Sine of 1 Degree

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Theorem

$\sin 1^\circ = \sin \dfrac {\pi} {180} = $

where $\sin$ denotes the sine function.


Proof

\(\displaystyle \sin \left({3 \times 1^\circ}\right)\) \(=\) \(\displaystyle 3 \sin 1^\circ - 4 \sin^3 1^\circ\) Triple Angle Formula for Sine
\(\displaystyle \sin 3^\circ\) \(=\) \(\displaystyle 3 \sin 1^\circ - 4 \sin^3 1^\circ\)
\(\displaystyle 4 \sin^3 1^\circ - 3 \sin 1^\circ + \sin 3^\circ\) \(=\) \(\displaystyle 0\)


This is in the form:

$a x^3 + b x^2 + c x + d = 0$

where:

$x = \sin 1^\circ$
$a = 4$
$b = 0$
$c = -3$
$d = \sin 3^\circ$


From Cardano's Formula:

$x = S + T$

where:

$S = \sqrt [3] {R + \sqrt{Q^3 + R^2}}$
$T = \sqrt [3] {R - \sqrt{Q^3 + R^2}}$

where:

\(\displaystyle Q\) \(=\) \(\displaystyle \dfrac {3 a c - b^2} {9 a^2}\)
\(\displaystyle \) \(=\) \(\displaystyle \dfrac {3 \times 4 \times \left({-3}\right) - 0^2} {9 \times 4^2}\)
\(\displaystyle \) \(=\) \(\displaystyle -\dfrac 1 4\)

and:

\(\displaystyle R\) \(=\) \(\displaystyle \dfrac {9 a b c - 27 a^2 d - 2 b^3} {54 a^3}\)
\(\displaystyle \) \(=\) \(\displaystyle \dfrac {9 \times 4 \times 0 \times \left({-3}\right) - 27 \times 4^2 \times \sin 3^\circ - 2 \times 0^3} {54 \times 4^3}\)
\(\displaystyle \) \(=\) \(\displaystyle \dfrac {\sin 3^\circ} 8\)


Thus:

\(\displaystyle \sin 1^\circ\) \(=\) \(\displaystyle S + T\) putting $x = S + T$
\(\displaystyle \) \(=\) \(\displaystyle \sqrt[3] {R + \sqrt{Q^3 + R^2} } + \sqrt[3] {R - \sqrt{Q^3 + R^2} }\) substituting for $S$ and $T$
\(\displaystyle \) \(=\) \(\displaystyle \sqrt[3] {R + \sqrt{-\left({\dfrac 4 9}\right)^3 + R^2} } + \sqrt[3] {R - \sqrt {-\left({\dfrac 4 9}\right)^3 + R^2} }\) substituting for $Q$
\(\displaystyle \) \(=\) \(\displaystyle \sqrt[3] {-\dfrac{\sin 3^\circ} 8 + \sqrt{-\left({ \dfrac 4 9}\right)^3 + \dfrac {\sin^2 3^\circ} {64} } }\) substituting for $R$
\(\displaystyle \) \(\) \(\, \displaystyle + \, \) \(\displaystyle \sqrt[3] {-\dfrac {\sin 3^\circ} 8 - \sqrt{-\left({ \dfrac 4 9}\right)^3 + \dfrac {\sin^2 3^\circ} {64} } }\)



By Sine of 3 Degrees:

\(\displaystyle \sin 1^\circ\) \(=\) \(\displaystyle \sqrt[3] {-\dfrac{ \left({ \dfrac{ \sqrt{30} + \sqrt{10} - \sqrt 6 - \sqrt 2 - 2\sqrt {15 + 3 \sqrt5 } + 2\sqrt {5 + \sqrt5 } }{16} }\right) }{8} + \sqrt{-\left({ \dfrac{4}{9} }\right)^3 + \dfrac{ \left({ \dfrac{ \sqrt{30} + \sqrt{10} - \sqrt 6 - \sqrt 2 - 2\sqrt {15 + 3 \sqrt5 } + 2\sqrt {5 + \sqrt5 } }{16} }\right)^2 }{64} } }\)
\(\displaystyle \) \(\) \(\, \displaystyle + \, \) \(\displaystyle \sqrt[3] {-\dfrac{ \left({ \dfrac{ \sqrt{30} + \sqrt{10} - \sqrt 6 - \sqrt 2 - 2\sqrt {15 + 3 \sqrt5 } + 2\sqrt {5 + \sqrt5 } }{16} }\right) }{8} - \sqrt{-\left({ \dfrac{4}{9} }\right)^3 + \dfrac{\left({ \dfrac{ \sqrt{30} + \sqrt{10} - \sqrt 6 - \sqrt 2 - 2\sqrt {15 + 3 \sqrt5 } + 2\sqrt {5 + \sqrt5 } }{16} }\right)^2}{64} } }\)
\(\displaystyle \) \(=\) \(\displaystyle \sqrt[3] {-\dfrac{ \left({ \dfrac{ \sqrt{30} + \sqrt{10} - \sqrt 6 - \sqrt 2 - 2\sqrt {15 + 3 \sqrt5 } + 2\sqrt {5 + \sqrt5 } }{16} }\right) }{8} + \sqrt{-\left({ \dfrac{4}{9} }\right)^3 + \dfrac{ \left({ \dfrac{-8 + 24\sqrt3 - 16\sqrt5 - 8\sqrt{15} - 8\sqrt {50 + 10\sqrt5} + 8\sqrt {10 + 2 \sqrt5} - 8\sqrt{90 + 30\sqrt5} }{16^2} }\right) }{64} } }\)
\(\displaystyle \) \(\) \(\, \displaystyle + \, \) \(\displaystyle \sqrt[3] {-\dfrac{ \left({ \dfrac{ \sqrt{30} + \sqrt{10} - \sqrt 6 - \sqrt 2 - 2\sqrt {15 + 3 \sqrt5 } + 2\sqrt {5 + \sqrt5 } }{16} }\right) }{8} - \sqrt{-\left({ \dfrac{4}{9} }\right)^3 + \dfrac{\left({ \dfrac{-8 + 24\sqrt3 - 16\sqrt5 - 8\sqrt{15} - 8\sqrt {50 + 10\sqrt5} + 8\sqrt {10 + 2 \sqrt5} - 8\sqrt{90 + 30\sqrt5} }{16^2} }\right)}{64} } }\)
\(\displaystyle \) \(=\) \(\displaystyle \sqrt[3] {-\dfrac{ \sqrt{30} + \sqrt{10} - \sqrt 6 - \sqrt 2 - 2\sqrt {15 + 3 \sqrt5 } + 2\sqrt {5 + \sqrt5 } } {128} + \sqrt{-\dfrac{64}{729} + \dfrac{ -8 + 24\sqrt3 - 16\sqrt5 - 8\sqrt{15} - 8\sqrt {50 + 10\sqrt5} + 8\sqrt {10 + 2 \sqrt5} - 8\sqrt{90 + 30\sqrt5} }{16384} } }\)
\(\displaystyle \) \(\) \(\, \displaystyle + \, \) \(\displaystyle \sqrt[3] {-\dfrac{ \sqrt{30} + \sqrt{10} - \sqrt 6 - \sqrt 2 - 2\sqrt {15 + 3 \sqrt5 } + 2\sqrt {5 + \sqrt5 } } {128} - \sqrt{-\dfrac{64}{729} + \dfrac{ -8 + 24\sqrt3 - 16\sqrt5 - 8\sqrt{15} - 8\sqrt {50 + 10\sqrt5} + 8\sqrt {10 + 2 \sqrt5} - 8\sqrt{90 + 30\sqrt5} }{16384} } }\)
\(\displaystyle \) \(=\) \(\displaystyle \sqrt[3] {-\dfrac{ \sqrt{30} + \sqrt{10} - \sqrt 6 - \sqrt 2 - 2\sqrt {15 + 3 \sqrt5 } + 2\sqrt {5 + \sqrt5 } } {128} + \sqrt{-\dfrac{64}{729} + \dfrac{ -1 + 3\sqrt3 - 2\sqrt5 - \sqrt{15} - \sqrt {50 + 10\sqrt5} + \sqrt {10 + 2 \sqrt5} - \sqrt{90 + 30\sqrt5} }{2048} } }\)
\(\displaystyle \) \(\) \(\, \displaystyle + \, \) \(\displaystyle \sqrt[3] {-\dfrac{ \sqrt{30} + \sqrt{10} - \sqrt 6 - \sqrt 2 - 2\sqrt {15 + 3 \sqrt5 } + 2\sqrt {5 + \sqrt5 } } {128} - \sqrt{-\dfrac{64}{729} + \dfrac{ -1 + 3\sqrt3 - 2\sqrt5 - \sqrt{15} - \sqrt {50 + 10\sqrt5} + \sqrt {10 + 2 \sqrt5} - \sqrt{90 + 30\sqrt5} }{2048} } }\)

$\blacksquare$