\u25b3ABC \u306f, \u6761\u4ef6 $\\angle \\text{B} = 2 \\angle \\text{A}$ , $\\text{BC} = 1$ \u3092\u6e80\u305f\u3059\u4e09\u89d2\u5f62\u306e\u3046\u3061\u3067\u9762\u7a4d\u304c\u6700\u5927\u306e\u3082\u306e\u3067\u3042\u308b\u3068\u3059\u308b.\r\n\u3053\u306e\u3068\u304d, $\\cos \\angle \\text{B}$ \u3092\u6c42\u3081\u3088. <\/p>\r\n
$\\angle \\text{A} = \\theta \\ \\left( 0 \\lt \\theta \\lt \\dfrac{\\pi}{3} \\right) \\quad ... \\maru{1}$ \u3068\u304a\u304f.
\r\n\u6b63\u5f26\u5b9a\u7406\u3088\u308a\r\n$$\\begin{align}\r\n\\dfrac{1}{\\sin \\theta} & = \\dfrac{\\text{AC}}{\\sin 2 \\theta} \\\\\r\n\\therefore \\quad \\text{AC} & = 2 \\cos \\theta\r\n\\end{align}$$\r\n\u306a\u306e\u3067\r\n$$\\begin{align}\r\n\\triangle \\text{ABC} & = \\dfrac{1}{2} \\cdot 1 \\cdot 2 \\cos \\theta \\cdot \\sin \\left( 2 \\pi -3 \\theta \\right) \\\\\r\n& = \\cos \\theta \\sin 3 \\theta \\\\\r\n& = \\dfrac{1}{2} \\left\\{ \\sin ( 3 \\theta +\\theta ) +\\sin ( 3 \\theta -\\theta ) \\right\\} \\\\\r\n& = \\dfrac{1}{2} \\left( \\sin 4 \\theta +\\sin 2 \\theta \\right)\r\n\\end{align}$$\r\n\u3053\u308c\u3092 $f( \\theta )$ \u3068\u304a\u3051\u3070\r\n$$\\begin{align}\r\nf'( \\theta ) & = 2 \\cos 4 \\theta +\\cos 2 \\theta \\\\\r\n& = 2 \\left( 2 \\cos^2 2 \\theta -1 \\right) +\\cos 2 \\theta \\\\\r\n& = 4 \\cos^2 2 \\theta +\\cos 2 \\theta -2 \\\\\r\n& = 4 \\left( \\cos 2 \\theta -\\dfrac{-1 -\\sqrt{33}}{8} \\right) \\left( \\cos 2 \\theta -\\dfrac{-1 +\\sqrt{33}}{8} \\right)\r\n\\end{align}$$\r\n$\\maru{1}$ \u3088\u308a, $\\cos 2 \\theta \\gt \\cos \\dfrac{2 \\pi}{3} = -\\dfrac{1}{2}$ \u306a\u306e\u3067, $\\cos 2 \\alpha = \\dfrac{-1 +\\sqrt{33}}{8}$ \u306a\u308b $\\alpha \\ \\left( 0 \\lt \\alpha \\lt \\dfrac{\\pi}{3} \\right)$ \u3092\u304a\u3051\u3070, $f( \\theta )$ \u306e\u5897\u6e1b\u306f\u4e0b\u8868\u306e\u3088\u3046\u306b\u306a\u308b.\r\n$$\r\n\\begin{array}{c|ccccc} \\theta & (0) & \\cdots & \\alpha & \\cdots & \\left( \\dfrac{\\pi}{3} \\right) \\\\ \\hline f'( \\theta ) & & + & 0 & - & \\\\ \\hline f( \\theta ) & & \\nearrow & \\text{\u6700\u5927} & \\searrow & \\end{array}\r\n$$\r\n\u3088\u3063\u3066\r\n$$\r\n\\cos \\angle \\text{B} = \\cos 2 \\alpha = \\underline{\\dfrac{-1 +\\sqrt{33}}{8}}\r\n$$<\/p>","protected":false},"excerpt":{"rendered":"\u25b3ABC \u306f, \u6761\u4ef6 $\\angle \\text{B} = 2 \\angle \\text{A}$ , $\\text{BC} = 1$ \u3092\u6e80\u305f\u3059\u4e09\u89d2\u5f62\u306e\u3046\u3061\u3067\u9762\u7a4d\u304c\u6700\u5927\u306e\u3082\u306e\u3067\u3042\u308b\u3068\u3059\u308b. \u3053\u306e\u3068\u304d, $\\cos \\ang … \u7d9a\u304d\u3092\u8aad\u3080