Sine of 240 Degrees/Proof 1
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Theorem
- $\sin 240 \degrees = \sin \dfrac {4 \pi} 3 = -\dfrac {\sqrt 3} 2$
Proof
| \(\ds \sin 240 \degrees\) | \(=\) | \(\ds \map \sin {360 \degrees - 120 \degrees}\) | ||||||||||||
| \(\ds \) | \(=\) | \(\ds -\sin 120 \degrees\) | Sine of Conjugate Angle | |||||||||||
| \(\ds \) | \(=\) | \(\ds -\frac {\sqrt 3} 2\) | Sine of $120 \degrees$ |
$\blacksquare$