sin⁡−420∘cos⁡390∘+cos⁡−660∘sin⁡330∘ is

sin⁡−420∘cos⁡390∘+cos⁡−660∘sin⁡330∘ =−sin⁡90∘×4+60∘⋅cos⁡90∘×4+30∘+cos⁡90∘×7+30∘⋅sin⁡90∘×3+60∘ =−sin⁡60∘cos⁡30∘+sin⁡30∘−cos⁡60∘ =−sin⁡90∘=−1

sin⁡−420∘cos⁡390∘+cos⁡−660∘sin⁡330∘ is

sin⁡−420∘cos⁡390∘+cos⁡−660∘sin⁡330∘ =−sin⁡90∘×4+60∘⋅cos⁡90∘×4+30∘+cos⁡90∘×7+30∘⋅sin⁡90∘×3+60∘ =−sin⁡60∘cos⁡30∘+sin⁡30∘−cos⁡60∘ =−sin⁡90∘=−1

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$\sin (- 420^{\circ}) \cos (390^{\circ}) + \cos (- 660^{\circ}) \sin (330^{\circ})$ is

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answer is B.

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Detailed Solution

$\begin{array}{l} \sin (- 420^{\circ}) \cos (390^{\circ}) + \cos (- 660^{\circ}) \sin (330^{\circ}) \\ = - \sin (90^{\circ} \times 4 + 60^{\circ}) \cdot \cos (90^{\circ} \times 4 + 30^{\circ}) + \\ \cos (90^{\circ} \times 7 + 30^{\circ}) \cdot \sin (90^{\circ} \times 3 + 60^{\circ}) \\ = - (\sin 60^{\circ}) (\cos 30^{\circ}) + (\sin 30^{\circ}) (- \cos 60^{\circ}) \\ = - \sin 90^{\circ} = - 1 \end{array}$