\\(a\\) \u3092\u6b63\u306e\u5b9f\u6570\u3068\u3059\u308b.\r\n\u5ea7\u6a19\u5e73\u9762\u306b\u304a\u3044\u3066\u66f2\u7dda \\(y = \\sin x \\ \\left( 0 \\leqq x \\leqq \\pi \\right)\\) \u3068 \\(x\\) \u8ef8\u3068\u3067\u56f2\u307e\u308c\u305f\u56f3\u5f62\u306e\u9762\u7a4d\u3092 \\(S\\) \u3068\u3057, \u66f2\u7dda \\(y = \\sin x \\ \\left( 0 \\leqq x \\leqq \\dfrac{\\pi}{2} \\right)\\) , \u66f2\u7dda \\(y = a \\cos x \\ \\left( 0 \\leqq x \\leqq \\dfrac{\\pi}{2} \\right)\\) \u304a\u3088\u3073 \\(x\\) \u8ef8\u3068\u3067\u56f2\u307e\u308c\u305f\u56f3\u5f62\u306e\u9762\u7a4d\u3092 \\(T\\) \u3068\u3059\u308b.\r\n\u3053\u306e\u3068\u304d, \\(S : T = 3 : 1\\) \u3068\u306a\u308b\u3088\u3046\u306a \\(a\\) \u306e\u5024\u3092\u6c42\u3081\u3088.<\/p>\r\n
\\[\\begin{align}\r\nS & = \\displaystyle\\int _ 0^{\\pi} \\sin x \\, dx \\\\\r\n& = \\left[ -\\cos x \\right] _ 0^{\\pi} = 1 -(-1) = 2 \\\\\r\n\\text{\u2234} \\quad T & = \\dfrac{S}{3} = \\dfrac{2}{3} \\quad ... [1]\r\n\\end{align}\\]\r\n\\(0 \\leqq x \\leqq \\dfrac{\\pi}{2}\\) \u306b\u304a\u3051\u308b \\(y = \\sin x\\) \u3068 \\(y = a \\cos x\\) \u306e\u4ea4\u70b9\u306e \\(x\\) \u5ea7\u6a19\u3092 \\(\\alpha\\) \u3068\u304a\u304f\u3068\r\n\\[\\begin{align}\r\n\\sin \\alpha & = a \\cos \\alpha \\\\\r\n\\text{\u2234} \\quad a & = \\tan \\alpha\r\n\\end{align}\\]\r\n\\(0 \\leqq \\alpha \\leqq \\dfrac{\\pi}{2}\\) \u306a\u306e\u3067\r\n\\[\r\n\\sin \\alpha = \\dfrac{a}{\\sqrt{a^2+1}}, \\quad \\cos \\alpha = \\dfrac{1}{\\sqrt{a^2+1}} \\quad ... [2]\r\n\\]\r\n\\[\\begin{align}\r\nT & = \\displaystyle\\int _ 0^{\\alpha} \\sin x \\, dx +\\displaystyle\\int _ {\\alpha}^{\\frac{\\pi}{2}} a \\cos x \\, dx \\\\\r\n& = \\left[ -\\cos x \\right] _ 0^{\\alpha} + a \\left[ \\sin x \\right] _ {\\alpha}^{\\frac{\\pi}{2}} \\\\\r\n& = 1 -\\cos \\alpha +a \\left( 1 -\\sin \\alpha \\right) \\\\\r\n& = 1 -\\dfrac{1}{\\sqrt{a^2+1}} +a \\left( 1 -\\dfrac{a}{\\sqrt{a^2+1}} \\right) \\quad ( \\ \\text{\u2235} \\ [2] \\ ) \\\\\r\n& = 1 +a -\\sqrt{a^2+1}\r\n\\end{align}\\]\r\n[1] \u3088\u308a\r\n\\[\\begin{align}\r\n1 +a -\\sqrt{a^2+1} & = \\dfrac{2}{3} \\\\\r\na +\\dfrac{1}{3} & = \\sqrt{a^2+1}\r\n\\end{align}\\]\r\n\u8fba\u3005\u5e73\u65b9\u3057\u3066\r\n\\[\\begin{align}\r\n\\left( a +\\dfrac{1}{3} \\right)^2 & = a^2+1 \\\\\r\n\\dfrac{2}{3} a +\\dfrac{1}{9} & = 1 \\\\\r\n\\text{\u2234} \\quad a & = \\underline{\\dfrac{4}{3}}\r\n\\end{align}\\]\r\n","protected":false},"excerpt":{"rendered":"\\(a\\) \u3092\u6b63\u306e\u5b9f\u6570\u3068\u3059\u308b. \u5ea7\u6a19\u5e73\u9762\u306b\u304a\u3044\u3066\u66f2\u7dda \\(y = \\sin x \\ \\left( 0 \\leqq x \\leqq \\pi \\right)\\) \u3068 \\(x\\) \u8ef8\u3068\u3067\u56f2\u307e\u308c\u305f\u56f3\u5f62\u306e\u9762\u7a4d\u3092 \\(S\\) \u3068\u3057, … \u7d9a\u304d\u3092\u8aad\u3080