{"id":3096,"date":"2025-09-09T00:48:36","date_gmt":"2025-09-08T15:48:36","guid":{"rendered":"https:\/\/math-friend.com\/?p=3096"},"modified":"2025-09-09T00:51:46","modified_gmt":"2025-09-08T15:51:46","slug":"%e3%80%90%e4%bf%a1%e5%b7%9e%e5%a4%a7%e5%ad%a6%e5%85%a5%e8%a9%a6%e3%80%912022","status":"publish","type":"post","link":"https:\/\/math-friend.com\/?p=3096","title":{"rendered":"\u3010\u4fe1\u5dde\u5927\u5b66\u5165\u8a66\u3011tan\u306e\u500d\u89d2\u30fb\u4e09\u500d\u89d2\u3068\u4e09\u6b21\u65b9\u7a0b\u5f0f(2022)"},"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

\u4ee5\u4e0b\u306e\u554f\u3044\u306b\u7b54\u3048\u3088.
(1) \\(\\tan{2\\theta}\\), \\(\\tan{3\\theta}\\)\u3092\\(\\tan{\\theta}\\)\u3092\u7528\u3044\u3066\u8868\u305b.
(2) \\(\\displaystyle\\tan{\\frac{\\pi}{8}}\\), \\(\\displaystyle\\tan{\\frac{3\\pi}{8}}\\)\u306e\u5024\u3092\u6c42\u3081\u3088.
(3) \\(x^3-3(1+\\sqrt{2})x^2-3x+1+\\sqrt{2}=0\\)\u306e\u5b9f\u6570\u89e3\u3092\u3059\u3079\u3066\u6c42\u3081\u3088.
(2022 \u4fe1\u5dde\u5927\u5b66 \u6587\u7cfb [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) \\(\\tan\\)\u306e\u52a0\u6cd5\u5b9a\u7406\u3092\u7528\u3044\u3066,$$
\\begin{align}
\\tan{2\\theta}&=\\tan{(\\theta+\\theta)}\\\\[1.5ex]
&=\\frac{2\\tan{\\theta}}{1-\\tan^2{\\theta}}\\\\[2ex]
\\tan{3\\theta}&=\\tan{(\\theta+2\\theta)}\\\\[1.5ex]
&=\\frac{\\tan{\\theta}+\\tan{2\\theta}}{1-\\tan{\\theta}\\tan{2\\theta}}\\\\[1.5ex]
&=\\frac{\\tan{\\theta}+\\frac{2\\tan{\\theta}}{1-\\tan^2{\\theta}}}{1-\\tan{\\theta}\\cdot\\frac{2\\tan{\\theta}}{1-\\tan^2{\\theta}}}\\\\[1.5ex]
&=\\frac{3\\tan{\\theta}-\\tan^3{\\theta}}{1-3\\tan^2{\\theta}}
\\end{align}
$$<\/p>\n\n\n\n

(2) (1)\u3067\u6c42\u3081\u305f\\(\\tan{2\\theta}\\)\u306e\u5f0f\u3067\\(\\displaystyle \\theta=\\frac{\\pi}{8}\\)\u3068\u3059\u308b\u3068,$$
\\begin{align}
&\\tan{\\frac{\\pi}{4}}=\\frac{2\\tan{\\frac{\\pi}{8}}}{1-\\tan^2{\\frac{\\pi}{8}}}\\\\[1.5ex]
\\iff & 1-\\tan^2{\\frac{\\pi}{8}}=2\\tan{\\frac{\\pi}{8}}\\\\[1.5ex]
\\iff & \\tan^2{\\frac{\\pi}{8}}+2\\tan{\\frac{\\pi}{8}}-1=0
\\end{align}
$$\u3068\u306a\u308b\u304b\u3089, \\(\\displaystyle \\tan{\\frac{\\pi}{8}}\\)\u306f2\u6b21\u65b9\u7a0b\u5f0f\\(x^2+2x-1=0\\)\u306e\u89e3\u306b\u306a\u308b\u3053\u3068\u304c\u308f\u304b\u308b. \u3053\u306e2\u6b21\u65b9\u7a0b\u5f0f\u3092\u89e3\u304f\u3068,$$
x=-1\\pm\\sqrt{2}
$$\u3068\u306a\u308a, \\( \\displaystyle 0<\\frac{\\pi}{8}<\\frac{\\pi}{2}\\)\u3088\u308a, \\(\\displaystyle \\tan{\\frac{\\pi}{8}}>0\\)\u3067\u3042\u308b\u304b\u3089,$$
\\displaystyle \\tan{\\frac{\\pi}{8}}=\\sqrt{2}-1
$$\u3067\u3042\u308b\u3053\u3068\u304c\u308f\u304b\u308b.

\u6b21\u306b, (1)\u3067\u6c42\u3081\u305f\\(\\tan{3\\theta}\\)\u306e\u5f0f\u3067\\(\\displaystyle \\theta=\\frac{\\pi}{8}\\)\u3068\u3059\u308b\u3068,$$
\\begin{align}
\\tan{\\frac{3\\pi}{8}}&=\\frac{3\\tan{\\frac{\\pi}{8}}-\\tan^3{\\frac{\\pi}{8}}}{1-3\\tan^2{\\frac{\\pi}{8}}}\\\\[1.5ex]
&=\\frac{3(\\sqrt{2}-1)-(\\sqrt{2}-1)^3}{1-3(\\sqrt{2}-1)^2}\\\\[1.5ex]
&=\\frac{(\\sqrt{2}-1)(3-3+2\\sqrt{2}}{-8+6\\sqrt{2}}\\\\[1.5ex]
&=\\frac{4-2\\sqrt{2}}{-8+6\\sqrt{2}}\\\\[1.5ex]
&=\\frac{2-\\sqrt{2}}{-4+3\\sqrt{2}}\\\\[1.5ex]
&=\\frac{(2-\\sqrt{2})(-4-3\\sqrt{2})}{-2}\\\\[1.5ex]
&=\\frac{-2-2\\sqrt{2}}{-2}\\\\[1.5ex]
&=\\sqrt{2}+1
\\end{align}
$$\u3068\u6c42\u307e\u308b.<\/p>\n\n\n\n

(3) \\(\\tan{3\\theta}\\)\u306e\u5f0f\u3067\\(\\displaystyle \\theta=\\frac{\\pi}{8}\\)\u3068\u3057, \\(\\displaystyle \\tan{\\frac{3\\pi}{8}}=\\sqrt{2}+1\\)\u304b\u3089,
$$
\\begin{align}
&\\sqrt{2}+1 =\\frac{3\\tan{\\frac{\\pi}{8}}-\\tan^3{\\frac{\\pi}{8}}}{1-3\\tan^2{\\frac{\\pi}{8}}}\\\\[1.5ex]
\\iff & \\tan^3{\\frac{\\pi}{8}}-3(1+\\sqrt{2})\\tan^2{\\frac{\\pi}{8}}-3\\tan{\\frac{\\pi}{8}}+1+\\sqrt{2}=0
\\end{align}
$$\u3068\u306a\u308b. \u3053\u308c\u304b\u3089, \\(\\displaystyle \\tan{\\frac{\\pi}{8}}=\\sqrt{2}-1\\)\u306f\\(3\\)\u6b21\u65b9\u7a0b\u5f0f\\(x^3-3(1+\\sqrt{2})x^2-3x+1+\\sqrt{2}=0\\)\u306e\u89e3\u306b\u306a\u308b\u3053\u3068, \u305d\u3057\u3066, \u3053\u306e\\(3\\)\u6b21\u65b9\u7a0b\u5f0f\u306e\u5de6\u8fba\u306f\\(x-(\\sqrt{2}-1)\\)\u3067\u5272\u308a\u5207\u308c\u308b\u3053\u3068, \u304c\u308f\u304b\u308b.

\\(3\\)\u6b21\u65b9\u7a0b\u5f0f\u306e\u5de6\u8fba\u3092\u56e0\u6570\u5206\u89e3\u3059\u308b\u3068,
$$
\\left\\{x-(\\sqrt{2}-1)\\right\\}\\left\\{x^2-2(\\sqrt{2}+2)x-3-2\\sqrt{2}\\right\\}=0
$$\u3068\u306a\u308a. \\(2\\)\u6b21\u65b9\u7a0b\u5f0f\\(x^2-2(\\sqrt{2}+2)x-3-2\\sqrt{2}=0\\)\u3092\u89e3\u304f\u3068,$$
\\begin{align}
x&=\\sqrt{2}+2\\pm\\sqrt{(\\sqrt{2}+2)^2+3+2\\sqrt{2}}\\\\[1.5ex]
&=\\sqrt{2}+2\\pm\\sqrt{9+6\\sqrt{2}}\\\\[1.5ex]
&=\\sqrt{2}+2\\pm\\sqrt{9+2\\sqrt{18}}\\\\[1.5ex]
&=\\sqrt{2}+2\\pm\\sqrt{3+2\\sqrt{3\\cdot 6}+6}\\\\[1.5ex]
&=\\sqrt{2}+2\\pm\\sqrt{(\\sqrt{3}+\\sqrt{6})^2}\\\\[1.5ex]
&=\\sqrt{2}+2\\pm(\\sqrt{3}+\\sqrt{6})\\\\[1.5ex]
\\end{align}
$$\u3068\u306a\u308b.

\u3088\u3063\u3066\u4e0e\u3048\u3089\u308c\u305f\\(3\\)\u6b21\u65b9\u7a0b\u5f0f\u306e\u5b9f\u6570\u89e3\u306f,$$
x=\\sqrt{2}-1, 2+\\sqrt{2}+\\sqrt{3}+\\sqrt{6}, 2+\\sqrt{2}-\\sqrt{3}-\\sqrt{6}
$$\u306e\\(3\\)\u3064\u3067\u3042\u308b\u3053\u3068\u304c\u308f\u304b\u308b.<\/p>\n<\/div><\/div>\n\n\n\n

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