Sine of 3 Degrees

Theorem

$\sin 3^\circ = \sin \dfrac \pi {60} = \dfrac {\sqrt{30} + \sqrt{10} - \sqrt 6 - \sqrt 2 - 2 \sqrt {15 + 3 \sqrt 5} + 2 \sqrt {5 + \sqrt 5} } {16}$

where $\sin$ denotes the sine function.

Proof

\(\ds \sin 3^\circ\)

\(=\)

\(\ds \sin \left({75^\circ - 72^\circ}\right)\)

\(\ds \)

\(=\)

\(\ds \sin 75^\circ \cos 72^\circ - \cos 75^\circ \sin 72^\circ\)

Sine of Difference

\(\ds \)

\(=\)

\(\ds \dfrac {\sqrt 6 + \sqrt 2} 4 \times \dfrac {\sqrt 5 - 1} 4 - \cos 75^\circ \sin 72^\circ\)

Sine of $75^\circ$, Cosine of $72^\circ$ ...

\(\ds \)

\(=\)

\(\ds \dfrac {\sqrt 6 + \sqrt 2} 4 \times \dfrac {\sqrt 5 - 1} 4 - \dfrac {\sqrt 6 - \sqrt 2} 4 \times \dfrac {\sqrt {10 + 2 \sqrt 5} } 4\)

... Cosine of $75^\circ$, Sine of $72^\circ$

\(\ds \)

\(=\)

\(\ds \dfrac 1 {16} \times \left({\left({\sqrt 6 + \sqrt 2}\right) \times \left({\sqrt 5 - 1}\right) - \left({\sqrt 6 - \sqrt 2}\right) \times \left({\sqrt {10 + 2 \sqrt 5} }\right) }\right)\)

\(\ds \)

\(=\)

\(\ds \dfrac 1 {16} \times \left({\left({\sqrt{30} + \sqrt{10} - \sqrt 6 - \sqrt 2}\right) - \left({\sqrt {60 + 12 \sqrt 5} - \sqrt {20 + 4 \sqrt 5} }\right) }\right)\)

\(\ds \)

\(=\)

\(\ds \dfrac 1 {16} \times \left({\sqrt{30} + \sqrt{10} - \sqrt 6 - \sqrt 2 - \sqrt {60 + 12 \sqrt5 } + \sqrt {20 + 4 \sqrt 5} }\right)\)

\(\ds \)

\(=\)

\(\ds \dfrac 1 {16} \times \left({\sqrt{30} + \sqrt{10} - \sqrt 6 - \sqrt 2 - 2\sqrt {15 + 3 \sqrt5 } + 2 \sqrt {5 + \sqrt 5} }\right)\)

\(\ds \)

\(=\)

\(\ds \dfrac {\sqrt{30} + \sqrt{10} - \sqrt 6 - \sqrt 2 - 2 \sqrt {15 + 3 \sqrt 5} + 2\sqrt {5 + \sqrt 5} } {16}\)

$\blacksquare$