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Assume that we have been asked to find the value of $\sin (18^\circ)$. We know that there are many ways to find it out. However, I'll be going with golden ratio! Let's draw a triangle whose apical angle is $36^{\circ}$. Note that this is an isosceles triangle, otherwise we couldn't apply it.

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Consider $|BC| = 2$, from property of height, we have that $|BD| = 1$ and $|DC| = 1$. Hence I got a right triangle whose one angle is $18^{\circ}$. Now I almot found it.

$$\sin (18^{\circ}) = \dfrac{1}{2\varphi} = \dfrac{1}{\sqrt 5 + 1} = \dfrac{\sqrt 5 - 1}{4} $$

where $\varphi = \dfrac{1+\sqrt 5}{2}$.


Now consider we have been asked to find the value of $\cos (36^{\circ})$. Here I've to draw a triangle whose base angles are $36^{\circ}$.

enter image description here

From the triangle, we have that

$$\cos 36^{\circ} = \dfrac{\varphi}{2} = \dfrac{1+ \sqrt 5}{4}$$


Then what about $\sin (28^{\circ})$? I mean can we apply it for other trigonometric values?

Regards!

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    $\begingroup$ How did you determine $\phi$ ? $\endgroup$ – Nehal Samee Jun 8 '18 at 11:03
  • $\begingroup$ @NehalSamee I'm editing it now. $\endgroup$ – Busi Jun 8 '18 at 11:03
  • $\begingroup$ @projectilemotion My apologies, I had thought that it doesn't work. $\endgroup$ – Busi Jun 8 '18 at 11:05
  • $\begingroup$ It happens that $\cos\pi/5$ is half the golden ratio (but you have to prove that, here you didn't prove anything), and thus the golden ratio can be linked with some angles of the form $k\pi/(5\cdot2^p)$. (that's why the golden ratio appears in many places in a pentagon). However, you can't apply this to other angles in general. And most angles of the form $\pi/n$ do not lead to trigonometric functions with nice expressions with radicals. So basically, this is an exceptional situation. $\endgroup$ – Jean-Claude Arbaut Jun 8 '18 at 11:16
  • $\begingroup$ @Jean-ClaudeArbaut In pentagon, all the triangles you can see are favorable to apply golden ratio theorem. As stated in my perspective, there should be a way to find other trigonometric values. $\endgroup$ – Busi Jun 8 '18 at 11:17

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