Answer
true
Work Step by Step
reference angles:
$\left[\begin{array}{lllll}
Quadr.: & I & II & III & IV\\
\theta' & \theta & 180^{o}-\theta & \theta-180^{o} & 360^{o}-\theta
\end{array}\right]$
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$240^{o}$ is in quadrant III, and we find the reference angle with
$\theta'=\theta-180^{o}=240^{o}-180^{o}=60^{o}$ (special angle)
In quadrant III, sine is negative, so
LHS= $\displaystyle \sin 240^{o}=-\sin 60^{o}=-\frac{\sqrt{3}}{2}$
$120^{o}$ is in quadrant II, and we find the reference angle with
$\theta'=180^{o}-\theta=180^{o}-120^{o}=60^{o}$ (special angle)
In quadrant II, sine is positive, cosine is negative, so
$\displaystyle \sin 120^{o}=\sin 60^{o}=\frac{\sqrt{3}}{2},$
$\displaystyle \cos 120^{o}=-\cos 60^{o}=-\frac{1}{2}$
RHS= 2 $\displaystyle \sin 120^{o}\cos 120^{o}=2(\frac{\sqrt{3}}{2})(-\frac{1}{2})=-\frac{\sqrt{3}}{2}$
$LHS=RHS$
the statement is true.
