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Home » استخدم صيغة المحيط: C = 2πr. - ToelettAPP

استخدم صيغة المحيط: C = 2πr. - ToelettAPP

📅 March 9, 2026 👤 scraface

Mar 09, 2026

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📌 eq b $. Instead, note that $ rac{a + b}{a - b} + rac{a - b}{a + b} = rac{2(a^2 + b^2)}{a^2 - b^2} $. Let $ a = e^{i heta} $, $ b = e^{i\phi} $, then compute $ S = rac{2(e^{2i heta} + e^{2i\phi})}{e^{2i heta} - e^{2i\phi}} $. Multiply numerator and denominator by $ e^{-i heta} \overline{e^{i heta}} $:

📌 S = rac{2(e^{i heta} + e^{-i heta} + e^{i\phi} + e^{-i\phi})}{e^{i heta} - e^{-i heta} - e^{i\phi} + e^{-i\phi}}.

📌 This simplifies to $ rac{2(2\cos heta + 2\cos\phi)}{2i(\sin heta - \sin\phi)} = rac{2(\cos heta + \cos\phi)}{i(\sin heta - \sin\phi)} $, which is purely imaginary. However, the original expression simplifies directly:

📌 S = rac{(a + b)^2 + (a - b)^2}{a^2 - b^2} = rac{2a^2 + 2b^2}{a^2 - b^2}.

📌 But $ a^2 - b^2 = (a - b)(a + b) $, and $ a^2 + b^2 = (a + b)^2 - 2ab $. Instead, observe $ S = rac{2(a^2 + b^2)}{a^2 - b^2} $. Let $ a = 1 $, $ b = i $:

📌 S = rac{2(1 + (-1))}{1 - (-1)} = 0.

📌 But earlier steps suggest $ S = rac{2(a^2 + b^2)}{a^2 - b^2} $. For $ |a| = |b| = 1 $, $ a^2 \overline{a}^2 = 1 $, so $ a^2 = rac{1}{\overline{a}^2} $. This path is complex. Instead, let $ a = 1 $, $ b = -1 $: $ S = rac{0}{2} + rac{2}{0} $, invalid. Correct approach: Let $ a = e^{i heta} $, $ b = e^{i\phi} $, then

📌 S = rac{2(e^{2i heta} + e^{2i\phi})}{e^{2i heta} - e^{2i\phi}} = 2 \cdot rac{e^{i heta} + e^{i\phi}}{e^{i heta} - e^{i\phi}}.

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