Theorem
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\(\ds \sin 1 \degrees = \sin \dfrac \pi {180}\)
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\(=\)
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\(\ds \paren {\dfrac 1 8 + i \dfrac {\sqrt 3} 8} \paren {\sqrt [3] {\paren {\dfrac {\sqrt {30} + \sqrt {10} - \sqrt 6 - \sqrt 2 - 2 \sqrt {15 + 3 \sqrt 5} + 2 \sqrt {5 + \sqrt 5} } 2} + i \sqrt {32 - 6 \sqrt 3 + 2 \sqrt {15} + 2 \sqrt {50 + 10 \sqrt 5} - 2 \sqrt {10 + 2 \sqrt 5} + 2 \sqrt {90 + 30 \sqrt 5} } } }\)
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\(\ds \)
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\(\)
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\(\, \ds - \, \)
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\(\ds \paren {\dfrac 1 8 + i \dfrac {\sqrt 3} 8} \paren {\sqrt [3] {-\paren {\dfrac {\sqrt {30} + \sqrt {10} - \sqrt 6 - \sqrt 2 - 2 \sqrt {15 + 3 \sqrt 5} + 2 \sqrt {5 + \sqrt 5} } 2} + i \sqrt {32 - 6 \sqrt 3 + 2 \sqrt {15} + 2 \sqrt {50 + 10 \sqrt 5} - 2\sqrt {10 + 2 \sqrt 5} + 2\sqrt {90 + 30 \sqrt 5} } } }\)
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where $\sin$ denotes the sine function.
Proof
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\(\ds \map \sin {3 \times 1 \degrees}\)
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\(=\)
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\(\ds 3 \sin 1 \degrees - 4 \sin^3 1 \degrees\)
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Triple Angle Formula for Sine
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\(\ds \sin 3 \degrees\)
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\(=\)
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\(\ds 3 \sin 1 \degrees - 4 \sin^3 1 \degrees\)
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\(\ds 4 \sin^3 1 \degrees - 3 \sin 1 \degrees + \sin 3 \degrees\)
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\(=\)
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\(\ds 0\)
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This is in the form:
- $a x^3 + b x^2 + c x + d = 0$
where:
- $x = \sin 1 \degrees$
- $a = 4$
- $b = 0$
- $c = -3$
- $d = \sin 3 \degrees$
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:
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\(\ds Q\)
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\(=\)
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\(\ds \dfrac {3 a c - b^2} {9 a^2}\)
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\(\ds \)
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\(=\)
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\(\ds \dfrac {3 \times 4 \times \paren {-3} - 0^2} {9 \times 4^2}\)
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\(\ds \)
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\(=\)
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\(\ds -\dfrac 1 4\)
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and:
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\(\ds R\)
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\(=\)
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\(\ds \dfrac {9 a b c - 27 a^2 d - 2 b^3} {54 a^3}\)
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\(\ds \)
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\(=\)
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\(\ds \dfrac {9 \times 4 \times 0 \times \paren {-3} - 27 \times 4^2 \times \sin 3 \degrees - 2 \times 0^3} {54 \times 4^3}\)
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\(\ds \)
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\(=\)
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\(\ds -\dfrac {\sin 3 \degrees} 8\)
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Thus:
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\(\ds \sin 1 \degrees\)
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\(=\)
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\(\ds S + T\)
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putting $x = S + T$
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\(\ds \)
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\(=\)
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\(\ds \sqrt [3] {R + \sqrt {Q^3 + R^2} } + \sqrt[3] {R - \sqrt {Q^3 + R^2} }\)
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substituting for $S$ and $T$
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\(\ds \)
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\(=\)
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\(\ds \sqrt [3] {R + \sqrt {\paren {-\dfrac 1 4}^3 + R^2} } + \sqrt [3] {R - \sqrt {\paren {-\dfrac 1 4}^3 + R^2} }\)
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substituting for $Q$
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\(\ds \)
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\(=\)
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\(\ds \sqrt [3] {-\dfrac {\sin 3 \degrees} 8 + \sqrt {\paren {-\dfrac 1 4}^3 + \dfrac {\sin^2 3 \degrees} {64} } } + \sqrt [3] {-\dfrac {\sin 3 \degrees} 8 - \sqrt {\paren {-\dfrac 1 4}^3 + \dfrac {\sin^2 3 \degrees} {64} } }\)
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substituting for $R$
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\(\ds \)
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\(=\)
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\(\ds \sqrt [3] {-\dfrac {\sin 3 \degrees} 8 + \sqrt {\dfrac {\sin^2 3 \degrees - 1} {64} } } + \sqrt [3] {-\dfrac {\sin 3 \degrees} 8 - \sqrt {\dfrac {\sin^2 3 \degrees - 1} {64} } }\)
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\(\ds \)
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\(=\)
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\(\ds \sqrt [3] {-\dfrac {\sin 3 \degrees} 8 + \sqrt {\dfrac {-\cos^2 3 \degrees} {64} } } + \sqrt [3] {-\dfrac {\sin 3 \degrees} 8 - \sqrt {\dfrac {-\cos^2 3 \degrees} {64} } }\)
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Sum of Squares of Sine and Cosine
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\(\ds \)
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\(=\)
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\(\ds \sqrt [3] {-\dfrac {\sin 3 \degrees} 8 + i \dfrac {\cos 3 \degrees} 8 } + \sqrt [3] {-\dfrac {\sin 3 \degrees} 8 - i \dfrac {\cos 3 \degrees} 8}\)
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$i$ is a square root of $-1$, that is, $i = \sqrt {-1}$.
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\(\ds \)
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\(=\)
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\(\ds \sqrt [3] {-\dfrac {\sin 3 \degrees} 8 + i \dfrac {\cos 3 \degrees} 8 } + \sqrt [3] {-1 \paren {\dfrac {\sin 3 \degrees} 8 + i \dfrac {\cos 3 \degrees} 8} }\)
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\(\ds \)
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\(=\)
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\(\ds \dfrac 1 2 \sqrt [3] {-\sin 3 \degrees + i \cos 3 \degrees } - \dfrac 1 2 \sqrt [3] {\sin 3 \degrees + i \cos 3 \degrees}\)
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\(\ds \)
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\(=\)
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\(\ds \dfrac 1 2 \sqrt [3] {-\cos 87 \degrees + i \sin 87 \degrees } - \dfrac 1 2 \sqrt [3] {\cos 87 \degrees + i \sin 87 \degrees}\)
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Sine of Complement equals Cosine
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\(\ds \)
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\(=\)
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\(\ds \dfrac 1 2 \sqrt [3] {\cos 93 \degrees + i \sin 93 \degrees } - \dfrac 1 2 \sqrt [3] {\cos 87 \degrees + i \sin 87 \degrees}\)
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\(\ds \)
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\(=\)
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\(\ds \dfrac 1 2 \paren {\sqrt [3] {\cis 93 \degrees} - \sqrt [3] {\cis 87 \degrees} }\)
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KEY EQUATION
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- $\sqrt [3] {\cis 93 \degrees}$ has $3$ unique cube roots.
$\tuple {\cis 31 \degrees, \cis 151 \degrees, \cis 271 \degrees}$
- $\sqrt [3] {\cis 87 \degrees }$ also has $3$ unique cube roots.
$\tuple {\cis 29 \degrees, \cis 149 \degrees, \cis 269 \degrees}$
We need to investigate $9$ differences and find solutions where the complex part vanishes.
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\(\ds 1) \ \ \)
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\(\ds \dfrac 1 2 \paren {\sqrt [3] {\cis 93 \degrees } - \sqrt [3] {\cis 87 \degrees } }\)
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\(=\)
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\(\ds \dfrac 1 2 \paren { \paren {\cos 31 \degrees + i \sin 31 \degrees } - \paren {\cos 29 \degrees + i \sin 29 \degrees } }\)
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No. $\paren {\sin 31 \degrees - \sin 29 \degrees } > 0$
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\(\ds 2) \ \ \)
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\(\ds \)
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\(=\)
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\(\ds \dfrac 1 2 \paren { \paren {\cos 31 \degrees + i \sin 31 \degrees } - \paren {\cos 149 \degrees + i \sin 149 \degrees } }\)
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Yes. $\paren {\sin 31 \degrees - \sin 149 \degrees } = 0$
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\(\ds 3) \ \ \)
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\(\ds \)
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\(=\)
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\(\ds \dfrac 1 2 \paren { \paren {\cos 31 \degrees + i \sin 31 \degrees } - \paren {\cos 269 \degrees + i \sin 269 \degrees } }\)
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No. $\paren {\sin 31 \degrees - \sin 269 \degrees } > 0$
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\(\ds 4) \ \ \)
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\(\ds \)
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\(=\)
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\(\ds \dfrac 1 2 \paren { \paren {\cos 151 \degrees + i \sin 151 \degrees } - \paren {\cos 29 \degrees + i \sin 29 \degrees } }\)
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Yes. $\paren {\sin 151 \degrees - \sin 29 \degrees } = 0$
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\(\ds 5) \ \ \)
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\(\ds \)
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\(=\)
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\(\ds \dfrac 1 2 \paren { \paren {\cos 151 \degrees + i \sin 151 \degrees } - \paren {\cos 149 \degrees + i \sin 149 \degrees } }\)
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No. $\paren {\sin 151 \degrees - \sin 149 \degrees } < 0$
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\(\ds 6) \ \ \)
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\(\ds \)
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\(=\)
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\(\ds \dfrac 1 2 \paren { \paren {\cos 151 \degrees + i \sin 151 \degrees } - \paren {\cos 269 \degrees + i \sin 269 \degrees } }\)
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No. $\paren {\sin 151 \degrees - \sin 269 \degrees } > 0$
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\(\ds 7) \ \ \)
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\(\ds \)
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\(=\)
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\(\ds \dfrac 1 2 \paren { \paren {\cos 271 \degrees + i \sin 271 \degrees } - \paren {\cos 29 \degrees + i \sin 29 \degrees } }\)
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No. $\paren {\sin 271 \degrees - \sin 29 \degrees } < 0$
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\(\ds 8) \ \ \)
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\(\ds \)
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\(=\)
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\(\ds \dfrac 1 2 \paren { \paren {\cos 271 \degrees + i \sin 271 \degrees } - \paren {\cos 149 \degrees + i \sin 149 \degrees } }\)
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No. $\paren {\sin 271 \degrees - \sin 149 \degrees } < 0$
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\(\ds 9) \ \ \)
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\(\ds \)
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\(=\)
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\(\ds \dfrac 1 2 \paren { \paren {\cos 271 \degrees + i \sin 271 \degrees } - \paren {\cos 269 \degrees + i \sin 269 \degrees } }\)
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Yes. $\paren {\sin 271 \degrees - \sin 269 \degrees } = 0$
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In solution $2)$ above, we rotate $\sqrt [3] {\cis 87 \degrees }$ by $120 \degrees$ (multiply by $\paren {-\dfrac 1 2 + i \dfrac {\sqrt 3} 2}$)
In solution $4)$ above, we rotate $\sqrt [3] {\cis 93 \degrees }$ by $120 \degrees$ (multiply by $\paren {-\dfrac 1 2 + i \dfrac {\sqrt 3} 2}$)
In solution $9)$ above, we would rotate BOTH $\sqrt [3] {\cis 93 \degrees}$ and $\sqrt [3] {\cis 87 \degrees}$ by $240 \degrees$ (multiply by $\paren {-\dfrac 1 2 - i \dfrac {\sqrt 3} 2}$)
For the remainder of this proof, we will arbitrarily choose solution $9)$:
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\(\ds \sin 1 \degrees\)
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\(=\)
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\(\ds \paren { -\dfrac 1 2 - i \dfrac {\sqrt 3} 2 } \dfrac 1 2 \paren {\sqrt [3] {\cis 93 \degrees } - \sqrt [3] {\cis 87 \degrees } }\)
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Solution $9)$ above - rotating by $240 \degrees$
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\(\ds \)
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\(=\)
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\(\ds \paren { -\dfrac 1 4 - i \dfrac {\sqrt 3} 4 } \paren { \sqrt [3] {-\cos 87 \degrees + i \sin 87 \degrees } - \sqrt [3] {\cos 87 \degrees + i \sin 87 \degrees } }\)
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\(\ds \)
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\(=\)
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\(\ds \paren { -\dfrac 1 4 - i \dfrac {\sqrt 3} 4 } \paren { \sqrt [3] {-\sin 3 \degrees + i \cos 3 \degrees } - \sqrt [3] {\sin 3 \degrees + i \cos 3 \degrees } }\)
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Sine of Complement equals Cosine
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\(\ds \)
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\(=\)
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\(\ds \paren { -\dfrac 1 4 - i \dfrac {\sqrt 3} 4 } \paren { \sqrt [3] {-\sin 3 \degrees + i \sqrt { 1 - \sin^2 3 \degrees } } - \sqrt [3] {\sin 3 \degrees + i \sqrt {1 - \sin^2 3 \degrees } } }\)
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Sum of Squares of Sine and Cosine
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\(\ds \)
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\(=\)
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\(\ds \paren { -\dfrac 1 4 - i \dfrac {\sqrt 3} 4 } \paren { \sqrt [3] {-\paren {\dfrac {\sqrt {30} + \sqrt {10} - \sqrt 6 - \sqrt 2 - 2 \sqrt {15 + 3 \sqrt 5} + 2 \sqrt {5 + \sqrt 5} } {16} } + i \sqrt { 1 - \paren {\dfrac {\sqrt {30} + \sqrt {10} - \sqrt 6 - \sqrt 2 - 2 \sqrt {15 + 3 \sqrt 5} + 2 \sqrt {5 + \sqrt 5} } {16} }^2 } } }\)
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Sine of 3 Degrees
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\(\ds \)
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\(\)
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\(\, \ds - \, \)
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\(\ds \paren {-\dfrac 1 4 - i \dfrac {\sqrt 3} 4 } \paren {\sqrt [3] {\paren {\dfrac {\sqrt {30} + \sqrt {10} - \sqrt 6 - \sqrt 2 - 2 \sqrt {15 + 3 \sqrt 5} + 2 \sqrt {5 + \sqrt 5} } {16} } + i \sqrt { 1 - \paren {\dfrac {\sqrt {30} + \sqrt {10} - \sqrt 6 - \sqrt 2 - 2 \sqrt {15 + 3 \sqrt 5} + 2 \sqrt {5 + \sqrt 5} } {16} }^2 } } }\)
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Squaring the numerator of the Sine of 3 Degrees, we get:
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\(\ds \paren {\sqrt {30} + \sqrt {10} - \sqrt 6 - \sqrt 2 - 2 \sqrt {15 + 3 \sqrt 5} + 2 \sqrt {5 + \sqrt 5} }^2\)
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\(=\)
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\(\ds 30 + 10\sqrt 3 - 6\sqrt 5 - 2\sqrt {15} - 6 \sqrt {50 + 10 \sqrt 5} + 2 \sqrt {150 + 30\sqrt 5}\)
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\(\ds \)
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\(\)
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\(\, \ds + \, \)
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\(\ds 10\sqrt 3 + 10 - 2\sqrt {15} - 2\sqrt 5 - 2 \sqrt {150 + 30\sqrt 5} + 2 \sqrt {50 + 10 \sqrt 5}\)
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\(\ds \)
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\(\)
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\(\, \ds - \, \)
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\(\ds 6\sqrt 5 - 2\sqrt {15} + 6 + 2\sqrt 3 + 6 \sqrt {10 + 2 \sqrt 5} - 2 \sqrt {30 + 6\sqrt 5}\)
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\(\ds \)
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\(\)
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\(\, \ds - \, \)
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\(\ds 2\sqrt {15} - 2\sqrt 5 + 2\sqrt 3 + 2 + 2 \sqrt {30 + 6 \sqrt 5} - 2 \sqrt {10 + 2\sqrt 5}\)
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\(\ds \)
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\(\)
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\(\, \ds - \, \)
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\(\ds 6 \sqrt {50 + 10 \sqrt 5} - 2 \sqrt {150 + 30\sqrt 5} + 6 \sqrt {10 + 2 \sqrt 5} + 2 \sqrt {30 + 6 \sqrt 5} + \paren {60 + 12 \sqrt 5 } - 4 \sqrt {90 + 30\sqrt 5}\)
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\(\ds \)
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\(\)
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\(\, \ds + \, \)
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\(\ds 2 \sqrt {150 + 30\sqrt 5} + 2 \sqrt {50 + 10 \sqrt 5} - 2 \sqrt {30 + 6\sqrt 5} - 2 \sqrt {10 + 2\sqrt 5} - 4 \sqrt {90 + 30\sqrt 5} + \paren {20 + 4 \sqrt 5 }\)
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\(\ds \)
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\(=\)
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\(\ds 128 + 24 \sqrt 3 - 8 \sqrt {15} - 8 \sqrt {50 + 10 \sqrt 5} + 8 \sqrt {10 + 2 \sqrt 5} - 8 \sqrt {90 + 30\sqrt 5}\)
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Aggregating terms
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Bringing that result back into our main result, we get:
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\(\ds \)
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\(=\)
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\(\ds \paren {-\dfrac 1 4 - i \dfrac {\sqrt 3} 4 } \paren { \sqrt [3] {-\paren {\dfrac {\sqrt {30} + \sqrt {10} - \sqrt 6 - \sqrt 2 - 2 \sqrt {15 + 3 \sqrt 5} + 2 \sqrt {5 + \sqrt 5} } {16} } + i \sqrt { 1 - \paren {\dfrac {128 + 24 \sqrt 3 - 8 \sqrt {15} - 8 \sqrt {50 + 10 \sqrt 5} + 8 \sqrt {10 + 2 \sqrt 5} - 8 \sqrt {90 + 30 \sqrt 5} } {256} } } } }\)
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Sine of 3 Degrees
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\(\ds \)
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\(\)
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\(\, \ds - \, \)
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\(\ds \paren {-\dfrac 1 4 - i \dfrac {\sqrt 3} 4} \paren { \sqrt [3] {\paren {\dfrac {\sqrt {30} + \sqrt {10} - \sqrt 6 - \sqrt 2 - 2 \sqrt {15 + 3 \sqrt 5} + 2 \sqrt {5 + \sqrt 5} } {16} } + i \sqrt { 1 - \paren {\dfrac {128 + 24 \sqrt 3 - 8 \sqrt {15} - 8 \sqrt {50 + 10 \sqrt 5} + 8 \sqrt {10 + 2 \sqrt 5} - 8 \sqrt {90 + 30 \sqrt 5} } {256} } } } }\)
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\(\ds \)
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\(=\)
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\(\ds \paren {-\dfrac 1 4 - i \dfrac {\sqrt 3} 4} \paren { \sqrt [3] {-\paren {\dfrac {\sqrt {30} + \sqrt {10} - \sqrt 6 - \sqrt 2 - 2 \sqrt {15 + 3 \sqrt 5} + 2 \sqrt {5 + \sqrt 5} } {16} } + i \sqrt {\paren {\dfrac {128 - 24 \sqrt 3 + 8 \sqrt {15} + 8 \sqrt {50 + 10 \sqrt 5} - 8 \sqrt {10 + 2 \sqrt 5} + 8 \sqrt {90 + 30 \sqrt 5} } {256} } } } }\)
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$\paren {1 - \dfrac {128} {256} } = \dfrac {128} {256}$
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\(\ds \)
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\(\)
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\(\, \ds - \, \)
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\(\ds \paren { -\dfrac 1 4 - i \dfrac {\sqrt {3} } 4 } \paren { \sqrt [3] {\paren {\dfrac {\sqrt {30} + \sqrt {10} - \sqrt 6 - \sqrt 2 - 2 \sqrt {15 + 3 \sqrt 5} + 2 \sqrt {5 + \sqrt 5} } {16} } + i \sqrt {\paren {\dfrac {128 - 24 \sqrt 3 + 8 \sqrt {15} + 8 \sqrt {50 + 10 \sqrt 5} - 8 \sqrt {10 + 2 \sqrt 5} + 8 \sqrt {90 + 30\sqrt 5} } {256} } } } }\)
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Canceling $4$ from the numerator and denominator, we get:
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\(\ds \)
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\(=\)
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\(\ds \paren { -\dfrac 1 4 - i \dfrac {\sqrt 3} 4 } \paren { \sqrt [3] {-\paren {\dfrac {\sqrt {30} + \sqrt {10} - \sqrt 6 - \sqrt 2 - 2 \sqrt {15 + 3 \sqrt 5} + 2 \sqrt {5 + \sqrt 5} } {16} } + i \sqrt {\paren {\dfrac {32 - 6 \sqrt 3 + 2\sqrt {15} + 2\sqrt {50 + 10 \sqrt 5} - 2\sqrt {10 + 2 \sqrt 5} + 2\sqrt {90 + 30\sqrt 5} } {64} } } } }\)
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\(\ds \)
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\(\)
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\(\, \ds - \, \)
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\(\ds \paren {-\dfrac 1 4 - i \dfrac {\sqrt 3} 4} \paren { \sqrt [3] {\paren {\dfrac {\sqrt {30} + \sqrt {10} - \sqrt 6 - \sqrt 2 - 2 \sqrt {15 + 3 \sqrt 5} + 2 \sqrt {5 + \sqrt 5} } {16} } + i \sqrt {\paren {\dfrac {32 - 6 \sqrt 3 + 2\sqrt {15} + 2\sqrt {50 + 10 \sqrt 5} - 2\sqrt {10 + 2 \sqrt 5} + 2\sqrt {90 + 30\sqrt 5} } {64} } } } }\)
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Moving an $8$ to the outside, we get:
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\(\ds \)
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\(=\)
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\(\ds \paren {-\dfrac 1 8 - i \dfrac {\sqrt 3} 8} \paren { \sqrt [3] {-\paren {\dfrac {\sqrt {30} + \sqrt {10} - \sqrt 6 - \sqrt 2 - 2 \sqrt {15 + 3 \sqrt 5} + 2 \sqrt {5 + \sqrt 5} } 2} + i \sqrt {32 - 6 \sqrt 3 + 2 \sqrt {15} + 2 \sqrt {50 + 10 \sqrt 5} - 2 \sqrt {10 + 2 \sqrt 5} + 2 \sqrt {90 + 30 \sqrt 5} } } }\)
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\(\ds \)
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\(\)
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\(\, \ds - \, \)
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\(\ds \paren {-\dfrac 1 8 - i \dfrac {\sqrt 3} 8} \paren { \sqrt [3] {\paren {\dfrac {\sqrt {30} + \sqrt {10} - \sqrt 6 - \sqrt 2 - 2 \sqrt {15 + 3 \sqrt 5} + 2 \sqrt {5 + \sqrt 5} } 2} + i \sqrt {32 - 6 \sqrt 3 + 2 \sqrt {15} + 2 \sqrt {50 + 10 \sqrt 5} - 2 \sqrt {10 + 2 \sqrt 5} + 2 \sqrt {90 + 30 \sqrt 5} } } }\)
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Flipping the top and bottom lines, we get our main result:
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\(\ds \)
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\(=\)
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\(\ds \paren {\dfrac 1 8 + i \dfrac {\sqrt 3} 8} \paren {\sqrt [3] {\paren {\dfrac {\sqrt {30} + \sqrt {10} - \sqrt 6 - \sqrt 2 - 2 \sqrt {15 + 3 \sqrt 5} + 2 \sqrt {5 + \sqrt 5} } 2} + i \sqrt {32 - 6 \sqrt 3 + 2 \sqrt {15} + 2 \sqrt {50 + 10 \sqrt 5} - 2 \sqrt {10 + 2 \sqrt 5} + 2 \sqrt {90 + 30 \sqrt 5} } } }\)
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\(\ds \)
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\(\)
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\(\, \ds - \, \)
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\(\ds \paren { \dfrac 1 8 + i \dfrac {\sqrt {3} } 8 } \paren { \sqrt [3] {-\paren {\dfrac {\sqrt {30} + \sqrt {10} - \sqrt 6 - \sqrt 2 - 2 \sqrt {15 + 3 \sqrt 5} + 2 \sqrt {5 + \sqrt 5} } 2} + i \sqrt {32 - 6 \sqrt 3 + 2 \sqrt {15} + 2 \sqrt {50 + 10 \sqrt 5} - 2 \sqrt {10 + 2 \sqrt 5} + 2 \sqrt {90 + 30 \sqrt 5} } } }\)
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We can verify that this is correct by noting that:
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\(\ds 8 \sin 3 \degrees\)
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\(=\)
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\(\ds \paren {\dfrac {\sqrt {30} + \sqrt {10} - \sqrt 6 - \sqrt 2 - 2 \sqrt {15 + 3 \sqrt 5} + 2 \sqrt {5 + \sqrt 5} } 2}\)
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\(\ds 8 \cos 3 \degrees\)
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\(=\)
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\(\ds \sqrt {32 - 6 \sqrt 3 + 2 \sqrt {15} + 2 \sqrt {50 + 10 \sqrt 5} - 2\sqrt {10 + 2 \sqrt 5} + 2\sqrt {90 + 30 \sqrt 5} }\)
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\(\ds \paren {\dfrac 1 8 + i \dfrac {\sqrt 3} 8}\)
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\(=\)
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\(\ds \dfrac 1 4 \cis 60 \degrees\)
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Plugging these into the main result, we see:
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\(\ds \)
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\(=\)
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\(\ds \dfrac 1 4 \cis 60 \degrees \paren {\paren { \sqrt [3] {8 \sin 3 \degrees + i 8 \cos 3 \degrees } } - \paren { \sqrt [3] {-8 \sin 3 \degrees + i 8 \cos 3 \degrees } } }\)
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\(\ds \)
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\(=\)
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\(\ds \dfrac 1 2 \cis 60 \degrees \paren {\paren { \sqrt [3] {\sin 3 \degrees + i \cos 3 \degrees } } - \paren { \sqrt [3] {-\sin 3 \degrees + i \cos 3 \degrees } } }\)
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moving the $8$ outside
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\(\ds \)
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\(=\)
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\(\ds \dfrac 1 2 \cis 60 \degrees \paren {\paren { \sqrt [3] {\cos 87 \degrees + i \sin 87 \degrees } } - \paren { \sqrt [3] {\cos 93 \degrees + i \sin 93 \degrees } } }\)
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Sine of Complement equals Cosine
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\(\ds \)
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\(=\)
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\(\ds \dfrac 1 2 \cis 60 \degrees \paren { \cis 29 \degrees - \cis 31 \degrees }\)
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Roots of Complex Number
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\(\ds \)
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\(=\)
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\(\ds \dfrac 1 2 \paren { \cis 89 \degrees - \cis 91 \degrees }\)
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Complex Multiplication as Geometrical Transformation
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\(\ds \)
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\(=\)
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\(\ds \dfrac 1 2 \paren { \paren {\cos 89 \degrees + i \sin 89 \degrees } - \paren {\cos 91 \degrees + i \sin 91 \degrees } }\)
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\(\ds \)
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\(=\)
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\(\ds \dfrac 1 2 \paren { 2 \paren {\cos 89 \degrees } }\)
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Sine of Supplementary Angle and Cosine of Supplementary Angle
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\(\ds \)
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\(=\)
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\(\ds \dfrac 1 2 \paren { 2 \paren {\sin 1 \degrees } }\)
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Sine of Complement equals Cosine
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\(\ds \)
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\(=\)
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\(\ds \sin 1 \degrees\)
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$\blacksquare$