Since it seems no one has risen to the challenge of the trigonometry problem I posted a few days ago I suppose the time is probably ripe to maybe end the suspense and now give the solution. But first, here is the problem statement again.
Find cos(π/5)(=cos(36°)), cos(2π/5)(=cos(72°)), cos(3π/5)(=cos(108°)), & cos(4π/5)(=cos(144°)) explicitly in terms of the Golden Mean, φ = (√(5) + 1)/2, and show your work.
Solution: Consider the 5 complex 5th roots of unity, i.e. 1, exp(2πi/5), exp(4πi/5), exp(6πi/5), exp(8πi/5) as 2-d vectors in the complex plane. Note exp(6πi/5) = exp(-4πi/5) = (exp(4πi/5))^*, and exp(8πi/5) = exp(-2πi/5)= (exp(2πi/5))^*. Each of these numbers has unit absolute value and each subsequent one is rotated by 2π/5 rad = 72° CCW relative to the previous number/vector. This means if we start at the origin in the complex plane and sequentially add these roots up in the order given and follow along the sum of vectors we see that these vectors form an equilateral pentagon, which is a closed planar figure. Since it is closed we end up back at the complex origin after adding them all up. Thus all 5 of these roots sum to zero. IOW
0 = 1 + exp(2πi/5) + exp(4πi/5) + exp(6πi/5) + exp(8πi/5) = 1 + (exp(2πi/5) + exp(-2πi/5)) + (exp(4πi/5) + exp(-4πi/5))= 1 + 2*cos(2π/5) + 2*cos(4π/5) = 0. Now let's define c ≡ cos(2π/5). Using the double angle formula for cosines means cos(4π/5) = 2*c^2 - 1. substituting into the previous equation of the roots summing to zero gives
0 = 1 + 2*c + 2*(2*c^2 - 1) = 4*c^2 + 2*c - 1 = 0.
This quadratic equation has the roots: c = (-1 ± √(5))/4. Since the angle 2π/5 rad = 72° is in the 1st quadrant we know the value of c must be positive and we know that c must be the positive root of the quadratic equation. Thus c = (√(5) - 1)/4 = ((√(5) + 1)/2 - 1)/2 = (φ - 1)/2 where φ ≡ (√(5) + 1)/2 is the value of the Golden Mean. Now the Golden Mean itself obeys the quadratic equation φ^2 - φ - 1 = 0. Equivalently (dividing by φ & subtracting the last term from both sides) we know φ - 1 = 1/φ. Thus cos(2π/5) = (φ - 1)/2 = 1/(2*φ). Using the facts that the cosine function is even in its argument and that adding or subtracting π radians to/from the argument of a cosine negates its value means
cos(3π/5) = cos(π - 2π/5) = -cos(-2π/5) = -cos(2π/5) = (1 - φ)/2 = -1/(2*φ). We can find cos(4π/5) by using the double angle formula for cosines on the now known c = cos(2π/5). Thus cos(4π/5) = 2*c^2 - 1 = ((φ - 1)^2)/2 - 1 = (φ^2 - 2*φ - 1)/2 = ((φ^2 - φ - 1) - φ)/2 = (0 - φ)/2 = -φ/2.
Thus cos(4π/5) = -φ/2.
We get cos(π/5) by again doing some argument shifting on cos(4π/5). IOW, cos(π/5) = cos(π - 4π/5) = -cos(-4π/5) = -cos(4π/5) = -(-φ/2) = φ/2.
This above trick can be generalized to formally find the cosines of integer multiples of any angle of the form π/(2k+1) rad where k is a natural number. The special case of the 5th roots of unity above simply uses k = 2. Applying the above trick for any arbitrary natural number, k, generates a closed equilateral (2*k+1)-gon which (with multiple angle formulas) generates a polynomial equation (with integer coefficients) in c ≡ cos(2π/(2*k+1)) of degree k. The roots of polynomial equations with integer coefficients are, by definition, algebraic numbers. For the values of k = 1,2,3,4 the corresponding cosines are solvable by radicals because there exist general standard formulae for the roots of such polynomial equations of those degrees in terms of radicals. Unfortunately for k = 3 & 4 the formulas involve taking square, cube & quartic roots of complex numbers such that the imaginary parts cancel in the end. If the formulas are reduced to find real solutions in terms of only real arithmetic then we need to be able to take trig functions of fractions of arc trig functions in order to evaluate the trig function of interest. And that essentially defeats the whole point of the exercise. For instance, more specifically, if we set k = 3 we can find cosines of integer multiples of π/7 rad. If we set c = cos(2π/7). Setting the sum of the 7th roots of unity to zero gives the cubic equation
0 = 8*c^3 + 4*c^2 - 4*c - 1.
This cubic equation has 3 real roots & the relevant real root for c = cos(2π/7) (using only real arithmetic) is
c = cos(2π/7) = (√(28)*cos((1/3)*arccos(1/√(28))) - 1)/6
which doesn't seems of be of much help when evaluating the exact value of cos(2π/7) without already knowing how to evaluate trig functions and their inverses. And when k is a natural number of at least 5 there aren't any general formulas which solve such a polynomial equation, anyway. But, be that as it may, even if we can't evaluate cos(n*π/(2k+1)) by radicals we are still assured that they are algebraic numbers because they can each be shown to obey a polynomial equation with integer coefficients. So we can deduce that the trig functions of rational multiples of π radians are necessarily algebraic numbers. This means the inverse trig functions of transcendental numbers are necessarily irrational multiples of π radians.