#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#

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