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Home » First, simplify the equation of the ellipse: - Veritas Home Health

First, simplify the equation of the ellipse: - Veritas Home Health

📅 March 1, 2026 👤 scraface

Mar 01, 2026

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📌 Somit ist der Umfang \(\boxed{8\sqrt{2}\pi}\) Sekundenlängen.

📌 Frage: Berechne \(\tan 75^\circ\) und gib deine Antwort in einfachster Radikalform.

📌 Um \(\tan 75^\circ\) zu berechnen, verwenden wir die Additionstheorem für Tangens:

📌 \tan(75^\circ) = \tan(45^\circ + 30^\circ) = \frac{\tan 45^\circ + \tan 30^\circ}{1 - \tan 45^\circ \tan 30^\circ}.

📌 Wir wissen, dass \(\tan 45^\circ = 1\) und \(\tan 30^\circ = \frac{1}{\sqrt{3}}\). Einsetzen:

📌 \tan 75^\circ = \frac{1 + \frac{1}{\sqrt{3}}}{1 - 1 \cdot \frac{1}{\sqrt{3}}} = \frac{\frac{\sqrt{3} + 1}{\sqrt{3}}}{\frac{\sqrt{3} - 1}{\sqrt{3}}} = \frac{\sqrt{3} + 1}{\sqrt{3} - 1}.

📌 Rationalisieren des Nenners:

📌 \frac{\sqrt{3} + 1}{\sqrt{3} - 1} \cdot \frac{\sqrt{3} + 1}{\sqrt{3} + 1} = \frac{(\sqrt{3} + 1)^2}{(\sqrt{3})^2 - (1)^2} = \frac{3 + 2\sqrt{3} + 1}{3 - 1} = \frac{4 + 2\sqrt{3}}{2} = 2 + \sqrt{3}.

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ight)^2 = rac{1}{2} - rac{1}{2} \cdot rac{1}{4} = rac{1}{2} - rac...
b_4 = F\left(rac{3}{8}
ight) = rac{3}{8} - rac{1}{2} \left(rac{3}{8}
ight)^2 = rac{3}{8} - rac{1}{2} \cdot rac{9}{64} = rac{3}{8} - ra...
b_5 = F\left(rac{39}{128}
ight) = rac{39}{128} - rac{1}{2} \left(rac{39}{128}
ight)^2 = rac{39}{128} - rac{1}{2} \cdot rac{1521}{16384} = rac{39...
Convert $ rac{39}{128} $ to denominator 32768:
rac{39}{128} = rac{39 \cdot 256}{128 \cdot 256} = rac{9984}{32768}
Now subtract:

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