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: 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
$$1 - 4s_1 - 4s_1^2 + 8s_1^3 = 0.$$
A: Try using the Identities
$$\sin^4\theta = \dfrac{3-4\cos2\theta +\cos 4\theta}{8}$$
$$\sin^2\theta = \dfrac{1-\cos2\theta}{2}$$
$$\sin{\dfrac{\pi}{14}}+6\sin^2{\dfrac{\pi}{14}}-8\sin^4{\dfrac{\pi}{14}}$$
Let $\theta = \dfrac{\pi}{14}$
$$\sin{\theta}+6\dfrac{1-\cos2\theta}{2}-8\dfrac{3-4\cos2\theta +\cos 4\theta}{8}$$
$$\sin{\theta}+3 - 3\cos 2\theta-3+4\cos2\theta-\cos 4\theta$$
$$=\sin{\theta}+\cos 2\theta-\cos4\theta\tag1$$
Replacing $\theta$ with $\dfrac{\pi}{14}$ in $(1)$ we get
$$= \cos\left(\dfrac{\pi}{7}\right) - \cos\left(\dfrac{2\pi}{7}\right) + \cos\left(\dfrac{\pi}{2} - \dfrac{\pi}{14}\right)$$
$$= \cos\left(\dfrac{\pi}{7}\right) - \cos(\dfrac{2\pi}{7}) + \cos\left(\dfrac{3\pi}{7}\right)$$
$$= \cos\left(\dfrac{\pi}{7}\right) + \cos\left(\dfrac{3\pi}{7}\right)+ \cos\left(\dfrac{5\pi}{7}\right)=\dfrac{1}{2}$$
\begin{cases}
Known\\
\sum_{i=0}^{n-1} \cos((2i+1)x) = \dfrac{\sin2nx}{2\sin x}\\
\text{for n = 3} \\
\sum_{i=0}^{2} \cos((2i+1)x) = \dfrac{\sin2\cdot 3 \cdot x}{2\sin x}\\
= \dfrac{\sin6x}{2\sin x} \\
\text{Let } x = \dfrac{\pi}{7}\\
= \dfrac{\sin \dfrac{6\pi}{7}}{2\sin \dfrac{\pi}{7}} = \dfrac{\sin (\pi -\dfrac{6\pi}{7})}{2\sin \dfrac{\pi}{7}} = \dfrac{\sin (\dfrac{\pi}{7})}{2\sin \dfrac{\pi}{7}}\\
=\dfrac{1}{2}
\end{cases}
