Sine of 72 Degrees

Theorem

$\sin 72 \degrees = \sin \dfrac {2 \pi} 5 = \dfrac {\sqrt{10 + 2 \sqrt 5} } 4$

where $\sin$ denotes the sine function.


Proof

\(\ds \sin 72 \degrees\) \(=\) \(\ds \sqrt {1 - \cos^2 72 \degrees}\) Sum of Squares of Sine and Cosine
\(\ds \) \(=\) \(\ds \sqrt {1 - \paren {\dfrac{\sqrt 5 - 1} 4}^2}\) Cosine of $72 \degrees$
\(\ds \) \(=\) \(\ds \sqrt {1 - \dfrac {6 - 2 \sqrt 5} {16} }\)
\(\ds \) \(=\) \(\ds \dfrac {\sqrt {10 + 2 \sqrt 5} } 4\)

$\blacksquare$


Sources

  • 1997: Donald E. Knuth: The Art of Computer Programming: Volume 1: Fundamental Algorithms (3rd ed.) ... (previous) ... (next): $\S 1.2.8$: Fibonacci Numbers: Exercise $19$