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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

Bart Michels
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    i am not trying to be funny, but why do you need to prove it? You can easily verify via a calculator. the quality standard of this site requires you make some of effort on the question or at least give the context from which the problem arises. it would make it easier for others to help you. – Lost1 Jan 09 '14 at 02:38
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    Approximate sin and cos by Taylor series. Both $\dfrac\pi7$ and $\dfrac\pi{14}$ are small enough for their fourth and fifth powers to be irrelevant. – Lucian Jan 09 '14 at 04:24

2 Answers2

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As suggested by Lucian, use Taylor series for each piece and expand around zero. Then
$$ 4\sin^2\left(x\right) + \frac{1}{4\cos^2\left(2x\right)} $$

gives 1/4 + 5 x^2 + (4 x^4)/3 + (56 x^6)/9 + (3964 x^8)/315 + ...

Replacing x by Pi/7 and taking into account that Pi^2 is almost 10, you notice that terms above x^4 are perfectly negligible. So, for x=Pi/7, you end with 1/4 + 5 Pi^2/196 + Pi^4/28812. Still assuming that Pi^2 is almost 10, the value you obtain is 14653/28812 which is 0.508573 which is in your bounds (0.5 , 0.585786). If you do not approximate Pi^2 by 10, you would obtain 0.505156

The exact value of your expression is 0.506041

Claude Leibovici
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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 $0.506$.

Jack D'Aurizio
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