If $A= 2 \cdot \pi/7$ then show that $ \sec A+ \sec 2A+ \sec 4A=-4$ 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.
 A: 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 $\displaystyle \large -4$
A: $$\frac1{\cos a}+\frac1{\cos2a}+\frac1{\cos4a}=\frac{\cos2a\cos4a+\cos a\cos4a+\cos a\cos2a}{\cos a\cos2a\cos4a}=$$
$$=\frac12\frac{\cos6a+\cos2a+\cos5a+\cos3a+\cos3a+\cos a}{\frac12\left[\cos a+\cos3a\right]\cos4a}=$$
$$2\frac{\cos a+\cos2a+2\cos3a+\cos5a+\cos6a}{\cos5a+\cos3a+\cos7a+\cos a}\;(*)$$
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
$$(***)=4\frac{-\frac12}{1-\frac12}=-4$$
A: $$\sec\frac{2\pi}{7}+\sec\frac{4\pi}{7}+\sec\frac{8\pi}{7}=\frac{1}{\cos\frac{2\pi}{7}}+\frac{1}{\cos\frac{4\pi}{7}}+\frac{1}{\cos\frac{8\pi}{7}}=$$
$$=\frac{\cos\frac{2\pi}{7}\cos\frac{4\pi}{7}+\cos\frac{2\pi}{7}\cos\frac{8\pi}{7}+\cos\frac{4\pi}{7}\cos\frac{8\pi}{7}}{\cos\frac{2\pi}{7}\cos\frac{4\pi}{7}\cos\frac{8\pi}{7}}=$$
$$=\frac{\cos\frac{6\pi}{7}+\cos\frac{2\pi}{7}+\cos\frac{4\pi}{7}+\cos\frac{6\pi}{7}+\cos\frac{2\pi}{7}+\cos\frac{4\pi}{7}}{2\cos\frac{2\pi}{7}\cos\frac{4\pi}{7}\cos\frac{8\pi}{7}}=$$
$$=\frac{\cos\frac{2\pi}{7}+\cos\frac{4\pi}{7}+\cos\frac{6\pi}{7}}{\cos\frac{2\pi}{7}\cos\frac{4\pi}{7}\cos\frac{8\pi}{7}}=\frac{2\sin\frac{\pi}{7}\left(\cos\frac{2\pi}{7}+\cos\frac{4\pi}{7}+\cos\frac{6\pi}{7}\right)\cdot8\cos\frac{\pi}{7}}{8\sin\frac{2\pi}{7}\cos\frac{2\pi}{7}\cos\frac{4\pi}{7}\cos\frac{8\pi}{7}}=$$
$$=\frac{\left(\sin\frac{3\pi}{7}-\sin\frac{\pi}{7}+\sin\frac{5\pi}{7}-\sin\frac{3\pi}{7}+\sin\frac{7\pi}{7}-\sin\frac{5\pi}{7}\right)\cdot8\cos\frac{\pi}{7}}{\sin\frac{16\pi}{7}}=$$
$$=\frac{-\sin\frac{\pi}{7}\cdot8\cos\frac{\pi}{7}}{\sin\frac{2\pi}{7}}=-4.$$
Done!
A: 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.
A: 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 $\pi$.
