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If $A= 2 \times \pi/7$ then how to show, $$\sec A+ \sec 2A+ \sec 4A=-4$$ I have tried using formula for $\cos 2A$ but I failed.

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$$\sec\frac{2\pi}{7}+\sec\frac{4\pi}{7}+\sec\frac{8\pi}{7}=\frac{1}{\cos\frac{2\pi}{7}}+\frac{1}{\cos\frac{4\pi}{7}}+\frac{1}{\cos\frac{8\pi}{7}}=$$ $$=\frac{\cos\frac{2\pi}{7}\cos\frac{4\pi}{7}+\cos\frac{2\pi}{7}\cos\frac{8\pi}{7}+\cos\frac{4\pi}{7}\cos\frac{8\pi}{7}}{\cos\frac{2\pi}{7}\cos\frac{4\pi}{7}\cos\frac{8\pi}{7}}=$$ $$=\frac{\cos\frac{6\pi}{7}+\cos\frac{2\pi}{7}+\cos\frac{4\pi}{7}+\cos\frac{6\pi}{7}+\cos\frac{2\pi}{7}+\cos\frac{4\pi}{7}}{2\cos\frac{2\pi}{7}\cos\frac{4\pi}{7}\cos\frac{8\pi}{7}}=$$ $$=\frac{\cos\frac{2\pi}{7}+\cos\frac{4\pi}{7}+\cos\frac{6\pi}{7}}{\cos\frac{2\pi}{7}\cos\frac{4\pi}{7}\cos\frac{8\pi}{7}}=\frac{2\sin\frac{\pi}{7}\left(\cos\frac{2\pi}{7}+\cos\frac{4\pi}{7}+\cos\frac{6\pi}{7}\right)\cdot8\cos\frac{\pi}{7}}{8\sin\frac{2\pi}{7}\cos\frac{2\pi}{7}\cos\frac{4\pi}{7}\cos\frac{8\pi}{7}}=$$ $$=\frac{\left(\sin\frac{3\pi}{7}-\sin\frac{\pi}{7}+\sin\frac{5\pi}{7}-\sin\frac{3\pi}{7}+\sin\frac{7\pi}{7}-\sin\frac{5\pi}{7}\right)\cdot8\cos\frac{\pi}{7}}{\sin\frac{16\pi}{7}}=$$ $$=\frac{-\sin\frac{\pi}{7}\cdot8\cos\frac{\pi}{7}}{\sin\frac{2\pi}{7}}=-4.$$ Done!

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  • $\begingroup$ Hi. I'm confused about the third line how does exactly $\cos\frac{2\pi}{7}\cos\frac{8\pi}{7}=\cos\frac{4\pi}{7}+\cos\frac{6\pi}{7}$? Shouldn't be $\cos\frac{2\pi}{7}\cos\frac{8\pi}{7}=\cos\frac{10\pi}{7}+\cos\frac{6\pi}{7}$? Perhaps can you offer some clarification? $\endgroup$ – Chris Steinbeck Bell Jul 29 '19 at 18:35
  • $\begingroup$ @Chris Steinbeck Bell I think you missed $2$ in the denominator. I used $\cos{x}\cos{y}=\frac{1}{2}(\cos(x+y)+\cos(x-y)).$ $\endgroup$ – Michael Rozenberg Jul 29 '19 at 18:37
  • $\begingroup$ I missed in my earlier comment that I ommited $2$ for brevity I could notice that it was a prosthaphaeresis formula. $\endgroup$ – Chris Steinbeck Bell Jul 29 '19 at 18:59
  • $\begingroup$ But I could not find a way to transform $\cos\frac{10\pi}{7}=\cos\frac{4\pi}{7}$. My best guess was that to ''obtain'' that desired angle in two adjacent quadrants for a cosine you must subtract $\pi$ from that angle. I did obtained that from obtaining the 'distance' between $\frac{3\pi}{2}$ and $\frac{10\pi}{7}$ and sum that to $\frac{\pi}{2}$ which will yield $\frac{\pi}{2}+\left(\frac{3\pi}{2}-\frac{10\pi}{7}\right)=\frac{4\pi}{2}-\frac{10\pi}{7}=\frac{28-20}{14}$. And this will become into $\frac{8\pi}{14}=\frac{4\pi}{7}$. I couldn't find an easier method though. $\endgroup$ – Chris Steinbeck Bell Jul 29 '19 at 19:08
  • $\begingroup$ The method which I described above I could replicate it for the sine function but again. Does it exist an easier method?. $\endgroup$ – Chris Steinbeck Bell Jul 29 '19 at 19:09
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The equation: $\displaystyle \large \cos{4\theta} = \cos{3\theta}$

has distinct roots $\displaystyle \large \theta = 0,\frac{2\pi}{7},\frac{4\pi}{7},\frac{8\pi}{7}$

Expanding the equation into a polynomial in terms of $\displaystyle \large c = \cos{\theta}$, using multiple angle identities, the equation becomes:

$\displaystyle \large 8c^4 - 4c^3 + 3c + 1 = 0$

By inspection, or otherwise, the irreducible factorisation of the polynomial is:

$\displaystyle \large (8c^3 + 4c^2 - 4c - 1)(c-1) = 0$

We only focus on the cubic, as the linear factor provides no meaningful information to this problem.

If the roots of the cubic are $\displaystyle \large \alpha, \beta, \gamma$, the roots we are after are $\displaystyle \large \frac{1}{\alpha}, \frac{1}{\beta}, \frac{1}{\gamma}$

By simple transformation of the cubic through an invertible substitution, the polynomial equation with the desired roots is:

$\displaystyle \large c^3 + 4c^2 - 4c - 8 = 0$

Hence, by Vieta's Formulae for Sums and Products of roots, the value of the desired sum is $\displaystyle \large -4$

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  • $\begingroup$ Very neat, @Jack Lam. I'm wondering whether one can do in the same way the corresponding result for $31=32-1$ in place of $7=8-1$. $\endgroup$ – ancientmathematician Apr 24 '17 at 13:41
  • $\begingroup$ I think the relationship is derived from the number of distinct, non-trivial roots of the generalised trigonometric equation $\displaystyle \large \cos{m\theta} = \cos{n\theta}$ for distinct $\displaystyle \large m,n$... And of course, the exact form of said roots..... $\endgroup$ – Jack Tiger Lam Apr 24 '17 at 13:44
  • $\begingroup$ Yes @Jack. I myself would have done it via Demoivre, which I guess the OP doesn't know, and it's reasonably easy to get (in the case here) the cubic satisfied by the three cosines without remembering any multiple angle formulas. It seemed to me that your way somehow captured the same info. $\endgroup$ – ancientmathematician Apr 24 '17 at 14:50
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$$\frac1{\cos a}+\frac1{\cos2a}+\frac1{\cos4a}=\frac{\cos2a\cos4a+\cos a\cos4a+\cos a\cos2a}{\cos a\cos2a\cos4a}=$$

$$=\frac12\frac{\cos6a+\cos2a+\cos5a+\cos3a+\cos3a+\cos a}{\frac12\left[\cos a+\cos3a\right]\cos4a}=$$

$$2\frac{\cos a+\cos2a+2\cos3a+\cos5a+\cos6a}{\cos5a+\cos3a+\cos7a+\cos a}\;(*)$$

But $\;\cos6a=\cos a\,,\,\,\cos5a=\cos2a\,,\,\,\cos7a=1\;$ , so

$$(*)=2\frac{2\left(\cos a+\cos 2a+\cos3a\right)}{1+\cos a+\cos2a+\cos3a}\;(***)$$

And

$$\sum_{n=1}^N\cos nt=-\frac12+\frac{\sin\left(N+\frac12\right)t}{2\sin\frac t2}\implies\cos a+\cos 2a+\cos3a=-\frac12+\frac{\sin\frac72a}{2\sin\frac a2}=$$$${}$$

$$=-\frac12+\frac{0}{2\sin\frac\pi7}=-\frac12$$

so finally

$$(***)=4\frac{-\frac12}{1-\frac12}=-4$$

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  • $\begingroup$ Can it be solved without using that summation formula? Or can you explain that part? $\endgroup$ – Fawad Apr 24 '17 at 12:48
  • $\begingroup$ The easiest way to prove that summation formula is by means of Euler's Formula for complex numbers and geometric series. $\endgroup$ – DonAntonio Apr 24 '17 at 13:28
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Let us consider $\Phi_7(x)=\frac{x^7-1}{x-1}$. It is a palyndromic polynomial, hence $\frac{\Phi_7(x)}{x^3}$ can be written as a polynomial of $x+\frac{1}{x}$. Given that $\Phi_7(x)$ vanishes at the primitive seventh roots of unity, we get a polynomial that vanishes at $\left\{\cos\frac{2\pi}{7},\cos\frac{4\pi}{7},\cos\frac{6\pi}{7}\right\}=\left\{\cos\frac{2\pi}{7},\cos\frac{4\pi}{7},\cos\frac{8\pi}{7}\right\}=A$: $$ f(x)=8x^3+4x^2-4x-1 = 8\prod_{\zeta\in A}(x-\zeta) \tag{1}$$ from which: $$ \frac{f'(x)}{f(x)} = \frac{d}{dx}\log f(x) =\sum_{\zeta\in A}\frac{1}{x-\zeta}\tag{2}$$ and: $$ \sum_{\zeta\in A}\frac{1}{\zeta} = -\frac{f'(0)}{f(0)} = -\frac{-4}{-1} = \color{red}{-4} \tag{3}$$ as wanted.

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First of all we can write your equation as $$1/cosA + 1/cos2A + 1/ cos4A$$ Using basic mathematics we can write $$\frac{cosA cos2A + cos2A cos 4A + cos4A cosA}{cosAcos2Acos4A}$$ We will further try to tackle with the denominator using the trig identity $$cosA \cdot cos2A \cdot cos4A \cdot......cos2^{n-1}A = \frac{sin(2^nA)}{2^n(sinA)}$$ which in your case becomes $$\Rightarrow \frac{sin(2^3A)}{2^3(sinA)}$$ $$\rightarrow \frac{sin(16 \pi/7 )}{2^3(sin2\pi/7)}$$

We can further write $sin(16 \pi/7)=sin(2 \pi+ 2\pi/7)=sin2\pi/7$

Thus finally $$\rightarrow \frac{sin(16 \pi/7 )}{2^3(sin2\pi/7)}= 1/8$$

Again rewriting your function $$8(cosA cos2A + cos2A cos 4A + cos4A cosA)$$ Factioring out "4" we can write $$4(2cosA cos2A + 2cos2A cos 4A + 2cos4A cosA)$$ Now you must be aware of the product-sum trig identity$$2cosx \cdot cosy=cos(x+y)+cos(x-y)$$ Similary writing our expression through this idenitity can yield $$4(cos3A + cos A+ cos6 A+ cos 2A+ cos 5A+ cos3 A)$$ Rewriting them with the value of $A$ $$4(cos6 \pi /7 + cos 2\pi /7+ cos12 \pi /7 + cos 4\pi /7+ cos 10\pi /7+ cos6 \pi /7)$$ Now see we can have complete symmetry up here if one of the $cos6\pi /7 $ 's is replaced by $cos8\pi /7$ and fortunnately we have $$cos(\pi -\pi /7)=-cos(\pi /7)=cos(\pi +\pi /7)$$

Thus our expression becomes $$4(cos 2\pi /7+cos 4\pi /7+cos6 \pi /7 +cos 8\pi /7 +cos 10\pi /7+cos12 \pi /7)$$

Now comes the role of another trig identity $$\cos(\alpha) + \cos(\alpha + \beta) + \cos(\alpha + 2\beta) + \dots + \cos[\alpha + (n-1)\beta] = \frac{\cos(\alpha + \frac{n-1}{2}\beta) \cdot \sin\frac{n\beta}{2}}{\sin\frac{\beta}{2}}$$ Realize the fact that here $\alpha = 2 \pi/7$ and $\beta$ also corresponds to the same value. You have $n=6$

I hope you can complete this last step, just put in values and don't forget to cancel the sines in denominator and numerator by correctly changing the angles as sum or difference from $\pi$.

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    $\begingroup$ Write \cos x , \sin x so that the names of the trigonometric functions will appear nicely. $\endgroup$ – DonAntonio Apr 24 '17 at 13:29

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