<s>[INST] 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
 [/INST] 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 
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