The attached figure almost offers a "proof without words." By similar triangles we have $\frac{x}{1}=\frac{x+1}{x}$ , so that $x=\phi$ , the golden section. Bisecting the angle at point $A$ , we see that $\sin(18^o)=\frac{\frac{1}{2}}{\phi}=\frac{1}{\sqrt{5}+1}=\frac{\sqrt{5}-1}{4}$

We can make use of the identities $\sin(3x) = 3\sin(x) - 4\sin^{3}(x)$ and $\cos(2x) = 1 - 2\sin^{2}(x).$ With $x = 18^{\circ}$ and letting $t = \sin(18^{\circ})$ we have that $\sin(54^{\circ}) = \sin(3*18^{\circ}) = 3t - 4t^{3}$ and $\cos(36^{\circ}) = 1 - 2t^{2}.$

But $\sin(54^{\circ}) = \cos(36^{\circ}),$ so

$3t - 4t^{3} = 1 - 2t^{2} \Longrightarrow 4t^{3} - 2t^{2} - 3t + 1 = 0 \Longrightarrow (t - 1)(4t^{2} + 2t - 1) = 0.$

Now clearly $t \ne 1,$ so $t$ must be a root of $4t^{2} + 2t - 1$ , i.e.,

$t = \dfrac{-2 \pm \sqrt{4 + 16}}{8} = \dfrac{-1 \pm \sqrt{5}}{4}.$

Now clearly $t = \sin(18^{\circ}) \gt 0,$ so we must have $\sin(18^{\circ}) = \boxed{\dfrac{-1 + \sqrt{5}}{4}}.$

I used Chebyshev Polynomials .

$\sin(18^\circ ) = \cos(72^\circ )$ . With $T_5 (x)= 16x^5 - 20x^3 + 5x$ which could be easily verified. Letting $x = 72^\circ \Rightarrow 5x = 360^\circ \Rightarrow \cos(5x) = 1$ and let $y = \cos (x)$ :

We have $16y^5 - 20y^3 + 5y = -1$ . By Rational Root Theorem, we can factor out $(y+1)$ , leaving: $16y^4 - 16y^3 -4y^2 + 4y + 1 = (4y^2 - 2y - 1)^2$ . By quadratic formula and taking only the positive root: $\sin(18^\circ ) = \cos(72^\circ ) = \boxed{\frac{ -1 + \sqrt 5}{4 }}$ .

Let x = 18

Then 5 x = 90

Then

sin 2 x = cos 3 x .................. .....................(1)

sin 2 x = 2 sin x cos x ............................. (2)

cos 3 x = cos (2 x + x) = cos 2 x cos x - sin 2 x sin x =

[1 - 2 (sin x)^2] cos x - 2 cos x (sin x)^2

Then

cos 3 x = cos x[1 - 4 (sin x)^2] ............. (3)

Substitute from (2) , (3) in (1)

4 (sin x)^2 + 2 sin x - 1 = 0

Solving for sin x, we get the solution

Consider the equation $x^{5}-1=0$ .

The complex roots of the above equation are the roots of $x^{4}+x^{3}+x^{2}+x+1=0$

Hence, $x^{2}+\frac{1}{x^{2}}+x+\frac{1}{x}+1=0$

Letting, $z=x+\frac{1}{x}$ the equation becomes $z^{2}+z-1=0$ giving $z=\frac{-1_{-}^{+}\sqrt{5}}{2}$

The values of $x$ are $cos(\frac{2rπ}{5})_{-}^{+}isin(\frac{2rπ}{5})$ for $r=1,2$

Hence, the values of $z$ are $2cos(\frac{2π}{5})$ and $2cos(\frac{4π}{5})$

Now, $cos(\frac{4π}{5})=-cos(\frac{π}{5})$ .

Hence, $cos(\frac{π}{5})=cos36(degrees)=\frac{\sqrt{5}+1}{4}$ .

Now using the formula, $cos2A=1-2sin^{2}A$ one can easily find the value of $sin18$

This could be solved geometrically, but I solved it with the small angle approximation. First convert to radians $18^\circ = \frac{18}{180}\pi = \frac{\pi}{10} \\ \approx .314$ $\frac{\pi}{10}$ is a sufficiently small angle, so by the small angle approximation, $\sin\frac{\pi}{10} \approx \frac{\pi}{10}$ Note that $4\times .314 = 1.256 \approx \sqrt{5} - 1.$ So a possible closed form is $\frac{-1 + \sqrt{5}}{4}$

Since $\sin18^\circ=\cos72^\circ$ , we will determine value of $\cos72^\circ$

Set $z=\cos72^\circ+i\sin72^\circ$ , we have then $z+\frac{1}{z}=2\cos72^\circ$

Now, using De Moivre's Therome, we obtain

$z^5=\cos360^\circ+i\sin360^\circ=1\\(z-1)(z^4+z^3+z^2+z+1)=0$

It gives two condition. First, $z-1=0\implies{z=1}$ , for $z=1$ , then $z+\frac{1}{z}=1+1=2\cos72^\circ\text{or}\cos72^\circ=1$ . This result is obviously wrong since $1=\cos0^\circ$ .

Second,

$z^4+z^3+z^2+z+1)=0\\\left(z^2+\dfrac{1}{z^2}\right)+\left(z+\dfrac{1}{z}\right)+1=0\\\left(z+\dfrac{1}{z}\right)^2+\left(z+\dfrac{1}{z}\right)-1=0\\\left(z+\dfrac{1}{z}\right)=\dfrac{-1\pm\sqrt{5}}{2}\\2\cos72^\circ=\dfrac{-1\pm\sqrt{5}}{2}\\\cos72^\circ=\dfrac{-1\pm\sqrt{5}}{4}$ .

We know that $\cos72^\circ$ is positive, then the required answer is $\cos72^\circ=\dfrac{-1+\sqrt{5}}{4}$

Note that $\sin 18^\circ = \sin \frac \pi {10} = \cos \left(\frac \pi 2-\frac \pi {10}\right) = \cos \frac {2\pi}5$ . Using the following identity:

$\begin{aligned} \cos \frac {2\pi}5 + \cos \frac {4\pi}5 & = - \frac 12 \\ \cos \frac {2\pi}5 + 2\cos^2 \frac {2\pi}5 - 1 & = - \frac 12 \\ 4 \cos^2 \frac {2\pi}5 + 2\cos \frac {2\pi}5 - 1 & = 0 \\ \implies \cos \frac {2\pi}5 & = \frac {-2+\sqrt{2^2+16}}8 & \small \color{#3D99F6} \text{Note that }\cos \frac {2\pi}5 > 0 \\ \sin 18^\circ & = \boxed{\dfrac {-1+\sqrt 5}4} \end{aligned}$

Sin 18 always positive ,,,1-√5\4&1-√5\2 are wrong answers,,,,assume a right angle triangle ,assume sin 18=(-1+√5)\4=opposite \hypotenuse, so adjacent side to angle =√(16-(-1+√5)^2)=√(10+2√5),so cos(18)=√(10+2√5)\4,sin^2(18)+cos^2(18) must be =1,sin^2(18)=(-1+√5)^2\16=6-2√5\16,,,cos^2(18)=√(10+2√5)^2\16=10+2√5\16,sin^2(18)+cos^2 (18)=6-2√5+10+2√5\16=16\16=1###-1+√5\4 is the right answer .