How to find the value of $\sin{\dfrac{\pi}{14}}+6\sin^2{\dfrac{\pi}{14}}-8\sin^4{\dfrac{\pi}{14}}$ 
Determine
  $$
\sin\left(\pi \over 14\right) + 6\sin^{2}\left(\pi \over 14\right)
-8\sin^{4}\left(\pi \over 14\right)
$$

My idea: Let $\displaystyle{\sin\left(\pi \over 14\right)} = x$.

A non-trivial result is that $\sin \dfrac{\pi}{14}$ is a root to the equation $^{\color{blue}{(**)}}$ 
$$1 - 4x - 4x^2 + 8x^3 = 0. \tag{1}$$
If you believe this, then you can simply apply polynomial division to obtain 
$$x + 6x^2 - 8x^4 = -\left(x + \frac{1}{2}\right)(8x^3 - 4x^2 - 4x + 1) + \frac{1}{2}.$$
So then, filling in $x = \sin \dfrac{\pi}{14}$ gives you the answer:
$$\boxed{\sin{\dfrac{\pi}{14}}+6\sin^2{\dfrac{\pi}{14}}-8\sin^4{\dfrac{\pi}{14}} = \dfrac{1}{2}.}$$

$^{\color{blue}{(**)}}$ For a proof that $\sin \dfrac{\pi}{14}$ is indeed a root of equation $(1)$, you need some patience and careful bookkeeping. Let us write $c_k = \cos \dfrac{k \pi}{14}$ and $s_k = \sin \dfrac{k \pi}{14}$. Then, using the triple angle formula for the cosine, we have:
$$c_3 = 4 c_1^3 - 3c_1 = 4 c_1 (1 - s_1^2) - 3c_1 = c_1 (1 - 4 s_1^2).$$
On the other hand, using the double angle formula for the sine twice, we also have:
$$c_3 = s_4 = 2s_2 c_2 = 4 s_1 c_1 (1 - 2s_1^2) = c_1 (4 s_1 - 8 s_1^2)).$$
Equating these two expressions and dividing by $c_1 \neq 0$, we get
$$1 - 4s_1^2 = 4s_1 - 8s_1^3.$$
Bringing all terms to one side, this leads to
$$1 - 4s_1 - 4s_1^2 + 8s_1^3 = 0.$$
