Question

Prove that Sin 600. Cos 330 + Cos 120 Sin 150
+ Sin 180 Cos 180 = -1

Answer 1

Step-by-step explanation:

this is the solution.

As far as i know.

Answer 2

$Given \: \sin(600) . \cos(330) + \cos(120) \sin(150) = - 1 \\ LHS = \sin(3 \times 180 + 60) . \cos(2 \times 180 - 30) + \cos(180 - 60) . \sin(180 - 30) \\ = \sin(60) \times - \cos(30) + \cos(60) \times - \sin(30) \\ = \frac{ \sqrt{3} }{2} \times \frac{ - \sqrt{3} }{2} + \frac{1}{2} \times \frac{ - 1}{2} \\ = \frac{ - 3}{4} + \frac{ - 1}{4} \\ = \frac{ - 3 - 1}{4} = \frac{ - 4}{4} = - 1 = RHS$