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 $\cos18^{\circ}$

I think you're going at this backwards.  Quickly, $\arg \mathrm{i} = \pi/2 = 90^\circ$, so fifth roots of $\mathrm{i}$ are points on the unit circle at small multiples of $18^\circ$.  If the given values of cosine and sine are to be believed, then \begin{align*}
\mathrm{i} &= \left( \cos 18^\circ + \mathrm{i} \sin 18^\circ \right)^5  \\
    &= \left( \frac{\sqrt{5+2\sqrt5}}{\sqrt[5]{176+80\sqrt5}} + \mathrm{i} \frac{1}{\sqrt[5]{176+80\sqrt5}} \right)^5  \text{.}
\end{align*}
Let $x$ be the expression we are raising to the fifth power.  Then, squaring twice, (and leaving the denominator largely unchanged) \begin{align*}
    x^2 &= \frac{2(2 + \sqrt{5}) + 2\mathrm{i}\sqrt{5+2\sqrt{5}}}{(\sqrt[5]{176+80\sqrt{5}})^2}  \text{,}  \\
    x^4 &= \frac{8(2 + \sqrt{5}) + 8\mathrm{i}((2+\sqrt{5})\sqrt{5+2\sqrt{5}})}{(\sqrt[5]{176+80\sqrt{5}})^4}  \text{, and}  \\
    x^5 = x^4 \cdot x &= \frac{16 \mathrm{i}(11+5\sqrt{5})}{176+80\sqrt{5}} = \mathrm{i}  \text{.}
\end{align*}
This confirms that the given values of cosine and sine are a fifth root of $\mathrm{i}$.  The other roots also have unit magnitude and arguments multiples of $360^\circ/5 = 72^\circ$ greater than the given root, so have arguments $90^\circ$, $162^\circ$, $234^\circ$, and $306^\circ$.  We see that $x$ is the only one in the first quadrant, so is the desired root.
For a trigonometry method:
$$ 0 = \cos(90^\circ) = \cos(5 \cdot 18^\circ) = \cos( 2 \cdot 2 \cdot 18^\circ + 18^\circ)  \text{.}  $$
Then use sum of angles and half angle, twice.  Do the same thing to sine.  You'll get a pair of algebraic equations in $\sin 18^\circ$ and $\cos 18^\circ$.  Then check that the given sine and cosine values actually satisfy them.
