Question

The value of $\sin \left(600°\right)\cos \left(330°\right)+\cos \left(120°\right)\sin \left(150°\right)$ is

• A. $-1$
• B. $1$
• C. $\frac{1}{\sqrt{2}}$
• D. $\frac{\sqrt{3}}{2}$

Answer 1

The correct option is A

$-1$

The explanation for the correct option

The given trigonometric expression: $\sin \left(600°\right)\cos \left(330°\right)+\cos \left(120°\right)\sin \left(150°\right)$.

$$\begin{gathered} \sin \left(600°\right)\cos \left(330°\right)+\cos \left(120°\right)\sin \left(150°\right) \\ =\sin \left(540°+60°\right)\cos \left(360°-30°\right)+\cos \left(180°-60°\right)\sin \left(180°-30°\right) \\ =\sin \left[\left(3\times 180°\right)+60°\right]\cos \left[\left(2\times 180°\right)-30°\right]-\cos \left(60°\right)\sin \left(30°\right) \\ =-\sin \left(60°\right)\cos \left(30°\right)-\cos \left(60°\right)\sin \left(30°\right) \\ =-\left[\sin \left(60°\right)\cos \left(30°\right)+\cos \left(60°\right)\sin \left(30°\right)\right] \\ =-\sin \left(60°+30°\right)\mathbf{[}\mathbf{\because }\mathbf{s}\mathbf{i}\mathbf{n}\left(A+B\right)\mathbf{=}\mathbf{s}\mathbf{i}\mathbf{n}\left(A\right)\mathbf{c}\mathbf{o}\mathbf{s}\left(B\right)\mathbf{+}\mathbf{c}\mathbf{o}\mathbf{s}\left(A\right)\mathbf{s}\mathbf{i}\mathbf{n}\left(B\right)\mathbf{]} \\ =-\sin \left(90°\right) \\ =-1 \end{gathered}$$

Therefore, the value of $\sin \left(600°\right)\cos \left(330°\right)+\cos \left(120°\right)\sin \left(150°\right)$ is $-1$.

Hence, the correct option is (A).