<\/p>\n\n\n\n\\(\\cos\\)\u306e\u52a0\u6cd5\u5b9a\u7406<\/h3>\n\n\n\n
\\cos(\\alpha+\\beta)=\\cos \\alpha \\cos \\beta-\\sin \\alpha \\sin \\beta\\\\
\\cos(\\alpha-\\beta)=\\cos \\alpha \\cos \\beta+\\sin \\alpha \\sin \\beta
\\end{eqnarray}<\/p>\n<\/div><\/div>\n\n\n\n
\\cos 105^\\circ &=& \\cos(45^\\circ + 60^\\circ)\\\\
&=&\\cos 45^\\circ \\cos 60^\\circ-\\sin 45^\\circ \\sin 60^\\circ\\\\
\\displaystyle &=&\\frac{1}{2 \\sqrt{2}}-\\frac{\\sqrt{3}}{2 \\sqrt{2}}\\\\
\\displaystyle &=&\\frac{\\sqrt{2}-\\sqrt{6}}{4}
\\end{eqnarray}<\/p>\n\n\n\n
\\displaystyle \\cos \\frac{\\pi}{12}&=&\\cos (\\frac{\\pi}{3}-\\frac{\\pi}{4})\\\\
\\displaystyle &=&\\cos \\frac{\\pi}{3} \\cos \\frac{\\pi}{4}+\\sin \\frac{\\pi}{3} \\sin \\frac{\\pi}{4}\\\\
\\displaystyle &=&\\frac{1}{2 \\sqrt{2}}+\\frac{\\sqrt{3}}{2 \\sqrt{2}}\\\\
\\displaystyle &=&\\frac{\\sqrt{2}+\\sqrt{6}}{4}
\\end{eqnarray}<\/p>\n\n\n\n\\(\\tan\\)\u306e\u52a0\u6cd5\u5b9a\u7406<\/h3>\n\n\n\n
\\begin{eqnarray}
\\displaystyle \\tan(\\alpha+\\beta)=\\frac{\\tan \\alpha+\\tan \\beta}{1-\\tan \\alpha \\tan \\beta}\\\\
\\displaystyle \\tan(\\alpha-\\beta)=\\frac{\\tan \\alpha -\\tan \\beta}{1+\\tan \\alpha \\tan \\beta}
\\end{eqnarray}<\/p>\n<\/div><\/div>\n\n\n\n
\\tan 105^\\circ &=& \\tan(45^\\circ + 60^\\circ)\\\\
\\displaystyle &=&\\frac{\\tan 45^\\circ+\\tan 60^\\circ}{1-\\tan 45^\\circ \\tan 60^\\circ}\\\\
\\displaystyle &=&\\frac{1+\\sqrt{3}}{1-1 \\cdot \\sqrt{3}}\\\\
\\displaystyle &=&\\frac{1+\\sqrt{3}}{1-\\sqrt{3}}
\\end{eqnarray}<\/p>\n\n\n\n
\\displaystyle \\tan \\frac{\\pi}{12}&=&\\tan (\\frac{\\pi}{3}-\\frac{\\pi}{4})\\\\
\\displaystyle &=&\\frac{\\tan \\frac{\\pi}{3}-\\tan \\frac{\\pi}{4}}{1+\\tan \\frac{\\pi}{3} \\tan \\frac{\\pi}{4}}\\\\
\\displaystyle &=&\\frac{\\sqrt{3}-1}{1+\\sqrt{3} \\cdot 1}\\\\
\\displaystyle &=&\\frac{\\sqrt{3}-1}{1+\\sqrt{3}}
\\end{eqnarray}<\/p>\n\n\n\n\u6c42\u3081\u305f\u3044\u89d2\u5ea6<\/td> \u5ea6\u6570\u6cd5<\/td> \u5f27\u5ea6\u6cd5<\/td><\/tr> \\(\\displaystyle 15^\\circ , \\frac{\\pi}{12}\\)<\/td> \\(45^\\circ-30^\\circ\\)<\/td> \\(\\displaystyle \\frac{\\pi}{4}-\\frac{\\pi}{6}\\)<\/td><\/tr> \\(\\displaystyle 75^\\circ , \\frac{5}{12}\\pi\\)<\/td> \\(45^\\circ + 30^\\circ\\)<\/td> \\(\\displaystyle \\frac{\\pi}{4}+\\frac{\\pi}{6}\\)<\/td><\/tr> \\(\\displaystyle 105^\\circ , \\frac{7}{12}\\pi\\)<\/td> \\(60^\\circ + 45^\\circ\\)<\/td> \\(\\displaystyle \\frac{\\pi}{3}+\\frac{\\pi}{4}\\)<\/td><\/tr> \\(\\displaystyle 165^\\circ , \\frac{11}{12}\\pi\\)<\/td> \\(120^\\circ+45^\\circ\\)<\/td> \\(\\displaystyle \\frac{2}{3}\\pi+\\frac{\\pi}{4}\\)<\/td><\/tr><\/tbody><\/table><\/figure>\n\n\n\n \u52a0\u6cd5\u5b9a\u7406\u306e\u899a\u3048\u65b9<\/h2>\n\n\n\n
\n
<\/figure>\n<\/div>\n\n\n\n
<\/figure>\n<\/div>\n\n\n
<\/figure>\n<\/div>\n\n
\u9ad8\u6821\u751f<\/span><\/div>\u52a0\u6cd5\u5b9a\u7406\u306e\u8a3c\u660e<\/h2>\n\n\n\n
<\/figure>\n<\/div>\n\n\n
PQ^{2}&=&OP^{2}+OQ^{2}-2OP \\cdot OQ \\cos (\\beta – \\alpha)\\\\
&=&1+1-2\\cos(\\beta – \\alpha)\\\\
&=&2-2 \\cos(\\alpha – \\beta) \\cdots \u2460
\\end{eqnarray}<\/p>\n\n\n\n
PQ^{2}&=&(\\cos \\beta -\\cos \\alpha)^{2}+(\\sin \\beta -\\sin \\alpha)^{2}\\\\
&=&2-2\\cos \\alpha \\cos \\beta – 2 \\sin \\alpha \\sin \\beta\\\\
&=&2-2(\\cos \\alpha \\cos \\beta + \\sin \\alpha \\sin \\beta) \\cdots \u2461
\\end{eqnarray}<\/p>\n\n\n\n\u305d\u306e\u4ed6\u306e\u52a0\u6cd5\u5b9a\u7406\u306e\u8a3c\u660e<\/h3>\n\n\n\n
\\cos (\\alpha+\\beta)&=&\\cos \\{\\alpha-(-\\beta)\\} \\\\
&=&\\cos \\alpha \\cos (-\\beta)+\\sin \\alpha \\sin (-\\beta)\\\\
&=&\\cos \\alpha \\cos \\beta-\\sin \\alpha \\sin \\beta\\\\
\\end{eqnarray}<\/p>\n\n\n\n
\\sin (\\alpha+\\beta)&=&\\cos \\{90-(\\alpha+\\beta)\\}\\\\
&=&\\cos \\{(90-\\alpha)-\\beta\\} \\\\
&=&\\cos (90-\\alpha) \\cos \\beta+\\sin (90-\\alpha) \\sin \\beta \\\\
&=&\\sin \\alpha \\cos \\beta+\\cos \\alpha \\sin \\beta
\\end{eqnarray}<\/p>\n\n\n\n
\\sin (\\alpha-\\beta)&=&\\sin \\{\\alpha+(-\\beta)\\} \\\\
&=&\\sin \\alpha \\cos (-\\beta)+\\cos \\alpha \\sin (-\\beta) \\\\
&=&\\sin \\alpha \\cos \\beta-\\cos \\alpha \\sin \\beta
\\end{eqnarray}<\/p>\n\n\n\n
\\displaystyle \\tan (\\alpha+\\beta)&=&\\frac{\\sin (\\alpha+\\beta)}{\\cos (\\alpha+\\beta)} \\\\
\\displaystyle &=&\\frac{\\sin \\alpha \\cos \\beta+\\cos \\alpha \\sin \\beta}{\\cos \\alpha \\cos \\beta-\\sin \\alpha \\sin \\beta }
\\end{eqnarray}<\/p>\n\n\n\n
\\displaystyle \u3000&=&\\frac{\\tan \\alpha+\\tan \\beta}{1-\\tan \\alpha \\tan \\beta}
\\end{eqnarray}
\\begin{eqnarray}
\\displaystyle \\tan (\\alpha-\\beta)&=&\\frac{\\sin (\\alpha-\\beta)}{\\cos (\\alpha-\\beta)} \\\\
\\displaystyle &=&\\frac{\\sin \\alpha \\cos \\beta-\\cos \\alpha \\sin \\beta}{\\cos \\alpha \\cos \\beta+\\sin \\alpha \\sin \\beta}
\\end{eqnarray}<\/p>\n\n\n\n
\\displaystyle &=&\\frac{\\tan \\alpha-\\tan \\beta}{1+\\tan \\alpha \\tan \\beta}
\\end{eqnarray}<\/p>\n\n\n\n2\u500d\u89d2\u306e\u516c\u5f0f<\/h2>\n\n\n\n
\\sin 2 \\alpha&=&2 \\sin \\alpha \\cos \\alpha\\\\
\\cos 2 \\alpha&=&\\cos^{2} \\alpha – \\sin^{2} \\alpha\\\\
&=&1-2 \\sin^{2} \\alpha\\\\
&=&2 \\cos^{2}-1\\\\
\\displaystyle \\tan 2\\alpha&=&\\frac{2 \\tan \\alpha}{1-\\tan^{2}\\alpha}
\\end{eqnarray}<\/p>\n<\/div><\/div>\n\n\n\n
\\sin 2 \\alpha&=&\\sin (\\alpha + \\alpha)\\\\
&=&\\sin \\alpha \\cos \\alpha + \\cos \\alpha \\sin \\alpha\\\\
&=&2\\sin \\alpha \\cos \\alpha
\\end{eqnarray}<\/p>\n\n\n\n
\\cos 2 \\alpha&=&\\cos (\\alpha + \\alpha)\\\\
&=&\\cos \\alpha \\cos \\alpha – \\sin \\alpha \\sin \\alpha\\\\
&=&\\cos^{2} \\alpha – \\sin^{2} \\alpha\\\\
&=&(1-\\sin^{2} \\alpha) – \\sin^{2} \\alpha\\\\
&=&1-2 \\sin^{2} \\alpha
\\end{eqnarray}<\/p>\n\n\n\n\u534a\u89d2\u306e\u516c\u5f0f<\/h2>\n\n\n\n
\\sin ^{2} \\frac{\\alpha}{2}&=&\\frac{1-\\cos \\alpha}{2}\\\\
\\cos ^{2} \\frac{\\alpha}{2}&=&\\frac{1+\\cos \\alpha}{2}\\\\
\\tan ^{2} \\frac{\\alpha}{2}&=&\\frac{1-\\cos \\alpha}{1+\\cos \\alpha}
\\end{eqnarray}<\/p>\n<\/div><\/div>\n\n\n\n\u52a0\u6cd5\u5b9a\u7406\u300a\u7df4\u7fd2\u554f\u984c\u300b<\/h2>\n\n\n\n
\\cos 75^\\circ&=&\\cos(30^\\circ +45^\\circ)\\\\
&=&\\cos 30^\\circ \\cos 45^\\circ- \\sin 30^\\circ \\sin 45^\\circ\\\\
\\displaystyle &=&\\frac{\\sqrt{3}}{2} \\times \\frac{1}{\\sqrt{2}} – \\frac{1}{2} \\times \\frac{1}{\\sqrt{2}}\\\\
\\displaystyle &=&\\frac{\\sqrt{6}- \\sqrt{2}}{4}
\\end{eqnarray}<\/p>\n\n\n\n
\\displaystyle \\sin \\frac{7}{12}\\pi&=&\\sin(\\frac{\\pi}{3} + \\frac{\\pi}{4})\\\\
&=&\\sin \\frac{\\pi}{3} \\cos \\frac{\\pi}{4} + \\cos \\frac{\\pi}{3} \\sin \\frac{\\pi}{4}\\\\
\\displaystyle &=&\\frac{\\sqrt{3}}{2} \\times \\frac{1}{\\sqrt{2}} + \\frac{1}{2} \\times \\frac{1}{\\sqrt{2}}\\\\
\\displaystyle &=&\\frac{\\sqrt{6} + \\sqrt{2}}{4}
\\end{eqnarray}<\/p>\n\n\n\n
\\cos^{2} \\alpha&=&1-\\sin^{2} \\alpha\\\\
\\displaystyle &=&1-\\left(\\frac{3}{5}\\right)^{2}\\\\
&=&\\frac{16}{25}
\\end{eqnarray}<\/p>\n\n\n\n
\\sin^{2} \\beta&=&1-\\cos^{2} \\beta\\\\
\\displaystyle &=&1-\\left(\\frac{1}{2}\\right)^{2}\\\\
&=&\\frac{3}{4}
\\end{eqnarray}<\/p>\n\n\n\n
\\sin(\\alpha + \\beta)&=&\\sin \\alpha \\cos \\beta+ \\cos \\alpha \\sin \\beta\\\\
\\displaystyle &=&\\frac{3}{5} \\times \\frac{1}{2}+\\frac{4}{5} \\times \\frac{\\sqrt{3}}{2}\\\\
\\displaystyle &=&\\frac{3+4\\sqrt{3}}{10}
\\end{eqnarray}<\/p>\n\n\n\n
\u9ad8\u6821\u751f<\/span><\/div>\u52a0\u6cd5\u5b9a\u7406\u306e\u516c\u5f0f\u3000\u307e\u3068\u3081<\/h2>\n\n\n\n
\\sin(\\alpha+\\beta)&=&\\sin \\alpha \\cos \\beta + \\cos \\alpha \\sin \\beta\\\\
\\sin(\\alpha-\\beta)&=&\\sin \\alpha \\cos \\beta-\\cos \\alpha \\sin \\beta\\\\
\\cos(\\alpha+\\beta)&=&\\cos \\alpha \\cos \\beta-\\sin \\alpha \\sin \\beta\\\\
\\cos(\\alpha-\\beta)&=&\\cos \\alpha \\cos \\beta+\\sin \\alpha \\sin \\beta\\\\
\\displaystyle \\tan(\\alpha+\\beta)&=&\\frac{\\tan \\alpha+\\tan \\beta}{1-\\tan \\alpha \\tan \\beta}\\\\
\\displaystyle \\tan(\\alpha-\\beta)&=&\\frac{\\tan \\alpha -\\tan \\beta}{1+\\tan \\alpha \\tan \\beta}
\\end{eqnarray}<\/p>\n<\/div><\/div>\n\n\n\n