خماسي أضلاع: الفرق بين النسختين

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[[ملف:Regular Pentagon Using Carlyle Circle.gif|thumb|رسم خماسي باستعمال دوائر كارليل]]
 
==== البرهان على أن cos 36° = <math>\tfrac{1+\sqrt{5}}{4}</math>====
 
:<math>0 = \cos 90^\circ</math>
::<math> = \cos (72^\circ+18^\circ)</math>
::<math> = \cos 72^\circ\cos 18^\circ - \sin 72^\circ\sin 18^\circ</math> (using the [[List of trigonometric identities#Angle sum and difference identities|angle addition formula for cosine]])
::<math> = (2\cos ^2 36^\circ -1)\sqrt{\tfrac{1+\cos 36^\circ}{2}}-2\sin 36^\circ\cos 36^\circ\sqrt{\tfrac{1-\cos 36^\circ}{2}}</math> (using [[List of trigonometric identities#Double-angle, triple-angle, and half-angle formulae|double and half angle formulas]])
 
:Let ''u'' = cos 36°. First, note that 0 < ''u'' < 1 (which will help us simplify as we work). Now,
 
:<math>\begin{align}
0 & {} = (2u^2 -1)\sqrt{\tfrac{1+u}{2}}-2\sqrt{1-u^2}\cdot u\sqrt{\tfrac{1-u}{2}} \\
2\sqrt{1-u^2}\cdot u\sqrt{\tfrac{1-u}{2}} & {} = (2u^2 -1)\sqrt{\tfrac{1+u}{2}} \\
2\sqrt{1+u}\sqrt{1-u}\cdot u\sqrt{1-u} & {} = (2u^2 -1)\sqrt{1+u} \\
2u(1-u) & {} = 2u^2-1 \\
2u-2u^2 & {} = 2u^2-1 \\
0 & {} = 4u^2-2u-1 \\
u & {} = \frac{2+\sqrt{(-2)^2-4(4)(-1)}}{2(4)} \\
u & {} = \frac{1+\sqrt{5}}{4}
\end{align}</math>
 
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