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