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 $\cos18^{\circ}$

Trigonometrical way: One has 
$$\sin 36^\circ = \cos 54^\circ = 4\cos^3 18^\circ - 3\cos 18^\circ$$
$$2\sin 18^\circ\cos 18^\circ = 4\cos^3 18^\circ - 3\cos 18^\circ$$
$$2\sin 18^\circ = 4\cos 18^\circ -3 = 1-4\sin 18^\circ$$
So, $\sin 18^\circ$ is positive root of equation $4x^2+2x-1=0$, or $\sin 18^\circ = \frac{-1+\sqrt{5}}{4}$.
