How to find trigonometric values that = radical values?

I have always hoped to rearrange a trigonometric expression into another expression without trigonometric functions. For example, it is common known that \(\sin 60^\circ =\dfrac{\sqrt{3}}{2}\), but what about other ones?
I have found some already (below) and I wish to find more as this may be helpful when I am to simplify an expression.

instructions

If you find any typos or improvements comment under my comment that reads: report
The angles I am looking for are integers in range [0,180][0^\circ ,180^\circ].
Please do not comment straight away as things will get messy.
If you have found a new solution, comment under my comment below this note in format:
<equation>
<proof———>

The Transformation table

Formulae used:
(1)   sin(πA)=sinA,  cos(πA)=cosA,  tan(πA)=tanA.(1)~~~\sin (\pi -A)=\sin A,~~\cos (\pi -A)=-\cos A,~~\tan (\pi -A)=-\tan A.
(2)   sin2A+cos2A=1,  tanA=sinAcosA.(2)~~~\sin ^2 A+\cos ^2 A=1,~~\tan A=\dfrac{\sin A}{\cos A}.
(3)   sin(π2+A)=cosA,  cos(π2+A)=sinA.(3)~~~\sin (\dfrac{\pi}{2}+A)=\cos A,~~\cos (\dfrac{\pi}{2}+A)=-\sin A.
(4)   sin2A=2sinAcosA.(4)~~~\sin 2A=2\sin A\cos A.
(5)   cos2A=cos2Asin2A=2cos2A1=12sin2A.(5)~~~\cos 2A=\cos ^2 A-\sin ^2 A=2\cos ^2 A-1 =1-2\sin ^2 A.
(6)   tan2A=2tanA1tan2A.(6)~~~\tan 2A=\dfrac{2\tan A}{1-\tan ^2 A}.
Table 1: integer values in degrees:

angle θ\theta in degreessinθ\sin \thetacosθ\cos \thetatanθ\tan \thetaProof
1515^\circ624\dfrac{\sqrt{6}-\sqrt{2}}{4}6+24\dfrac{\sqrt{6}+\sqrt{2}}{4}232-\sqrt{3}
1818^\circ514\dfrac{\sqrt{5}-1}{4}10+254\dfrac{\sqrt{10+2\sqrt{5}}}{4}251055\dfrac{\sqrt{25-10\sqrt{5}}}{5}
2020^\circ
3030^\circ12\dfrac{1}{2}32\dfrac{\sqrt{3}}{2}33\dfrac{\sqrt{3}}{3}N/A
3636^\circ10254\dfrac{\sqrt{10-2\sqrt{5}}}{4}5+14\dfrac{\sqrt{5}+1}{4}803254\dfrac{\sqrt{80-32\sqrt{5}}}{4}
4545^\circ22\dfrac{\sqrt{2}}{2}22\dfrac{\sqrt{2}}{2}11N/A
5454^\circ5+14\dfrac{\sqrt5+1}{4}
6060^\circ32\dfrac{\sqrt{3}}{2}12\dfrac{1}{2}3\sqrt{3}N/A
7272^\circ514\dfrac{\sqrt{5}-1}{4}
7575^\circ6+24\dfrac{\sqrt{6}+\sqrt{2}}{4}624\dfrac{\sqrt{6}-\sqrt{2}}{4}2+32+\sqrt{3}
9090^\circ1100undefinedN/A
105105^\circ6+24\dfrac{\sqrt6 +\sqrt2}{4}264\dfrac{\sqrt2 -\sqrt6}{4}23-2-\sqrt3
108108^\circ154\dfrac{1-\sqrt{5}}{4}
120120^\circ32\dfrac{\sqrt3}{2}12-\dfrac{1}{2}3-\sqrt3
126126^\circ
135135^\circ22\dfrac{\sqrt{2}}{2}22-\dfrac{\sqrt{2}}{2}1-1
144144^\circ10254\dfrac{\sqrt{10-2\sqrt{5}}}{4}514\dfrac{-\sqrt{5}-1}{4}803254-\dfrac{\sqrt{80-32\sqrt{5}}}{4}
150150^\circ12\dfrac{1}{2}32-\dfrac{\sqrt3}{2}33-\dfrac{\sqrt3}{3}
162162^\circ514\dfrac{\sqrt5-1}{4}10+254-\dfrac{\sqrt{10+2\sqrt{5}}}{4}251055-\dfrac{\sqrt{25-10\sqrt{5}}}{5}
165165^\circ624\dfrac{\sqrt6-\sqrt2}{4}624\dfrac{-\sqrt6-\sqrt2}{4}32\sqrt3-2

Table 2: fraction values in radians:

Note by Jeff Giff
1 week, 1 day ago

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Jeff Giff - 1 week, 1 day ago

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Jeff Giff - 1 week, 1 day ago

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If any picture isn’t functioning, tell me below this comment :)

Jeff Giff - 1 week, 1 day ago

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I don’t see any picture in this note at all?!! , I don’t even see any space for any picture, is there a picture supposed to be there for the proof

Jason Gomez - 1 week, 1 day ago

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Well in the buttons there are pics I think?

Jeff Giff - 1 week, 1 day ago

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Previously I saw no pic in the first button, but now I can, but I can’t see any pic in the second button

Jason Gomez - 1 week ago

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@Jason Gomez And I really like that proof, I used to always run around quadratic formulae if I forgot the value of sin\cos 15 and it’s multiples

Jason Gomez - 1 week ago

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@Jason Gomez You’ll also like the proof for sin18\sin 18^\circ :)

Jeff Giff - 1 week ago

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@Jason Gomez It’s very nice to see it has a clear geometry approach, takes a cosine law to find ϕ\phi and the rest is easy, but it does have the same complexity as sin36°=cos54°\sin 36\degree = \cos 54\degree approach

Jason Gomez - 1 week ago

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I also knew this one before: sin(1.5°)=22422+3+1+52+3(1+52)5228..........[1]\sin(1.5 \degree)=\dfrac{\sqrt{2\sqrt[4]{2}-\sqrt{2\sqrt{2}+\sqrt{3+\frac{1+\sqrt{5}}{2}+\sqrt{3(\frac{1+\sqrt{5}}{2})\sqrt{5}}}}}}{2\sqrt[8]{2}}..........[1]

Proof: (Skipping some of the calculations) sin5θ=(5+16sin4θ20sin2θ)sinθ\sin 5\theta=(5+16\sin^4\theta-20\sin^2\theta)\sin\theta let θ=36°\theta=36\degree (5+16sin436°20sin236°)sin36°=sin180°=0(5+16\sin^436\degree-20\sin^236\degree)\sin 36\degree=\sin 180\degree=0 sin36°=0\because \sin 36\degree\cancel{=}0 5+16sin436°20sin236°=0\therefore 5+16\sin^436\degree-20\sin^236\degree=0 let x=sin236°<0.5x=\sin^236\degree<0.5 16x220x+5=0\Rightarrow 16x^2-20x+5=0 x=5+58\Rightarrow x=\frac{5^+_-\sqrt{5}}{8} 5+58>0.5\because \frac{5+\sqrt{5}}{8}>0.5 x=558\therefore x=\frac{5-\sqrt{5}}{8} sin36°=5522\Rightarrow \sin 36\degree=\frac{\sqrt{5-\sqrt{5}}}{2\sqrt{2}} cos36°=3+522\Rightarrow \cos 36\degree=\frac{\sqrt{3+\sqrt{5}}}{2\sqrt{2}} sin6°=sin(36°30°)\Rightarrow \sin 6\degree=\sin(36\degree-30\degree) =sin36°cos36°sin30°cos36°=\sin 36\degree\cos 36\degree-\sin 30\degree\cos 36\degree =15353+542=\dfrac{\sqrt{15-3\sqrt{5}}-\sqrt{3+\sqrt{5}}}{4\sqrt{2}} Now using the identity : sinθ2=+11sin2θ2\sin\frac{\theta}{2}=^+_-\dfrac{\sqrt{1-\sqrt{1-\sin^2\theta}}}{\sqrt{2}} and 0<sin1.5°<sin3°0<\sin 1.5\degree<\sin 3\degree we can find [1][1]

Zakir Husain - 1 week, 1 day ago

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You can use the angle addition equations to obtain more that are multiples of 15°15° and 18°18°:

sin45°=22\sin 45° = \cfrac{\sqrt{2}}{2}, cos45°=22\cos 45° = \cfrac{\sqrt{2}}{2}, tan45°=1\tan 45° = 1

sin60°=32\sin 60° = \cfrac{\sqrt{3}}{2}, cos60°=12\cos 60° = \cfrac{1}{2}, tan60°=3\tan 60° = \sqrt{3}

sin75°=2+64\sin 75° = \cfrac{\sqrt{2} + \sqrt{6}}{4}, cos75°=624\cos 75° = \cfrac{\sqrt{6} - \sqrt{2}}{4}, tan75°=2+3\tan 75° = 2 + \sqrt{3}

sin90°=1\sin 90° = 1, cos90°=0\cos 90° = 0

sin105°=2+62\sin 105° = \cfrac{\sqrt{2} + \sqrt{6}}{2}, cos105°=262\cos 105° = \cfrac{\sqrt{2} - \sqrt{6}}{2}, tan105°=23\tan 105° = -2 - \sqrt{3}

sin120°=32\sin 120° = \cfrac{\sqrt{3}}{2}, cos120°=12\cos 120° = -\cfrac{1}{2}, tan120°=3\tan 120° = -\sqrt{3}

sin135°=22\sin 135° = \cfrac{\sqrt{2}}{2}, cos135°=22\cos 135° = -\cfrac{\sqrt{2}}{2}, tan135°=1\tan 135° = -1

sin150°=12\sin 150° = \cfrac{1}{2}, cos150°=32\cos 150° = -\cfrac{\sqrt{3}}{2}, tan150°=33\tan 150° = -\cfrac{\sqrt{3}}{3}

sin165°=622\sin 165° = \cfrac{\sqrt{6} - \sqrt{2}}{2}, cos165°=264\cos 165° = \cfrac{-\sqrt{2} - \sqrt{6}}{4}, tan165°=32\tan 165° = \sqrt{3} - 2

cos36°=1+54\cos 36° = \cfrac{1 + \sqrt{5}}{4}

sin54°=1+54\sin 54° = \cfrac{1 + \sqrt{5}}{4}

cos72°=514\cos 72° = \cfrac{\sqrt{5} - 1}{4}

cos108°=154\cos 108° = \cfrac{1 - \sqrt{5}}{4}

sin126°=1+54\sin 126° = \cfrac{1 + \sqrt{5}}{4}

cos144°=154\cos 144° = \cfrac{-1 - \sqrt{5}}{4}

sin162°=514\sin 162° = \cfrac{\sqrt{5} - 1}{4}

If you allow for radicals inside radicals, there are several more that can be obtained using half angle formulas (for example, sin22.5°=222\sin 22.5° = \cfrac{\sqrt{2 - \sqrt{2}}}{2})

Other angles are roots of polynomials, for example, cos(180°7)\cos\bigg(\cfrac{180°}{7}\bigg) is a root of 8x34x24x+18x^3 - 4x^2 - 4x + 1 (see here), but this doesn't come out to a clean rational number in radical form.

David Vreken - 1 week, 1 day ago

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Could you give a hint on how to prove the cosπ7\cos{\frac π 7} part, should I be using the cos3θ\cos 3 θ formula or should I be using (eiπ7)7(e^{\frac {iπ}7})^7

Jason Gomez - 1 week, 1 day ago

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Check out this website here.

Looks like either approach could work.

David Vreken - 1 week, 1 day ago

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@David Vreken Thank you, never expected both would work

Jason Gomez - 1 week ago

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Well luckily I have a book including one tedious formula for cubic functions :D only it gives nested radicals :(

Jeff Giff - 1 week ago

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Are you talking about The Cubic Formula?

Zakir Husain - 1 week ago

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@Zakir Husain Yes :)

Jeff Giff - 1 week ago

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@Zakir Husain the cardano’s method seems better

Jeff Giff - 1 week ago

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@Jeff Giff Best is the Newton–Raphson method it is applicable for all polynomial equations of any finite degree (mostly gives approximated value not exact)

Zakir Husain - 1 week ago

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@Zakir Husain Ah yes, but again it only works for approximations :) to require precise results things get tedious :) I suppose I like Cardano’s method better :)

Jeff Giff - 1 week ago

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Check this

Zakir Husain - 1 week, 1 day ago

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Thanks!

Jeff Giff - 1 week, 1 day ago

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