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 $\cos18^{\circ}$

This problem seems a bit wonky to me because there are simpler expressions for these functions and easier ways to prove them, but let's roll with it.
Once you render $c^2=5\pm2\sqrt5$, take its square roots to get four roots
$c=a/b=\pm\sqrt{5\pm2\sqrt5}$
with the $\pm$ signs independent.  Now from DeMoivre's Theorem we know that the complex root we are seeking is $\cos18°+i\sin18°$, so $a/b$ must be the cotangent of $18°$, therefore greater than $1$. Only by choosing both $\pm$ signs as positive can we get that result; $\sqrt{5-2\sqrt5}$ is too small with $2\sqrt5=\sqrt{20}$ lying between $4$ and $5$.
So, $a=(\sqrt{5+2\sqrt5})b$.  Plug that into the imaginary part of $(a+bi)^5$ you computed above and equate this to $1$, which gives $1/b^5=176+80\sqrt5$.  This will lead easily to the final claims.
