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$$\;\;$ -- eqn. (c'''), with factors as : d =$5\pm 2\sqrt5$
finding value of c for the two values, get square roots of the two values for d.
//Unable to proceed any further with (c''').
Only root of significance, from eqn. (c) is $a = 0$. 
Taking eqn.(d), and substituting $a = 0$, we get:$\;\;\;b^5$=1 => $b =1$ 
//Unable to prove any of the two values for $\sin18^{\circ}$, or $\cos18^{\circ}$
 A: Let's start with equation $x^{5}-1=0$ whose one root is $x=\cos(2\pi/5)+i\sin(2\pi/5)$. The equation can be written as $$(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).
A: 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.
A: 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 $\sin 18^\circ = \frac{-1+\sqrt{5}}{4}$.
A: 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.
