# If, θ=2π/7 So Prove That? Secθ+Sec2θ+Sec4θ = -4

May 29, 2018

#### Explanation:

Secant is reciprocal cosine.

Let's start with the equation $\cos 3 a = \cos 4 a$.

$3 a = \pm 4 a + 2 \pi k \quad$ integer $k$

Minus sign subsumes plus:

$7 a = 2 \pi k$

$a = \frac{2 \pi k}{7}$

So this equation has solutions $0 , \frac{2 \pi}{7} , \frac{4 \pi}{7} , \ldots$ There are four unique cosines in the bunch: $\cos 0 , \cos \left(\frac{2 \pi}{7}\right) , \cos \left(\frac{4 \pi}{7}\right) , \cos \left(\frac{6 \pi}{7}\right) .$ That's pretty close to what we want to add up.

What we're aiming for is a polynomial whose roots are the secants we seek; then the sum of the roots is given by Viete.

Let's expand $\cos 3 a = \cos 4 a$ using the triple and quadruple angle formulas. We let $x = \cos a$.

$\cos 3 a = 4 {\cos}^{3} a - 3 \cos a = 4 {x}^{3} - 3 x$

$\cos 2 a = 2 {\cos}^{2} x - 1 = 2 {x}^{2} - 1$

$\cos 4 a = 2 {\cos}^{2} \left(2 a\right) - 1 = 2 {\left(2 {x}^{2} - 1\right)}^{2} - 1 = 8 {x}^{4} - 8 {x}^{2} + 1$

So our equation with those cosines as roots is

$4 {x}^{3} - 3 x = 8 {x}^{4} - 8 {x}^{2} + 1$

$0 = 8 {x}^{4} - 4 {x}^{3} - 8 {x}^{2} + 3 x + 1$

The fourth degree equation verifies that there are (at most) four unique cosines here.

We actually want the equation with the secants as roots.

Let $y = \frac{1}{x} = \frac{1}{\cos} a = \sec a$

$0 = \frac{8}{y} ^ 4 - \frac{4}{y} ^ 3 - \frac{8}{y} ^ 2 + \frac{3}{y} + 1$

Multiply both sides by ${y}^{4} ,$

$0 = {y}^{4} + 3 {y}^{3} - 8 {y}^{2} - 4 y + 8$

If we imagine this factored $0 = \left(y - {r}_{1}\right) \left(y - {r}_{2}\right) \left(y - {r}_{3}\right) \left(y - {r}_{4}\right)$ we see the ${y}^{3}$ coefficient is $\left(- {r}_{1} - {r}_{2} - {r}_{3} - {r}_{4}\right) = - \left({r}_{1} + {r}_{2} + {r}_{3} + {r}_{4}\right)$. That's of course one of Viete's formulas.

Since the sum of the roots are the sum of the four secants mentioned, we have

$\sec 0 + \sec \left(\frac{2 \pi}{7}\right) + \sec \left(\frac{4 \pi}{7}\right) + \sec \left(\frac{6 \pi}{7}\right) = - \left(3\right)$

 sec({2pi}/7) + sec({4pi}/7) + sec ({6pi}/7) = -3 - 1/cos 0 = -4 quad sqrt

May 29, 2018

Given that, $\rightarrow x = \frac{2 \pi}{7}$ then $7 x = 2 \pi$

$L H S = \sec x + \sec 2 x + \sec 4 x$

$= \frac{1}{\cos} x + \frac{1}{\cos 2 x} + \frac{1}{\cos 4 x}$

$= \frac{\sin x \left[2 \cos 4 x \cdot \cos 2 x + 2 \cos 4 x \cdot \cos x + 2 \cos 2 x \cdot \cos x\right]}{2 \sin x \cdot \cos x \cdot \cos 2 x \cdot \cos 4 x}$

$= \frac{2 \sin x \left[\cos 6 x + \cos 2 x + \cos 4 x + \cos 2 x + \cos 3 x + \cos x\right]}{2 \sin 2 x \cdot \cos 2 x \cdot \cos 4 x}$

$= \frac{4 \sin x \left[\cos \left(7 x - x\right) + 2 \cos 2 x + \cos 4 x + \cos \left(7 x - 4 x\right) + \cos x\right]}{2 \sin 4 x \cdot \cos 4 x}$

$= \frac{4 \left[2 \cos x \cdot \sin x + 2 \cos 2 x \cdot \sin x + 2 \cos 4 x \cdot \sin x\right]}{\sin 8 x}$

=(4[sin2xcancel(+sin3x)-sinx+sin5xcancel(-sin3x)])/(sin(7x+x)

$= 4 \cdot \left[\frac{- \sin x + \sin 2 x + \sin \left(7 x - 2 x\right)}{\sin} x\right]$

$= 4 \cdot \left[\frac{- \sin x \cancel{+ \sin 2 x} \cancel{- \sin 2 x}}{\sin x}\right] = - 4 = R H S$