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Home » "Unlock the Hidden Stranger Strangeness of Vecna – It’ll Blow Your Mind! - ToelettAPP

"Unlock the Hidden Stranger Strangeness of Vecna – It’ll Blow Your Mind! - ToelettAPP

📅 March 10, 2026 👤 scraface

Mar 10, 2026

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📌 \sin^2 x \cos^2 x = \left(\frac{1}{2} \sin 2x\right)^2 = \frac{1}{4} \sin^2 2x

📌 \sin^4 x + \cos^4 x = 1 - 2 \left(\frac{1}{4} \sin^2 2x\right) = 1 - \frac{1}{2} \sin^2 2x

📌 \(\sin^2 2x\) の範囲は \([0, 1]\) です。したがって、\(\frac{1}{2} \sin^2 2x\) の最大値は \(\frac{1}{2}\) であり、\(1 - \frac{1}{2} \sin^2 2x\) の最小値は、\(\sin^2 2x = 1\) のときに達成されます:

📌 - \frac{1}{2} \cdot 1 = \frac{1}{2}

📌 したがって、\(\sin^4 x + \cos^4 x\) の最小値は \(\boxed{\frac{1}{2}}\) です。

📌 問題: 点 \((1, 2)\) に最も近い直線 \(y = 2x + 3\) 上の点を求めよ。

📌 解: 直線 \(y = 2x + 3\) 上の点 \((1, 2)\) に最も近い点を見つけるには、点から直線までの垂直距離を最小化する必要があります。直線 \(y = 2x + 3\) の傾きは 2 なので、垂直な直線の傾きは \(-\frac{1}{2}\) です。

📌 点 \((1, 2)\) を通り傾き \(-\frac{1}{2}\) の直線の式は次のとおりです:

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