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Simplify the following expression: $$y =\sec^2 \frac{2\pi}{7} + \sec^2 \frac{4\pi}{7} + \sec^2 \frac{8\pi}{7}$$

Note. The source asks the value of $y/3$, which, according to the instructions, has to be an integer from $0$ to $9$.

My Attempt:

Man! I tried everything I could. Converted it into sin's and cosine's, tan's and sec's. Applied every identity I could find in my book. But to no avail.

All I was aiming is to somehow make the squares disappear and converting everything in sin's and cosine's since we only know to simplify such expressions like ($\cos x + \cos 2x + \cos 3x+\cdots$ and $\cos x\cos 2x\cos 4x\cdots$ etc) in tems of sin's and cosine's. (Sorry for all the sin's and cosine's being repeated too many times :) )

Any help would be appreciated.

Blue
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Tony
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1 Answers1

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Like If $\alpha = \frac{2\pi}{7}$ then the find the value of $\tan\alpha .\tan2\alpha +\tan2\alpha \tan4\alpha +\tan4\alpha \tan\alpha.$,

the roots of $$t^3-21t^2+35t-7=0$$ are $\tan^2\dfrac{2n\pi}7, n=1,2,4$

Using Vieta's Formula,

$$\tan^2\dfrac{2\pi}7+\tan^2\dfrac{4\pi}7+\tan^2\dfrac{8\pi}7=\dfrac{21}1$$

lab bhattacharjee
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  • So, how did you guess that this cubic ($x^3 - 21x^2 + 35x -7$) has the given roots ($\tan^2 \frac{2n\pi}{7}, n = 1,2,4$)? – Tony May 04 '19 at 15:30
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    @Tony, Observe from https://math.stackexchange.com/questions/823819/if-alpha-frac2-pi7-then-the-find-the-value-of-tan-alpha-tan2-alpha/824178#824178, the roots of $$x^6-21x^4+35x^2-7=0$$ are $\tan\dfrac{n\pi}7, n\equiv\pm1,\pm2,\pm3\pmod7$ and $$\tan(\pi\pm y)=\pm\tan y$$ So, $\tan\dfrac{8\pi}7=\tan\dfrac\pi7$ – lab bhattacharjee May 04 '19 at 15:33