The value of sin 600^circ cos 330^circ + cos | vedclass

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Question

(A) We use the reduction formulas: $\sin 600^\circ = \sin(720^\circ - 120^\circ) = -\sin 120^\circ = -\sin(180^\circ - 60^\circ) = -\sin 60^\circ = -\

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$-1$

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(A) We use the reduction formulas: $\sin 600^\circ = \sin(720^\circ - 120^\circ) = -\sin 120^\circ = -\sin(180^\circ - 60^\circ) = -\sin 60^\circ = -\frac{\sqrt{3}}{2}$ $\cos 330^\circ = \cos(360^\circ - 30^\circ) = \cos 30^\circ = \frac{\sqrt{3}}{2}$ $\cos 120^\circ = \cos(180^\circ - 60^\circ) = -\cos 60^\circ = -\frac{1}{2}$ $\sin 150^\circ = \sin(180^\circ - 30^\circ) = \sin 30^\circ = \frac{1}{2}$ Substituting these values into the expression: $(- \frac{\sqrt{3}}{2})(\frac{\sqrt{3}}{2}) + (- \frac{1}{2})(\frac{1}{2}) = -\frac{3}{4} - \frac{1}{4} = -\frac{4}{4} = -1$.

The value of $\sin 600^\circ \cos 330^\circ + \cos 120^\circ \sin 150^\circ$ is

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