Sine of 1 Degree
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This article needs to be tidied, namely:
This is insane.
There is believed to be a mistake here, possibly a typo, namely:
I believe the substituted value for $Q$ above may be wrong. It appears that $Q = -\dfrac 4 9$ has been used, but in the calculation above it may be the case that it should be $Q = -\dfrac 1 4$. The algebra has gone wrong somewhere and needs to be reviewed.
You can help ProofWiki by reviewing it, and either correcting it or adding some explanatory material as to why you believe it is actually correct after all.
This is insane.
It may also need to be brought up to our standard house style.
If you have done this or know it has been done, please remove {{Tidy}} from the code.
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 $R$ | ||||||||||
| \(\displaystyle \) | \(\) | \(\, \displaystyle + \, \) | \(\displaystyle \sqrt[3] {-\dfrac {\sin 3^\circ} 8 - \sqrt{-\left({ \dfrac 4 9}\right)^3 + \dfrac {\sin^2 3^\circ} {64} } }\) |
I believe the substituted value for $Q$ above may be wrong. It appears that $Q = -\dfrac 4 9$ has been used, but in the calculation above it may be the case that it should be $Q = -\dfrac 1 4$. The algebra has gone wrong somewhere and needs to be reviewed.
You can help ProofWiki by reviewing it, and either correcting it or adding some explanatory material as to why you believe it is actually correct after all.
To discuss this point in more detail, feel free to use the talk page.
If you are able to resolve this, then when you have done so you can remove this instance of {{Mistake}} from the code.
If you would welcome a second opinion as to whether your explanation is correct, add a call to {{Proofread}} the page (see the proofread template for usage).
| \(\displaystyle \sin 1^\circ\) | \(=\) | \(\displaystyle \sqrt[3] {-\dfrac{ \left({ \dfrac{ \sqrt{30} + \sqrt{10} - \sqrt 6 - \sqrt 2 - 2\sqrt {15 + 3 \sqrt5 } + 2\sqrt {5 + \sqrt5 } }{16} }\right) }{8} + \sqrt{-\left({ \dfrac{4}{9} }\right)^3 + \dfrac{ \left({ \dfrac{ \sqrt{30} + \sqrt{10} - \sqrt 6 - \sqrt 2 - 2\sqrt {15 + 3 \sqrt5 } + 2\sqrt {5 + \sqrt5 } }{16} }\right)^2 }{64} } }\) | |||||||||||
| \(\displaystyle \) | \(\) | \(\, \displaystyle + \, \) | \(\displaystyle \sqrt[3] {-\dfrac{ \left({ \dfrac{ \sqrt{30} + \sqrt{10} - \sqrt 6 - \sqrt 2 - 2\sqrt {15 + 3 \sqrt5 } + 2\sqrt {5 + \sqrt5 } }{16} }\right) }{8} - \sqrt{-\left({ \dfrac{4}{9} }\right)^3 + \dfrac{\left({ \dfrac{ \sqrt{30} + \sqrt{10} - \sqrt 6 - \sqrt 2 - 2\sqrt {15 + 3 \sqrt5 } + 2\sqrt {5 + \sqrt5 } }{16} }\right)^2}{64} } }\) | ||||||||||
| \(\displaystyle \) | \(=\) | \(\displaystyle \sqrt[3] {-\dfrac{ \left({ \dfrac{ \sqrt{30} + \sqrt{10} - \sqrt 6 - \sqrt 2 - 2\sqrt {15 + 3 \sqrt5 } + 2\sqrt {5 + \sqrt5 } }{16} }\right) }{8} + \sqrt{-\left({ \dfrac{4}{9} }\right)^3 + \dfrac{ \left({ \dfrac{-8 + 24\sqrt3 - 16\sqrt5 - 8\sqrt{15} - 8\sqrt {50 + 10\sqrt5} + 8\sqrt {10 + 2 \sqrt5} - 8\sqrt{90 + 30\sqrt5} }{16^2} }\right) }{64} } }\) | |||||||||||
| \(\displaystyle \) | \(\) | \(\, \displaystyle + \, \) | \(\displaystyle \sqrt[3] {-\dfrac{ \left({ \dfrac{ \sqrt{30} + \sqrt{10} - \sqrt 6 - \sqrt 2 - 2\sqrt {15 + 3 \sqrt5 } + 2\sqrt {5 + \sqrt5 } }{16} }\right) }{8} - \sqrt{-\left({ \dfrac{4}{9} }\right)^3 + \dfrac{\left({ \dfrac{-8 + 24\sqrt3 - 16\sqrt5 - 8\sqrt{15} - 8\sqrt {50 + 10\sqrt5} + 8\sqrt {10 + 2 \sqrt5} - 8\sqrt{90 + 30\sqrt5} }{16^2} }\right)}{64} } }\) | ||||||||||
| \(\displaystyle \) | \(=\) | \(\displaystyle \sqrt[3] {-\dfrac{ \sqrt{30} + \sqrt{10} - \sqrt 6 - \sqrt 2 - 2\sqrt {15 + 3 \sqrt5 } + 2\sqrt {5 + \sqrt5 } } {128} + \sqrt{-\dfrac{64}{729} + \dfrac{ -8 + 24\sqrt3 - 16\sqrt5 - 8\sqrt{15} - 8\sqrt {50 + 10\sqrt5} + 8\sqrt {10 + 2 \sqrt5} - 8\sqrt{90 + 30\sqrt5} }{16384} } }\) | |||||||||||
| \(\displaystyle \) | \(\) | \(\, \displaystyle + \, \) | \(\displaystyle \sqrt[3] {-\dfrac{ \sqrt{30} + \sqrt{10} - \sqrt 6 - \sqrt 2 - 2\sqrt {15 + 3 \sqrt5 } + 2\sqrt {5 + \sqrt5 } } {128} - \sqrt{-\dfrac{64}{729} + \dfrac{ -8 + 24\sqrt3 - 16\sqrt5 - 8\sqrt{15} - 8\sqrt {50 + 10\sqrt5} + 8\sqrt {10 + 2 \sqrt5} - 8\sqrt{90 + 30\sqrt5} }{16384} } }\) | ||||||||||
| \(\displaystyle \) | \(=\) | \(\displaystyle \sqrt[3] {-\dfrac{ \sqrt{30} + \sqrt{10} - \sqrt 6 - \sqrt 2 - 2\sqrt {15 + 3 \sqrt5 } + 2\sqrt {5 + \sqrt5 } } {128} + \sqrt{-\dfrac{64}{729} + \dfrac{ -1 + 3\sqrt3 - 2\sqrt5 - \sqrt{15} - \sqrt {50 + 10\sqrt5} + \sqrt {10 + 2 \sqrt5} - \sqrt{90 + 30\sqrt5} }{2048} } }\) | |||||||||||
| \(\displaystyle \) | \(\) | \(\, \displaystyle + \, \) | \(\displaystyle \sqrt[3] {-\dfrac{ \sqrt{30} + \sqrt{10} - \sqrt 6 - \sqrt 2 - 2\sqrt {15 + 3 \sqrt5 } + 2\sqrt {5 + \sqrt5 } } {128} - \sqrt{-\dfrac{64}{729} + \dfrac{ -1 + 3\sqrt3 - 2\sqrt5 - \sqrt{15} - \sqrt {50 + 10\sqrt5} + \sqrt {10 + 2 \sqrt5} - \sqrt{90 + 30\sqrt5} }{2048} } }\) |
$\blacksquare$
