{"id":3080,"date":"2025-09-07T18:45:22","date_gmt":"2025-09-07T09:45:22","guid":{"rendered":"https:\/\/math-friend.com\/?p=3080"},"modified":"2025-09-07T18:46:19","modified_gmt":"2025-09-07T09:46:19","slug":"%e3%80%90%e6%9d%b1%e4%ba%ac%e9%83%bd%e7%ab%8b%e5%a4%a7%e5%ad%a6%e5%85%a5%e8%a9%a6%e3%80%91%e4%ba%8c%e6%ac%a1%e6%96%b9%e7%a8%8b%e5%bc%8f%e3%81%ae%e8%a7%a3%e3%81%8c%e4%b8%89%e8%a7%92%e6%af%94%e3%81%ae","status":"publish","type":"post","link":"https:\/\/math-friend.com\/?p=3080","title":{"rendered":"\u3010\u6771\u4eac\u90fd\u7acb\u5927\u5b66\u5165\u8a66\u3011\u4e8c\u6b21\u65b9\u7a0b\u5f0f\u306e\u89e3\u304c\u4e09\u89d2\u6bd4\u306e\u548c\u306b\u306a\u308b\u554f\u984c(2017)"},"content":{"rendered":"\n

\u4eca\u56de\u306f\u3053\u3061\u3089\u306e\u554f\u984c\u3092\u89e3\u3044\u3066\u3044\u304d\u307e\u3059.<\/p>\n\n\n\n

\\(k\\)\u3092\u6b63\u306e\u5b9f\u6570\u3068\u3057, \\(2\\)\u6b21\u65b9\u7a0b\u5f0f\\(8x^2-12kx+3k^2+8=0\\)\u306f\\(\\sin{\\theta}+2\\cos{\\theta}\\), \\(2\\sin{\\theta}+\\cos{\\theta}\\)\u3092\u89e3\u306b\u6301\u3064\u3068\u3059\u308b. \u305f\u3060\u3057, \\(\\displaystyle 0\\leq\\theta\\leq\\frac{\\pi}{4}\\)\u3068\u3059\u308b. \u3053\u306e\u3068\u304d, \u4ee5\u4e0b\u306e\u554f\u3044\u306b\u7b54\u3048\u3088.
(1) \\(\\sin{\\theta}+\\cos{\\theta}\\), \\(\\sin{\\theta}\\cos{\\theta}\\)\u3092\u305d\u308c\u305e\u308c\\(k\\)\u3092\u7528\u3044\u3066\u8868\u305b.
(2) \\(k\\)\u306e\u5024\u3092\u6c42\u3081\u3088.
(3) \\(\\sin{\\theta}\\), \\(\\cos{\\theta}\\)\u306e\u5024\u3092\u6c42\u3081\u3088.
(2017 \u6771\u4eac\u90fd\u7acb\u5927\u5b66 [1])<\/span><\/p>\n\n\n\n

\u305d\u308c\u3067\u306f\u89e3\u3044\u3066\u3044\u304d\u307e\u3057\u3087\u3046.<\/p>\n\n\n\n

\n

(1) \\(8x^2-12kx+3k^2+8=0\\)\u306e\\(2\\)\u89e3\u304c\\(\\sin{\\theta}+2\\cos{\\theta}\\), \\(2\\sin{\\theta}+\\cos{\\theta}\\)\u3067\u3042\u308b\u304b\u3089, \u89e3\u3068\u4fc2\u6570\u306e\u95a2\u4fc2\u3088\u308a,$$
\\begin{align}
(\\sin{\\theta}+2\\cos{\\theta})+(2\\sin{\\theta}+\\cos{\\theta})&=-\\frac{-12k}{8}\\\\[1.5ex]
(\\sin{\\theta}+2\\cos{\\theta})(2\\sin{\\theta}+\\cos{\\theta})&=\\frac{3k^2+8}{8}\\\\[1.5ex]
\\end{align}
$$\u3068\u306a\u308b. \u305d\u308c\u305e\u308c\u6574\u7406\u3059\u308b\u3068,$$
\\begin{align}
&(\\sin{\\theta}+2\\cos{\\theta})+(2\\sin{\\theta}+\\cos{\\theta})=-\\frac{-12k}{8}\\\\[1.5ex]
\\iff & 3\\sin{\\theta}+3\\cos{\\theta}=\\frac{3k}{2}\\\\[1.5ex]
\\iff & \\sin{\\theta}+\\cos{\\theta}=\\frac{k}{2}
\\end{align}
$$
$$\\begin{align}
&(\\sin{\\theta}+2\\cos{\\theta})(2\\sin{\\theta}+\\cos{\\theta})=\\frac{3k^2+8}{8}\\\\[1.5ex]
\\iff & 2\\sin^2{\\theta}+5\\sin{\\theta}\\cos{\\theta}+2\\cos^2{\\theta}=\\frac{3k^2+8}{8}\\\\[1.5ex]
\\iff & 2+5\\sin{\\theta}\\cos{\\theta}=\\frac{3k^2+8}{8}\\\\[1.5ex]
\\iff & 5\\sin{\\theta}\\cos{\\theta}=\\frac{3k^2-8}{8}\\\\[1.5ex]
\\iff & \\sin{\\theta}\\cos{\\theta}=\\frac{3k^2-8}{40}
\\end{align}
$$\u3068\u306a\u308b\u304b\u3089, \\(\\displaystyle\\sin{\\theta}+\\cos{\\theta}=\\frac{k}{2}\\), \\(\\displaystyle \\sin{\\theta}\\cos{\\theta}=\\frac{3k^2-8}{40}\\)\u3067\u3042\u308b.<\/p>\n\n\n\n

(2) \\(\\displaystyle\\sin{\\theta}+\\cos{\\theta}=\\frac{k}{2}\\)\u306e\u4e21\u8fba\u3092\\(2\\)\u4e57\u3059\u308b\u3068,$$
\\sin^2{\\theta}+2\\sin{\\theta}\\cos{\\theta}+\\cos^2{\\theta}=\\frac{k^2}{4}
$$\u3068\u306a\u308a, \\(\\displaystyle \\sin{\\theta}\\cos{\\theta}=\\frac{3k^2-8}{40}\\)\u3088\u308a,$$
\\begin{align}
& \\sin^2{\\theta}+2\\sin{\\theta}\\cos{\\theta}+\\cos^2{\\theta}=\\frac{k^2}{4}\\\\[1.5ex]
\\iff & 1+\\frac{3k^2-8}{20}=\\frac{k^2}{4}\\\\[1.5ex]
\\iff & \\left(\\frac{3}{20}-\\frac{1}{4}\\right)k^2=-1+\\frac{2}{5}\\\\[1.5ex]
\\iff & -\\frac{k^2}{10}=-\\frac{3}{5}\\\\[1.5ex]
\\iff & k^2=6
\\end{align}
$$\u3068\u306a\u308a, \\(k>0\\)\u3088\u308a, \\(k=\\sqrt{6}\\)\u3067\u3042\u308b.<\/p>\n\n\n\n

(3) (1), (2)\u3088\u308a,$$
\\begin{align}
\\sin{\\theta}+\\cos{\\theta}&=\\frac{\\sqrt{6}}{2}\\\\[1.5ex]
\\sin{\\theta}\\cos{\\theta}&=\\frac{18-8}{40}=\\frac{1}{4}
\\end{align}
$$\u3067\u3042\u308b\u304b\u3089, \u89e3\u3068\u4fc2\u6570\u306e\u95a2\u4fc2\u3088\u308a\\(\\sin{\\theta}\\), \\(\\cos{\\theta}\\)\u3092\\(2\\)\u89e3\u3068\u3059\u308b\\(x\\)\u306e\\(2\\)\u6b21\u65b9\u7a0b\u5f0f\u306f,$$
x^2-\\frac{\\sqrt{6}}{2}x+\\frac{1}{4}
$$\u3067\u3042\u308b. \u3053\u308c\u3092\u89e3\u306e\u516c\u5f0f\u3067\u89e3\u304f\u3068,$$
x=\\frac{\\frac{\\sqrt{6}}{2}\\pm\\sqrt{\\left(\\frac{\\sqrt{6}}{2}\\right)^2-1}}{2}=\\frac{\\sqrt{6}\\pm\\sqrt{2}}{4}
$$\u3068\u306a\u308b. \u3053\u3053\u3067, \\(\\displaystyle 0\\leq\\theta\\leq\\frac{\\pi}{4}\\)\u3088\u308a, \u3053\u306e\u7bc4\u56f2\u3067\u306f, \\(\\sin{\\theta}\\leq\\cos{
\\theta}\\)\u3067\u3042\u308b\u304b\u3089, \\(\\sin{\\theta}\\)\u306f\\(2\\)\u89e3\u306e\u3046\u3061\u5c0f\u3055\u3044\u65b9, \\(\\cos{\\theta}\\)\u306f\\(2\\)\u89e3\u306e\u3046\u3061\u5927\u304d\u3044\u65b9, \u3068\u306a\u308b\u304b\u3089,$$
\\begin{align}
\\sin{\\theta}&=\\frac{\\sqrt{6}-\\sqrt{2}}{4}\\\\[1.5ex]
\\cos{\\theta}&=\\frac{\\sqrt{6}+\\sqrt{2}}{4}
\\end{align}
$$\u3068\u306a\u308b.<\/p>\n<\/div><\/div>\n\n\n\n

youtube\u3067\u3082\u89e3\u8aac\u3057\u3066\u3044\u307e\u3059.<\/p>\n\n\n\n