## Trigonometry (11th Edition) Clone

reference angles: $\left[\begin{array}{lllll} Quadr.: & I & II & III & IV\\ \theta' & \theta & 180^{o}-\theta & \theta-180^{o} & 360^{o}-\theta \end{array}\right]$ ------------------ $240^{o}$ 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.