Sine of 240 Degrees/Proof 2
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Theorem
- $\sin 240 \degrees = \sin \dfrac {4 \pi} 3 = -\dfrac {\sqrt 3} 2$
Proof
When $240 \degrees$ is embedded in a Cartesian plane, it makes an angle of $60 \degrees$ with the $x$-axis.
$240 \degrees$ can be found in the third quadrant.
Hence by definition of sine function in the third quadrant, $\sin 240 \degrees$ is negative.
Thus:
- $\sin 240 \degrees = -\sin 60 \degrees = -\dfrac {\sqrt 3} 2$
$\blacksquare$
Sources
- 1953: L. Harwood Clarke: A Note Book in Pure Mathematics ... (previous) ... (next): $\text V$. Trigonometry: Angles larger than $90 \degrees$: Example