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sin
0
=
0
{\displaystyle \sin 0=0\,}
cos
0
=
1
{\displaystyle \cos 0=1\,}
tan
0
=
0
{\displaystyle \tan 0=0\,}
cot
0
=
∞
{\displaystyle \cot 0=\infty \,}
sin
π
60
=
sin
3
∘
=
2
(
1
−
3
)
5
+
5
+
2
(
5
−
1
)
(
3
+
1
)
16
{\displaystyle \sin {\frac {\pi }{60}}=\sin 3^{\circ }={\frac {2(1-{\sqrt {3}}){\sqrt {5+{\sqrt {5}}}}+{\sqrt {2}}({\sqrt {5}}-1)({\sqrt {3}}+1)}{16}}\,}
cos
π
60
=
cos
3
∘
=
2
(
1
+
3
)
5
+
5
+
2
(
5
−
1
)
(
3
−
1
)
16
{\displaystyle \cos {\frac {\pi }{60}}=\cos 3^{\circ }={\frac {2(1+{\sqrt {3}}){\sqrt {5+{\sqrt {5}}}}+{\sqrt {2}}({\sqrt {5}}-1)({\sqrt {3}}-1)}{16}}\,}
tan
π
60
=
tan
3
∘
=
(
(
2
−
3
)
(
3
+
5
)
−
2
)
(
2
−
2
(
5
−
5
)
)
4
{\displaystyle \tan {\frac {\pi }{60}}=\tan 3^{\circ }={\frac {\left((2-{\sqrt {3}})(3+{\sqrt {5}})-2\right)\left(2-{\sqrt {2(5-{\sqrt {5}})}}\right)}{4}}\,}
cot
π
60
=
cot
3
∘
=
(
(
2
+
3
)
(
3
+
5
)
−
2
)
(
2
+
2
(
5
−
5
)
)
4
{\displaystyle \cot {\frac {\pi }{60}}=\cot 3^{\circ }={\frac {\left((2+{\sqrt {3}})(3+{\sqrt {5}})-2\right)\left(2+{\sqrt {2(5-{\sqrt {5}})}}\right)}{4}}\,}
sin
π
30
=
sin
6
∘
=
6
(
5
−
5
)
−
5
−
1
8
{\displaystyle \sin {\frac {\pi }{30}}=\sin 6^{\circ }={\frac {{\sqrt {6(5-{\sqrt {5}})}}-{\sqrt {5}}-1}{8}}\,}
cos
π
30
=
cos
6
∘
=
2
(
5
−
5
)
+
3
(
5
+
1
)
8
{\displaystyle \cos {\frac {\pi }{30}}=\cos 6^{\circ }={\frac {{\sqrt {2(5-{\sqrt {5}})}}+{\sqrt {3}}({\sqrt {5}}+1)}{8}}\,}
tan
π
30
=
tan
6
∘
=
2
(
5
−
5
)
−
3
(
5
−
1
)
2
{\displaystyle \tan {\frac {\pi }{30}}=\tan 6^{\circ }={\frac {{\sqrt {2(5-{\sqrt {5}})}}-{\sqrt {3}}({\sqrt {5}}-1)}{2}}\,}
cot
π
30
=
cot
6
∘
=
3
(
3
+
5
)
+
2
(
25
+
11
5
)
2
{\displaystyle \cot {\frac {\pi }{30}}=\cot 6^{\circ }={\frac {{\sqrt {3}}(3+{\sqrt {5}})+{\sqrt {2(25+11{\sqrt {5}})}}}{2}}\,}
sin
π
20
=
sin
9
∘
=
2
(
5
+
1
)
−
2
5
−
5
8
{\displaystyle \sin {\frac {\pi }{20}}=\sin 9^{\circ }={\frac {{\sqrt {2}}({\sqrt {5}}+1)-2{\sqrt {5-{\sqrt {5}}}}}{8}}\,}
cos
π
20
=
cos
9
∘
=
2
(
5
+
1
)
+
2
5
−
5
8
{\displaystyle \cos {\frac {\pi }{20}}=\cos 9^{\circ }={\frac {{\sqrt {2}}({\sqrt {5}}+1)+2{\sqrt {5-{\sqrt {5}}}}}{8}}\,}
tan
π
20
=
tan
9
∘
=
5
+
1
−
5
+
2
5
{\displaystyle \tan {\frac {\pi }{20}}=\tan 9^{\circ }={\sqrt {5}}+1-{\sqrt {5+2{\sqrt {5}}}}\,}
cot
π
20
=
cot
9
∘
=
5
+
1
+
5
+
2
5
{\displaystyle \cot {\frac {\pi }{20}}=\cot 9^{\circ }={\sqrt {5}}+1+{\sqrt {5+2{\sqrt {5}}}}\,}
sin
π
15
=
sin
12
∘
=
2
(
5
+
5
)
−
3
(
5
−
1
)
8
{\displaystyle \sin {\frac {\pi }{15}}=\sin 12^{\circ }={\frac {{\sqrt {2(5+{\sqrt {5}})}}-{\sqrt {3}}({\sqrt {5}}-1)}{8}}\,}
cos
π
15
=
cos
12
∘
=
6
(
5
+
5
)
+
5
−
1
8
{\displaystyle \cos {\frac {\pi }{15}}=\cos 12^{\circ }={\frac {{\sqrt {6(5+{\sqrt {5}})}}+{\sqrt {5}}-1}{8}}\,}
tan
π
15
=
tan
12
∘
=
3
(
3
−
5
)
−
2
(
25
−
11
5
)
2
{\displaystyle \tan {\frac {\pi }{15}}=\tan 12^{\circ }={\frac {{\sqrt {3}}(3-{\sqrt {5}})-{\sqrt {2(25-11{\sqrt {5}})}}}{2}}\,}
cot
π
15
=
cot
12
∘
=
3
(
5
+
1
)
+
2
(
5
+
5
)
2
{\displaystyle \cot {\frac {\pi }{15}}=\cot 12^{\circ }={\frac {{\sqrt {3}}({\sqrt {5}}+1)+{\sqrt {2(5+{\sqrt {5}})}}}{2}}\,}
sin
π
12
=
sin
15
∘
=
2
(
3
−
1
)
4
{\displaystyle \sin {\frac {\pi }{12}}=\sin 15^{\circ }={\frac {{\sqrt {2}}({\sqrt {3}}-1)}{4}}\,}
cos
π
12
=
cos
15
∘
=
2
(
3
+
1
)
4
{\displaystyle \cos {\frac {\pi }{12}}=\cos 15^{\circ }={\frac {{\sqrt {2}}({\sqrt {3}}+1)}{4}}\,}
tan
π
12
=
tan
15
∘
=
2
−
3
{\displaystyle \tan {\frac {\pi }{12}}=\tan 15^{\circ }=2-{\sqrt {3}}\,}
cot
π
12
=
cot
15
∘
=
2
+
3
{\displaystyle \cot {\frac {\pi }{12}}=\cot 15^{\circ }=2+{\sqrt {3}}\,}
sin
π
10
=
sin
18
∘
=
5
−
1
4
=
φ
−
1
2
=
1
2
φ
{\displaystyle \sin {\frac {\pi }{10}}=\sin 18^{\circ }={\frac {{\sqrt {5}}-1}{4}}={\frac {\varphi -1}{2}}={\frac {1}{2\varphi }}\,}
cos
π
10
=
cos
18
∘
=
2
(
5
+
5
)
4
{\displaystyle \cos {\frac {\pi }{10}}=\cos 18^{\circ }={\frac {\sqrt {2(5+{\sqrt {5}})}}{4}}\,}
tan
π
10
=
tan
18
∘
=
5
(
5
−
2
5
)
5
{\displaystyle \tan {\frac {\pi }{10}}=\tan 18^{\circ }={\frac {\sqrt {5(5-2{\sqrt {5}})}}{5}}\,}
cot
π
10
=
cot
18
∘
=
5
+
2
5
{\displaystyle \cot {\frac {\pi }{10}}=\cot 18^{\circ }={\sqrt {5+2{\sqrt {5}}}}\,}
sin
7
π
60
=
sin
21
∘
=
2
(
3
+
1
)
5
−
5
−
2
(
3
−
1
)
(
1
+
5
)
16
{\displaystyle \sin {\frac {7\pi }{60}}=\sin 21^{\circ }={\frac {2({\sqrt {3}}+1){\sqrt {5-{\sqrt {5}}}}-{\sqrt {2}}({\sqrt {3}}-1)(1+{\sqrt {5}})}{16}}\,}
cos
7
π
60
=
cos
21
∘
=
2
(
3
−
1
)
5
−
5
+
2
(
3
+
1
)
(
1
+
5
)
16
{\displaystyle \cos {\frac {7\pi }{60}}=\cos 21^{\circ }={\frac {2({\sqrt {3}}-1){\sqrt {5-{\sqrt {5}}}}+{\sqrt {2}}({\sqrt {3}}+1)(1+{\sqrt {5}})}{16}}\,}
tan
7
π
60
=
tan
21
∘
=
(
2
−
(
2
+
3
)
(
3
−
5
)
)
(
2
−
2
(
5
+
5
)
)
4
{\displaystyle \tan {\frac {7\pi }{60}}=\tan 21^{\circ }={\frac {\left(2-(2+{\sqrt {3}})(3-{\sqrt {5}})\right)\left(2-{\sqrt {2(5+{\sqrt {5}})}}\right)}{4}}\,}
cot
7
π
60
=
cot
21
∘
=
(
2
−
(
2
−
3
)
(
3
−
5
)
)
(
2
+
10
5
)
4
{\displaystyle \cot {\frac {7\pi }{60}}=\cot 21^{\circ }={\frac {\left(2-(2-{\sqrt {3}})(3-{\sqrt {5}})\right)\left(2+{\sqrt {10{\sqrt {5}}}}\right)}{4}}\,}
sin
π
8
=
sin
22.5
∘
=
2
−
2
2
{\displaystyle \sin {\frac {\pi }{8}}=\sin 22.5^{\circ }={\frac {\sqrt {2-{\sqrt {2}}}}{2}}\,}
cos
π
8
=
cos
22.5
∘
=
2
+
2
2
{\displaystyle \cos {\frac {\pi }{8}}=\cos 22.5^{\circ }={\frac {\sqrt {2+{\sqrt {2}}}}{2}}\,}
tan
π
8
=
tan
22.5
∘
=
2
−
1
{\displaystyle \tan {\frac {\pi }{8}}=\tan 22.5^{\circ }={\sqrt {2}}-1\,}
cot
π
8
=
cot
22.5
∘
=
2
+
1
{\displaystyle \cot {\frac {\pi }{8}}=\cot 22.5^{\circ }={\sqrt {2}}+1\,}
sin
2
π
15
=
sin
24
∘
=
3
(
5
+
1
)
−
2
5
−
5
8
{\displaystyle \sin {\frac {2\pi }{15}}=\sin 24^{\circ }={\frac {{\sqrt {3}}({\sqrt {5}}+1)-{\sqrt {2}}{\sqrt {5-{\sqrt {5}}}}}{8}}\,}
cos
2
π
15
=
cos
24
∘
=
6
5
−
5
+
5
+
1
8
{\displaystyle \cos {\frac {2\pi }{15}}=\cos 24^{\circ }={\frac {{\sqrt {6}}{\sqrt {5-{\sqrt {5}}}}+{\sqrt {5}}+1}{8}}\,}
tan
2
π
15
=
tan
24
∘
=
50
+
22
5
−
3
(
3
+
5
)
2
{\displaystyle \tan {\frac {2\pi }{15}}=\tan 24^{\circ }={\frac {{\sqrt {50+22{\sqrt {5}}}}-{\sqrt {3}}(3+{\sqrt {5}})}{2}}\,}
cot
2
π
15
=
cot
24
∘
=
2
5
−
5
+
3
(
5
−
1
)
2
{\displaystyle \cot {\frac {2\pi }{15}}=\cot 24^{\circ }={\frac {{\sqrt {2}}{\sqrt {5-{\sqrt {5}}}}+{\sqrt {3}}({\sqrt {5}}-1)}{2}}\,}
sin
3
π
20
=
sin
27
∘
=
(
5
+
1
)
5
+
5
−
2
(
5
−
1
)
8
{\displaystyle \sin {\frac {3\pi }{20}}=\sin 27^{\circ }={\frac {({\sqrt {5}}+1){\sqrt {5+{\sqrt {5}}}}-{\sqrt {2}}({\sqrt {5}}-1)}{8}}\,}
cos
3
π
20
=
cos
27
∘
=
(
5
+
1
)
5
+
5
+
2
(
5
−
1
)
8
{\displaystyle \cos {\frac {3\pi }{20}}=\cos 27^{\circ }={\frac {({\sqrt {5}}+1){\sqrt {5+{\sqrt {5}}}}+{\sqrt {2}}({\sqrt {5}}-1)}{8}}\,}
tan
3
π
20
=
tan
27
∘
=
5
−
1
−
5
−
2
5
{\displaystyle \tan {\frac {3\pi }{20}}=\tan 27^{\circ }={\sqrt {5}}-1-{\sqrt {5-2{\sqrt {5}}}}\,}
cot
3
π
20
=
cot
27
∘
=
5
−
1
+
5
−
2
5
{\displaystyle \cot {\frac {3\pi }{20}}=\cot 27^{\circ }={\sqrt {5}}-1+{\sqrt {5-2{\sqrt {5}}}}\,}
sin
π
6
=
sin
30
∘
=
1
2
{\displaystyle \sin {\frac {\pi }{6}}=\sin 30^{\circ }={\frac {1}{2}}\,}
cos
π
6
=
cos
30
∘
=
3
2
{\displaystyle \cos {\frac {\pi }{6}}=\cos 30^{\circ }={\frac {\sqrt {3}}{2}}\,}
tan
π
6
=
tan
30
∘
=
3
3
{\displaystyle \tan {\frac {\pi }{6}}=\tan 30^{\circ }={\frac {\sqrt {3}}{3}}\,}
cot
π
6
=
cot
30
∘
=
3
{\displaystyle \cot {\frac {\pi }{6}}=\cot 30^{\circ }={\sqrt {3}}\,}
sin
11
π
60
=
sin
33
∘
=
2
(
3
−
1
)
5
+
5
+
2
(
1
+
3
)
(
5
−
1
)
16
{\displaystyle \sin {\frac {11\pi }{60}}=\sin 33^{\circ }={\frac {2({\sqrt {3}}-1){\sqrt {5+{\sqrt {5}}}}+{\sqrt {2}}(1+{\sqrt {3}})({\sqrt {5}}-1)}{16}}\,}
cos
11
π
60
=
cos
33
∘
=
2
(
3
+
1
)
5
+
5
+
2
(
1
−
3
)
(
5
−
1
)
16
{\displaystyle \cos {\frac {11\pi }{60}}=\cos 33^{\circ }={\frac {2({\sqrt {3}}+1){\sqrt {5+{\sqrt {5}}}}+{\sqrt {2}}(1-{\sqrt {3}})({\sqrt {5}}-1)}{16}}\,}
tan
11
π
60
=
tan
33
∘
=
(
2
−
(
2
−
3
)
(
3
+
5
)
)
(
2
+
2
(
5
−
5
)
)
4
{\displaystyle \tan {\frac {11\pi }{60}}=\tan 33^{\circ }={\frac {\left(2-(2-{\sqrt {3}})(3+{\sqrt {5}})\right)\left(2+{\sqrt {2(5-{\sqrt {5}})}}\right)}{4}}\,}
cot
11
π
60
=
cot
33
∘
=
(
2
−
(
2
+
3
)
(
3
+
5
)
)
(
2
−
2
(
5
−
5
)
)
4
{\displaystyle \cot {\frac {11\pi }{60}}=\cot 33^{\circ }={\frac {\left(2-(2+{\sqrt {3}})(3+{\sqrt {5}})\right)\left(2-{\sqrt {2(5-{\sqrt {5}})}}\right)}{4}}\,}
sin
π
5
=
sin
36
∘
=
2
(
5
−
5
)
4
{\displaystyle \sin {\frac {\pi }{5}}=\sin 36^{\circ }={\frac {\sqrt {2(5-{\sqrt {5}})}}{4}}\,}
cos
π
5
=
cos
36
∘
=
1
+
5
4
=
φ
2
{\displaystyle \cos {\frac {\pi }{5}}=\cos 36^{\circ }={\frac {1+{\sqrt {5}}}{4}}={\frac {\varphi }{2}}\,}
tan
π
5
=
tan
36
∘
=
5
−
2
5
{\displaystyle \tan {\frac {\pi }{5}}=\tan 36^{\circ }={\sqrt {5-2{\sqrt {5}}}}\,}
cot
π
5
=
cot
36
∘
=
5
(
5
+
2
5
)
5
{\displaystyle \cot {\frac {\pi }{5}}=\cot 36^{\circ }={\frac {\sqrt {5(5+2{\sqrt {5}})}}{5}}\,}
sin
13
π
60
=
sin
39
∘
=
2
(
1
−
3
)
5
−
5
+
2
(
3
+
1
)
(
5
+
1
)
16
{\displaystyle \sin {\frac {13\pi }{60}}=\sin 39^{\circ }={\frac {2(1-{\sqrt {3}}){\sqrt {5-{\sqrt {5}}}}+{\sqrt {2}}({\sqrt {3}}+1)({\sqrt {5}}+1)}{16}}\,}
cos
13
π
60
=
cos
39
∘
=
2
(
1
+
3
)
5
−
5
+
2
(
3
−
1
)
(
5
+
1
)
16
{\displaystyle \cos {\frac {13\pi }{60}}=\cos 39^{\circ }={\frac {2(1+{\sqrt {3}}){\sqrt {5-{\sqrt {5}}}}+{\sqrt {2}}({\sqrt {3}}-1)({\sqrt {5}}+1)}{16}}\,}
tan
13
π
60
=
tan
39
∘
=
(
(
2
−
3
)
(
3
−
5
)
−
2
)
(
2
−
2
(
5
+
5
)
)
4
{\displaystyle \tan {\frac {13\pi }{60}}=\tan 39^{\circ }={\frac {\left((2-{\sqrt {3}})(3-{\sqrt {5}})-2\right)\left(2-{\sqrt {2(5+{\sqrt {5}})}}\right)}{4}}\,}
cot
13
π
60
=
cot
39
∘
=
(
(
2
+
3
)
(
3
−
5
)
−
2
)
(
2
+
2
(
5
+
5
)
)
4
{\displaystyle \cot {\frac {13\pi }{60}}=\cot 39^{\circ }={\frac {\left((2+{\sqrt {3}})(3-{\sqrt {5}})-2\right)\left(2+{\sqrt {2(5+{\sqrt {5}})}}\right)}{4}}\,}
sin
7
π
30
=
sin
42
∘
=
6
5
+
5
−
5
+
1
8
{\displaystyle \sin {\frac {7\pi }{30}}=\sin 42^{\circ }={\frac {{\sqrt {6}}{\sqrt {5+{\sqrt {5}}}}-{\sqrt {5}}+1}{8}}\,}
cos
7
π
30
=
cos
42
∘
=
2
5
+
5
+
3
(
5
−
1
)
8
{\displaystyle \cos {\frac {7\pi }{30}}=\cos 42^{\circ }={\frac {{\sqrt {2}}{\sqrt {5+{\sqrt {5}}}}+{\sqrt {3}}({\sqrt {5}}-1)}{8}}\,}
tan
7
π
30
=
tan
42
∘
=
3
(
5
+
1
)
−
2
5
+
5
2
{\displaystyle \tan {\frac {7\pi }{30}}=\tan 42^{\circ }={\frac {{\sqrt {3}}({\sqrt {5}}+1)-{\sqrt {2}}{\sqrt {5+{\sqrt {5}}}}}{2}}\,}
cot
7
π
30
=
cot
42
∘
=
2
(
25
−
11
5
)
+
3
(
3
−
5
)
2
{\displaystyle \cot {\frac {7\pi }{30}}=\cot 42^{\circ }={\frac {{\sqrt {2(25-11{\sqrt {5}})}}+{\sqrt {3}}(3-{\sqrt {5}})}{2}}\,}
sin
π
4
=
sin
45
∘
=
2
2
{\displaystyle \sin {\frac {\pi }{4}}=\sin 45^{\circ }={\frac {\sqrt {2}}{2}}\,}
cos
π
4
=
cos
45
∘
=
2
2
{\displaystyle \cos {\frac {\pi }{4}}=\cos 45^{\circ }={\frac {\sqrt {2}}{2}}\,}
tan
π
4
=
tan
45
∘
=
1
{\displaystyle \tan {\frac {\pi }{4}}=\tan 45^{\circ }=1\,}
cot
π
4
=
cot
45
∘
=
1
{\displaystyle \cot {\frac {\pi }{4}}=\cot 45^{\circ }=1\,}
sin
π
3
=
sin
60
∘
=
3
2
{\displaystyle \sin {\frac {\pi }{3}}=\sin 60^{\circ }={\frac {\sqrt {3}}{2}}\,}
cos
π
3
=
cos
60
∘
=
1
2
{\displaystyle \cos {\frac {\pi }{3}}=\cos 60^{\circ }={\frac {1}{2}}\,}
tan
π
3
=
tan
60
∘
=
3
{\displaystyle \tan {\frac {\pi }{3}}=\tan 60^{\circ }={\sqrt {3}}\,}
cot
π
3
=
cot
60
∘
=
3
3
{\displaystyle \cot {\frac {\pi }{3}}=\cot 60^{\circ }={\frac {\sqrt {3}}{3}}\,}
មាឌនៃសូលីតដែលបង្កើតដោយពហុកោណមានជ្រុង៥(បញ្ចកោណ) ហើយ
a
{\displaystyle a\,}
សំដែងដោយ
V
=
5
a
3
cos
36
∘
tan
2
36
∘
{\displaystyle V={\frac {5a^{3}\cos {36^{\circ }}}{\tan ^{2}{36^{\circ }}}}}
ដោយប្រើ
cos
36
∘
=
5
+
1
4
{\displaystyle \cos 36^{\circ }={\frac {{\sqrt {5}}+1}{4}}\,}
tan
36
∘
=
5
−
2
5
{\displaystyle \tan 36^{\circ }={\sqrt {5-2{\sqrt {5}}}}\,}
វាក្លាយទៅជា
V
=
a
3
(
15
+
7
5
)
4
{\displaystyle V={\frac {a^{3}(15+7{\sqrt {5}})}{4}}\,}
