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

Secant is reciprocal cosine.

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

`3 a = pm 4 a + 2pi k quad` integer `k`

Minus sign subsumes plus:

`7 a = 2 pi k`

`a = {2 pi k}/7`

So this equation has solutions `0, {2pi}/7, {4pi}/7, ...` There are four unique cosines in the bunch: `cos 0, cos({2pi}/7), cos ({4pi}/7), cos({6pi}/7).` 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 3a=cos 4a` using the triple and quadruple angle formulas. We let `x=cos a`.

`cos 3a = 4cos^3 a - 3 cos a = 4 x^3 - 3x`

`cos2a = 2 cos ^2 x -1 = 2x^2-1`

`cos 4a = 2 cos^2 (2a) - 1 = 2(2x^2-1)^2-1 = 8x^4-8x^2+1`

So our equation with those cosines as roots is

`4x^3 - 3x = 8x^4 - 8x^2 + 1`

`0 = 8x^4 - 4x^3 - 8x^2 + 3x + 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=1/x=1/cos a = sec a`

`0 = 8/y^4 - 4/y^3 - 8/y^2 + 3/y + 1`

Multiply both sides by `y^4,`

`0 = y^4 + 3y^3 - 8y^2 - 4y + 8`

If we imagine this factored `0=(y-r_1)(y-r_2)(y-r_3)(y-r_4)` we see the `y^3` coefficient is `(-r_1-r_2-r_3-r_4)=-(r_1+r_2+r_3+r_4)`. 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

`sec0 + sec({2pi}/7) + sec({4pi}/7) + sec ({6pi}/7) = -(3)`

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

Given that, `rarrx=(2pi)/7` then `7x=2pi`

`LHS=secx+sec2x+sec4x`

`=1/cosx+1/(cos2x)+1/(cos4x)`

`=(sinx[2cos4x*cos2x+2cos4x*cosx+2cos2x*cosx])/(2sinx*cosx*cos2x*cos4x)`

`=(2sinx[cos6x+cos2x+cos4x+cos2x+cos3x+cosx])/(2sin2x*cos2x*cos4x)`

`=(4sinx[cos(7x-x)+2cos2x+cos4x+cos(7x-4x)+cosx])/(2sin4x*cos4x)`

`=(4[2cosx*sinx+2cos2x*sinx+2cos4x*sinx])/(sin8x)`

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

`=4*[(-sinx+sin2x+sin(7x-2x))/sinx]`

`=4*[(-sinxcancel(+sin2x)cancel(-sin2x))/(sinx)]=-4=RHS`