What is the value of cos 480^o sin 150^o + sin 600^o cos 390^o?

Given , cos 480° sin 150° + sin 600° cos 390°
= cos (360° + 120°) sin (180° – 30°) + sin (2 × 360° – 120°) cos (360° + 30°)
= cos 120° sin 30° – sin (120°) cos 30°
= – cos 60° sin 30° – sin (180° – 60°) cos 30°
= – cos 60° sin 30° – sin 60° cos 30°
= – (sin 30° cos 60° + sin 60° cos 30°)
= – sin (30° + 60°) ... (As, sin (A + B) = sin A cos B + cos A sin B)
= – sin 90° = – 1.