Help with Proof

Actually, I take it back; I based that on the idea that if your original answer was wrong, putting it through those formulas would produce a different value, which upon double-checking turned out not to be true; the "proof" is merely a tautology. Clearly I fail at logic today, so I went back and tried it again.

You can use angle complementation and half-angle formulas to make some progress;

cos[π/5] = +-√(1 + cos[2π/5]/2)

cos[2π/5] = sin[π/2 - 2π/5] = sin[π/10] = +-√(1 - cos[π/5]/2)

2(cos[π/5])^2 - 1 = cos[2π/5] = +-√(1 - cos[π/5]/2)

2(cos[π/5])^2 - 1 = +-√(1 - cos[π/5]/2)

8(cos[π/5])^4 - 8(cos[π/5])^2 + = -1 - cos[π/5]

which at least gets the entire thing in terms of cos[π/5].

Double edit: I can change it from a quartic into a cubic, but I doubt that the general solution for either is part of your coursework. The answer implies that it's possible to get the solution in quadratic form, but I can't see a way to do it. There is a way to get a sort-of proof, but I consider it semi-dishonest since the proof relies on plugging in the answer at a certain point (though I checked other answers this time and they don't check out; it will only work if the angle is correct, or with the correct angle times -1 as cosine an even function). I'll post it and let you see if you can divine anything from it, or pick out mistakes:

8(cos[π/5])^2[(cos [π/5])^2 - 1] = -1 - cos[π/5]

8(cos[π/5])^2 = -1 - cos[π/5]/{(cos[π/5])^2 - 1}

-8(cos[π/5])^2 = 1 + cos[π/5]/{cos[π/5] + 1}{cos[π/5] - 1}

8(cos[π/5])^2 = 1/(1 - cos[π/5])

8(sec[π/5])^2 = 1 - cos[π/5]

8(sec[π/5])^2 - 1 + cos[π/5] = 0

8/(cos[π/5])^2 - (cos[π/5])^2/(cos[π/5])^2 + (cos[π/5])^3/(cos[π/5])^2 = 0

8 + (cos[π/5])^3 - (cos[π/5])^2 = 0

(cos[π/5])^2 - (cos[π/5])^3 = 8

(cos[π/5])^2{1 - cos[π/5]} = 8

cos[π/5] = √(8/{1 - cos[π/5]})

At this point, you can substitute (1 + √5)/4 into the equation and it will check out. I've attempted to get a quadratic out of this jumble, but have met with no luck. All my attempts to set two equations for cos[π/5] equal to each other have resulted in aforementioned cubics.

I'm not sure if you still need help, but couldn't you use a triangle that has (pi/5) in it? Kind of like the right-angle triangle with the other 2 angles being (pi/4) and ((square root of 2)/2) for the adjacent and opposite and 1 as the hypotenuse.