Prove that $ \frac12 < 4\sin^2\left(\frac{\pi}{14}\right) + \frac{1}{4\cos^2\left(\frac{\pi}{7}\right)} < 2 - \sqrt{2} $ I can't figure out how to prove the following inequality:
$$
1/2 < 4\sin^2\left(\frac{\pi}{14}\right) + \frac{1}{4\cos^2\left(\frac{\pi}{7}\right)} < 2 - \sqrt{2}
$$
Thanks

Since $\cos\frac{\pi}{7}$ is a root of the Chebyshev polynomial $U_6(x)$
$$U_6(x) = 64x^6-80x^4+24x^2-1$$
we have
$$\frac{1}{4\cos^2\frac{\pi}{7}}=16\cos^4\frac{\pi}{7}-20\cos^2\frac{\pi}{7}+6,$$
so:
$$4\sin^2\frac{\pi}{14}+\frac{1}{\cos^2\frac{\pi}{7}}=4-2\cos\frac{\pi}{7}-2\cos\frac{2\pi}{7}+2\cos\frac{4\pi}{7},$$
or:
$$4\sin^2\frac{\pi}{14}+\frac{1}{4\cos^2\frac{\pi}{7}}=4-2\sum_{j=1}^{3}\cos\frac{\pi j}{7}=5-2\cdot\frac{\sin\frac{3\pi}{7}}{\sin\frac{\pi}{7}}=7-8\cos^2\frac{\pi}{7}=3-4\cos\frac{2\pi}{7}.$$
Now $\cos\frac{2\pi}{7}$ is a root of the third-degree polynomial
$$p(x)=U_6\left(\sqrt{\frac{x+1}{2}}\right)=8x^3+4x^2-4x-1,$$
so $3-4\cos\frac{2\pi}{7}$ is a root of 
$$q(x)=-8\cdot p\left(\frac{3-x}{4}\right)=x^3-11x^2+31x-13.$$
Now it is easy to check that $q(1/2)=-1/8<0$ and $q(\xi)>0$, where $\xi$ is the smallest positive root of $11x^2-31x+13$. Since $3-4\cos\frac{4\pi}{7}$ and $3-4\cos\frac{6\pi}{7}$, the other roots of $q(x)$, are bigger than $3$, and all the roots of $q(x)$ are positive, we have found:
$$4\sin^2\frac{\pi}{14}+\frac{1}{4\cos^2\frac{\pi}{7}}=3-4\cos\frac{2\pi}{7}\in\left(\frac{1}{2},\frac{20}{39}\right)=I.$$
Since $q(x)$ concave over $I$, we can further improve this bound with a step of the Newton's method with starting point $x=1/2$ and a step of the secant method with endpoints $\frac{1}{2},\frac{20}{39}$:
$$3-4\cos\frac{2\pi}{7}\in\left(\frac{42}{83},\frac{63512}{125503}\right).$$
The difference between the upper and lower bound is now just $3.552\cdot 10^{-5}$, so the first four figures of the LHS are $0.506$.
