Trig^1

Since it is given that $y$ is a rational root, we can use rational root theorem . The possible roots are $y = \pm 1, \pm 2, \pm 4, \pm 8$ and we find that the only root is $y=8$ . Then $\sin \left(\frac \pi y\right) = \sin \frac \pi 8$ . Consider the following:

$\begin{aligned} \cos \frac \pi 4 & = 1 - 2\sin^2 \frac \pi 8 \\ \frac 1{\sqrt 2} & = 1 - 2\sin^2 \frac \pi 8 \\ \sin^2 \frac \pi 8 & = \frac {2-\sqrt 2}4 \\ \implies \sin \frac \pi 8 & = \frac {\sqrt{2-\sqrt 2}}2 & \small \color{#3D99F6} \text{Multiply up and down by }\sqrt{2+\sqrt 2} \\ & = \frac {\sqrt 2}{2\sqrt{2+\sqrt 2}} \\ & = \frac 1{\sqrt 2\sqrt{2+\sqrt 2}} \\ & = \boxed{\frac 1{\sqrt{\sqrt 8+4}}} \end{aligned}$

$x^5 - 8x^4 + 2x^3 - 16x^2 + x - 8$ $(x - 8)(x^2 + 1)^2$ $(x^2 + 1)^2 = x^4 + 2x^2 + 1$ Let $u = x^2$ Then $u^2 + 2u + 1 = 0$ $u = -1$ $x^2 = -1$ Therefore $x = i,-i$ are the roots of this polynomial, so $8$ is the only rational root of the aforementioned polynomial. To find $\sin(\frac{\pi}{8})$ we will make use of the identity $\cos(2x) = \cos^2(x) - \sin^2(x) = 2\cos^2(x) - 1$ Setting $w = 2x$ : $\frac{\cos(w) + 1}{2} = \cos^2(\frac{w}{2})$ ; $\cos(\frac{w}{2}) = \sqrt{\frac{\cos(w) + 1}{2}}$ Next, we will use the identity $\sin(2x) = 2\cos(x)\sin(x)$ Setting $z = 2x$ : $\sin(z) = 2\cos(\frac{z}{2})\sin(\frac{z}{2})$ ; $\sin(z) = 2 \sqrt{\frac{\cos(z) + 1}{2}}\sin(\frac{z}{2})$ ; $\sin(\frac{z}{2}) = \frac{\sin(z)}{ 2\sqrt{\frac{\cos(z) + 1}{2}}}$ ; $\sin(\frac{z}{2}) = \frac{\sin(z)}{ \sqrt{2\cos(z) + 2}}$ ; Substituting $z = \frac{\pi}{4}$ : $\sin(\frac{\pi}{8}) = \frac{\sin(\frac{\pi}{4})}{ \sqrt{2\cos(\frac{\pi}{4}) + 2}} = \frac{\sqrt{2}}{2\sqrt{\sqrt{2} + 2}} = \frac{1}{\sqrt{2}\sqrt{\sqrt{2} + 2}} = \frac{1}{\sqrt{\sqrt{8} + 4}}$