Sine of 300 Degrees
Jump to navigation
Jump to search
Theorem
- $\sin 300 \degrees = \sin \dfrac {5 \pi} 3 = -\dfrac {\sqrt 3} 2$
where $\sin$ denotes the sine function.
Proof
\(\ds \sin 300 \degrees\) | \(=\) | \(\ds \map \sin {360 \degrees - 60 \degrees}\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds -\sin 60 \degrees\) | Sine of Conjugate Angle | |||||||||||
\(\ds \) | \(=\) | \(\ds -\dfrac {\sqrt 3} 2\) | Sine of $60 \degrees$ |
$\blacksquare$
Sources
- 1968: Murray R. Spiegel: Mathematical Handbook of Formulas and Tables ... (previous) ... (next): $\S 5$: Trigonometric Functions: Exact Values for Trigonometric Functions of Various Angles