The value of $\cos (480 °) \sin (150 °) + \sin (600 °) \cos (390 °)$ is

$0$

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$1$

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$\frac{1}{2}$

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$- 1$

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$- \frac{1}{2}$

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Solution

The correct option is D
$- 1$

Explanation for correct option
Given trigonometric function is $\cos (480 °) \sin (150 °) + \sin (600 °) \cos (390 °)$
$\cos (480 °) \sin (150 °) + \sin (600 °) \cos (390 °) = \cos (540 ° - 60 °) \sin (180 ° - 30 °) + \sin (540 ° + 60 °) \cos (360 ° + 30 °) = - \cos (60 °) \sin (30 °) + (- \sin (60 °) \cos (30 °)) = - \sin (60 ° + 30 °) [∵ \sin (A + B) + \sin (A) \cos (B) + \sin (B) \cos (A)] = - \sin (90 °) = - 1$
Hence, the correct option is $Option (D)$

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