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How do I solve $\sin(\frac{\pi}{5})$ analytically? I looked up here and in the first step I have to

Show that: $\cos \left(\frac{\pi }{5}\right)-\sin \left(\frac{\pi }{10}\right)=\frac{1}{2}$

My question is, why do I have to do that?

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By repeated application of angle sum formulas we may get,

$$\sin (5x)=\sin^5 x+5 \cos^4 x\sin x-10 \sin^3 x \cos^2 x$$

Let $x=\frac{\pi}{5}$ and let $\sin (\frac{\pi}{5})=u$ then we have,

$$0=u^5+5(1-u^2)^2 u-10(1-u^2)u^3$$

It is safe to say $\frac{\sqrt{2}}{2}>u>0$. So that we may divide by $u$ to get.

$$0=u^4+5(1-u^2)^2-10(1-u^2)u^2$$

$$0=16u^4-20u^2+5$$

By solving this for $u^2$ first and then $u$ you get the only root in $(0, \frac{\sqrt{2}}{2})$ to be,

$$u=\frac{1}{2} \sqrt{\frac{5}{2}-\frac{\sqrt{5}}{2}}$$

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    $\begingroup$ +1 for most generally applicable method. ( I was just about to write the same ). $\endgroup$ – mathreadler Apr 23 '17 at 17:55
  • $\begingroup$ Why do you apply the angle of sump formula repeatedly? Could you explain your first step? $\endgroup$ – Mark Read Apr 25 '17 at 0:23
  • $\begingroup$ I know to repeat it repeatedly because it's leave me with a polynomial in terms of $\sin x$ and $\cos x$, which if I'm lucky am able to solve. First $\sin(x+4x)=\sin x \cos 4x+\cos x \sin 4x$. Then $\sin (4x)=2\sin (2x) \cos (2x)$ etc etc. $\endgroup$ – Ahmed S. Attaalla Apr 25 '17 at 0:54
  • $\begingroup$ I'm still trying to understand your answer. Why do you use $\sin(5x)$ on your first step? I assume $x$ is $\frac{\pi}{5}$, so does your method work because $\sin(x)$ and $\sin(kx)$, k being an integer, are equivalent? $\endgroup$ – Mark Read Apr 28 '17 at 2:31
  • $\begingroup$ Yes $x$ is $\frac{\pi}{5}$ and I know then that $\sin (5x)=\sin (5\frac{\pi}{5})=\sin (\pi)=0$. $\endgroup$ – Ahmed S. Attaalla Apr 28 '17 at 3:13
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A simple way is provided by a geometric approach, rather than an analytic one.

It is well-known that if $ABCDE$ is a regular pentagon, $\frac{AC}{AB}=\varphi=\frac{1+\sqrt{5}}{2}$.
Since $\widehat{ABC}=\frac{3\pi}{5}$, that implies $\sin\frac{3\pi}{10}=\frac{\varphi}{2}$ and $$ \cos\frac{3\pi}{10}=\sin\frac{\pi}{5}=\sqrt{1-\left(\frac{\varphi}{2}\right)^2} = \color{red}{\sqrt{\frac{5-\sqrt{5}}{8}}}.$$

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