If sin 600^{circ} cos 30^{circ} + cos 120^{ci | vedclass

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(C) Given expression: $k = \sin 600^{\circ} \cos 30^{\circ} + \cos 120^{\circ} \sin 150^{\circ}$.First,simplify the trigonometric angles:$\sin 600^{\c

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$-1$

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(C) Given expression: $k = \sin 600^{\circ} \cos 30^{\circ} + \cos 120^{\circ} \sin 150^{\circ}$.First,simplify the trigonometric angles:$\sin 600^{\circ} = \sin(360^{\circ} + 240^{\circ}) = \sin 240^{\circ} = \sin(180^{\circ} + 60^{\circ}) = -\sin 60^{\circ} = -\frac{\sqrt{3}}{2}$.$\cos 120^{\circ} = \cos(180^{\circ} - 60^{\circ}) = -\cos 60^{\circ} = -\frac{1}{2}$.$\sin 150^{\circ} = \sin(180^{\circ} - 30^{\circ}) = \sin 30^{\circ} = \frac{1}{2}$.$\cos 30^{\circ} = \frac{\sqrt{3}}{2}$.Substitute these values into the expression:$k = \left(-\frac{\sqrt{3}}{2}\right) \left(\frac{\sqrt{3}}{2}\right) + \left(-\frac{1}{2}\right) \left(\frac{1}{2}\right)$.$k = -\frac{3}{4} - \frac{1}{4}$.$k = -\frac{4}{4} = -1$.

If $\sin 600^{\circ} \cos 30^{\circ} + \cos 120^{\circ} \sin 150^{\circ} = k$,then $k =$

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