#sin 24^circ#
Angles which are multiples of three degrees are constructible with a straightedge and compass, and thus have a closed form consisting of integers combined by addition, subtraction, multiplication, division and square roots (of positive numbers).
This one is doable if we know the trig functions of #144^circ# because #24^circ=144^circ - 120^circ# and #120^circ# is of course related to one of the Two Tired Triangles of Trig, 30/60/90.
I happen to know #cos 144^circ = -phi/2#, negative one-half the Golden Ratio. Let's derive that.
In general #cos 3 theta = cos 2 theta# has solutions
#3 theta = pm 2 theta + 360^circ k quad # integer #k#
The minus sign subsumes the plus, and we end up with
#5 theta = 360^circ k#
#theta = 72^circ k#
If we let #x=cos theta# then using the double and triple angle formulas #cos 3 theta = cos 2 theta # becomes
#4x^3 - 3 x = 2x^2 - 1 #
#4x^3 - 2x^2 - 3 x + 1 = 0#
Since #theta=72^circ k# we know #theta=0# works, so #x=cos 0 = 1 # is a root of this polynomial. We factor:
#(x-1) (4 x^2 + 2 x - 1)= 0#
The quadratic has solutions
# x = 1/4 (-1 pm sqrt{5}) #
#cos 144^circ # is the only negative root of the original cubic. #cos 0^circ# and #cos 72^circ# are the other two roots, both positive.
#cos 144^circ = -1/4(1 + sqrt{5})#
Like I said, negative one half the Golden Ratio #phi={1+sqrt{5}}/2.#
#sin 144^circ# is positive. Unfortunately it has a nested square root; we live with it.
#sin 144^circ = sqrt{1 - cos ^ 2 144^circ} = sqrt{1 - 1/16(6 + 2sqrt{5}) } = \sqrt{1/8(5-sqrt{5}) }#
Now we can finally apply the difference angle formula:
#sin 24^circ = sin(144^circ - 120^circ) #
#= sin 144^circ cos 120^circ - cos 144^circ sin 120^circ #
#= sqrt{1/8(5-sqrt{5}) } (-1/2) - (-1/4(1 + sqrt{5}))(sqrt{3}/2) #
#= -1/8 sqrt{16/8(5-sqrt{5}) } + 1/8(sqrt{3} + sqrt{15}))#
#sin 24^circ = 1/8 (sqrt(3) + sqrt(15) - sqrt(10 - 2 sqrt(5))) #
Check: Calculator.
# sin 24^circ - 1/8 (sqrt(3) + sqrt(15) - sqrt(10 - 2 sqrt(5))) =0 quad sqrt#
