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Find the value of sin 36 degree geometrically.

Note by Proff Snape 3 weeks, 6 days ago

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Use a regular pentagon!

As you can see, \(\triangle ABP \sim \triangle ABE.\)

Let \(\overline{AB}=1\) for convenience.

Let \(\overline{BE}=x,\) and you see that \(\overline{BP}=x-1.\)

\(x:1=1:x-1;\) solving this yields \(x=\dfrac{1+\sqrt{5}}{2},~\) and therefore,

\(\cos 36^{\circ}=\dfrac{\overline{BE}}{2\overline{AB}}=\dfrac{1+\sqrt{5}}{4}.\)

\(\sin 36^{\circ}=\dfrac{\sqrt{16-(1+\sqrt{5})^2}}{4}=\dfrac{\sqrt{10-2\sqrt{5}}}{4}.\) – H.M. 유 · 3 weeks, 6 days ago

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TopNewestUse a regular pentagon!

As you can see, \(\triangle ABP \sim \triangle ABE.\)

Let \(\overline{AB}=1\) for convenience.

Let \(\overline{BE}=x,\) and you see that \(\overline{BP}=x-1.\)

\(x:1=1:x-1;\) solving this yields \(x=\dfrac{1+\sqrt{5}}{2},~\) and therefore,

\(\cos 36^{\circ}=\dfrac{\overline{BE}}{2\overline{AB}}=\dfrac{1+\sqrt{5}}{4}.\)

\(\sin 36^{\circ}=\dfrac{\sqrt{16-(1+\sqrt{5})^2}}{4}=\dfrac{\sqrt{10-2\sqrt{5}}}{4}.\) – H.M. 유 · 3 weeks, 6 days ago

Log in to reply