Sine of 1 Degree

Theorem
where $\sin$ denotes the sine function.

Proof
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:

and:

Thus:


 * $\sqrt [3] {\cis 93 \degrees }$ 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.

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 $9)$

Squaring the Numerator of the Sine of 3 Degrees, we get...

Bringing that result back into our main result, we get...

Canceling $4$ from the numerator and denominator, we get...

Moving an $8$ to the outside, we get...

Flipping the top and bottom lines, we get our main result:

We can verify that this is correct by noting that...

Plugging these into the main result, we see...