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**The value of $$\sec\frac{\pi}{11}-\sec\frac{2\pi}{11}+\sec\frac{3\pi}{11}-\sec\frac{4\pi}{11}+\sec\frac{5\pi}{11}$$ is ...

My Approach

I used the fact that $$\sec (\pi-x)=-\sec x$$ to simplify the equation to $$\sec\frac{\pi}{11}+\sec\frac{3\pi}{11}+\sec\frac{5\pi}{11}+\sec\frac{7\pi}{11}+\sec\frac{9\pi}{11}$$

Now I tried to devise an equation whose roots are $$\sec\frac{\pi}{11}, \sec\frac{3\pi}{11}, \sec\frac{5\pi}{11}, \sec\frac{7\pi}{11}, \sec\frac{9\pi}{11}$$

Afterwards, I found that the equation $$\cos \frac{11x}{2}=0 $$ satisfy the condition. But the equation has infinite number of roots, so my plan to use Vieta's formula to calculate the required sum did not work. Please suggest how to proceed in this problem or share any other method.

Prime Mover
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S.M.
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  • You might want to investigate an approach using the complex 11th roots of unity (or minus unity) and expand the power series, and separate out real and imaginary parts. That usually works when you get one of these things in cosines and sines, I haven't tried playing around with it to see if it works for secants. – Prime Mover May 13 '21 at 15:04
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    @S.M.: You didn't find an equation whose roots are the secants, $\sec(\text{odd},\pi/11)$. You found an equation whose roots are the angles, $(\text{odd},\pi)/11$ (along with infinitely-many period-adjusted values). So, a Vieta-like analysis —whatever that might entail— wouldn't help you find the sum of the secants. ... That said, the minimum polynomial of $\sec(\pi/11)$ (ie, the smallest polynomial with the value as a root) is $-32+16x+32x^2-12x^3-6x^4+x^5$. This happens also have those other secants as roots, so that run-of-the-mill Vieta does indeed tell you the desired sum. – Blue May 13 '21 at 15:24
  • @Blue: I got my mistake, tysm. But how did you came up with the polynomial, please explain? – S.M. May 13 '21 at 15:31
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    @S.M.: "[H]ow did you came up with the polynomial [...]?" I used Mathematica*. :) ... To find it "by hand" typically isn't unlike what you did: find a trig equation solved by the angles; your $\cos(11 x/2)=0$ works. Then, the tedious part: expand $\cos(11 x/2)$ into a polynomial in terms of $\cos x$. (It's a little tricky in this case. You'd first expand in terms of $\cos(x/2)$, discard a factor of $\cos(x/2)$, then rewrite even powers of $\sin(x/2)$ & $\cos(x/2)$ in terms of $\cos x$.) Finally, replace $\cos x$ with $1/\sec x$ and clear denominators to get a polynomial in $\sec x$. Not fun! – Blue May 13 '21 at 15:55

2 Answers2

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Observe that $\cos(2n+1)\pi=-1$ for any integer $n$

If $11x=(2n+1)\pi,\cos6x=-\cos5x$

With $\cos x=c,$

$$\cos6x=2\cos^23x-1=2(4c^3-3c)^2-1=?$$

Using Prosthaphaeresis Formulas, $$\cos5x+\cos x=2\cos3x\cos2x=2(4c^3-3c)(2c^2-1)$$ to find

$$0=\cos6x+\cos5x=32c^6+16c^5-48c^4-20c^3+18c^2+5c-1$$

So, the roots of $$32c^6+16c^5-48c^4-20c^3+18c^2+5c-1=0$$ are $c_n=\cos\dfrac{(2n+1)\pi}{11},0\le n\le 5$

But $c_5=-1$

So, the roots of $$32c^5-16c^4-32c^3+12c^2+6c-1=0$$ are $c_n=\cos\dfrac{(2n+1)\pi}{11},0\le n\le 4$

Set $\dfrac1c=s$ to find the roots of $$s^5-6s^4+\cdots-32=0$$ to be $s_n=\sec\dfrac{(2n+1)\pi}{11},0\le n\le 4$

Can you take it from here?

lab bhattacharjee
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Let $\alpha = \dfrac{(2n+1)\pi}{11}$, which is the form of angles we are interested. Then $6\alpha = (2n+1)\pi - 5\alpha$ which implies $\cos 6\alpha +\cos 5\alpha = 0$. Now, expressing this in terms of $\cos \alpha$, we observe $$32\cos^6\alpha + 16\cos^5\alpha- 48\cos^4\alpha - 20\cos^3\alpha + 18\cos^2\alpha + 5\cos\alpha - 1 = 0$$ From this, we conclude that the equation $$32x^6 + 16x^5-48x^4-20x^3+18x^2+5x-1 = 0$$ has the roots of exactly same angles that you want (with $\cos$) and $1$, so divide by $(x-1)$ and substitute $1/x$ to get $\sec$ and use Vieta's formula.

Note: We partitioned $11\alpha$ to $5\alpha$ and $6\alpha$ since we needed a polynomial with (possibly exactly) $5$ roots (and $\lfloor 11 / 2 \rfloor = 5$).

VIVID
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