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# Trigeometry Rediscovered!

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,

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

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