Complex quintic equation Given the equation $x^5=i$, I need to show by both algebraic and trigonometrical approaches that
$$\cos18^{\circ}=\frac{\sqrt{5+2\sqrt5}}{\sqrt[5]{176+80\sqrt5}}$$
$$\sin18^{\circ}=\dfrac1{\sqrt[5]{176+80\sqrt5}}$$
Trying by trigonometric approach, 
$x^5$ = i     $\;\;\;\;$ -- eqn. (a)
=> x = $i\sin(\dfrac{\pi}{2} +2k\pi)$ => $i\sin(\pi\dfrac{4k + 1}{2}) $
Taking the value of k=0, for getting the principal root of 18$^{\circ}$, have x = $i\sin(\dfrac{\pi}{10}) $
Solving algebraically, the solution approach is : $(a+bi)^5$ = i $\;\;\;\;$ -- eqn. (b)
=> $a^5 + 5ia^4b -10a^3b^2 -10ia^2b^3 +5ab^4 +ib^5$
Separating the real & imaginary parts:
$a^5 -10a^3b^2 +5ab^4=0$$\;\;\;$ -- eqn. (c); $\;\;\;\;$$5a^4b -10ia^2b^3+b^5=1$$\;\;\;$ -- eqn. (d)
Solving (c), we have : $a(a^4 -10a^2b^2 +5b^4)=0$$\;\;\;$ -- eqn. (c);
Either $a$ = $0$, or $(a^4 -10a^2b^2 +5b^4)=0$$\;\;$ -- eqn. (c'), 
dividing both sides by $b^4$, and having c = a/b,  $(c^4 -10c^2 +5)=0$$\;\;$ -- eqn. (c''),
having d = $c^2$, get : $(d^2 -10d +5)=0$$\;\;$ -- eqn. (c'''), with factors as : d =$5\pm 2\sqrt5$
finding value of c for the two values, get square roots of the two values for d.
//Unable to proceed any further with (c''').
Only root of significance, from eqn. (c) is $a = 0$. 
Taking eqn.(d), and substituting $a = 0$, we get:$\;\;\;b^5$=1 => $b =1$ 
//Unable to prove any of the two values for $\sin18^{\circ}$, or $\cos18^{\circ}$

Let's start with equation $x^{5}-1=0$ whose one root is $x=\cos(2\pi/5)+i\sin(2\pi/5)$. The equation can be written as $$(x-1)(x^{4}+x^{3}+x^{2}+x+1)=0$$ The first factor gives the root $x=1$ and the second factor leads to the equation $$x^{2}+x^{-2}+x+x^{-1}+1=0$$ Putting $y=x+x^{-1}$ we get $$y^{2}+y-1=0$$ or $$y=\frac{-1\pm\sqrt{5}}{2}$$ Choosing the positive root we get $$2\cos(2\pi/5)=y=\frac{\sqrt{5}-1}{2}$$ Thus the value of $\sin 18^{\circ}$ is obtained as $(\sqrt{5}-1)/4$. Similarly we can find the value of $\cos 18^{\circ}$.

Observe that apart from $x=1$ there are $4$ distinct values of $x$ out of which we have to choose only one namely $x=\cos(2\pi/5)+i\sin(2\pi/5)$. The problem of choice is simplified considerably by using $y=x+x^{-1}$ which satisfies a quadratic equation and therefore has only two values. For our desired value of $x$ the expression $y>0$ and hence the positive root $y$ is chosen. And in reality we are interested in the value $\cos(2\pi/5)=(x+x^{-1})/2=y/2$ so the choice of $y$ completes our work.
The method can be generalized (thanks to Gauss) to solve higher degree equations of type $x^{n} =1$. For example we can solve $x^{17}=1$ and get the value of $\cos(2\pi/17)$ as $$\frac{-1 + \sqrt{17} + \sqrt{34 - 2 \sqrt{17}} + 2\sqrt{17 + 3 \sqrt{17} - \sqrt{34 - 2 \sqrt{17}} - 2\sqrt{34 + 2 \sqrt{17}}}}{16}$$ (see this post for more details).
