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

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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 $$1 - 4s_1 - 4s_1^2 + 8s_1^3 = 0.$$

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oh,Thank you my frend. – math110 Mar 31 '13 at 1:37
+1. Where can I see a proof of the first fact you mentioned and other similar results ? – Amr Mar 31 '13 at 12:03
@Amr: See the second half of my answer. – TMM Mar 31 '13 at 12:40

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}

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Then what?..... – Ron Gordon Mar 30 '13 at 18:19
@RonGordon: Please review the answer as it stands now :-) – Abhijit Mar 31 '13 at 8:54
Good job (+1).. – Ron Gordon Mar 31 '13 at 11:25
@RonGordon: Part help from OP though (collaborative effort :-) – Abhijit Mar 31 '13 at 11:28