Question

what is the value of sin600 degrees?​

Answer 1

Answer:

Solution

Let's first ascertain in which quadrant those angles lie.

600° = (3×180°+ 60°); angle will lie in 3rd quadrant

330° = (360°- 30°); angle will lie in 4th quadrant

120° = (180° - 60°); angle will lie in 2nd quadrant

150° = (180° - 30°); angle will lie in 2nd quadrant

Applying ASTC Rule

sin (600°) = sin (3×180°+ 60°) = - sin (60°); sine is -ive in III Q

cos (330°) = cos (360° - 60°) = cos (60°); cosine is +ive in IV Q

cos (120°) = cos (180° - 60°) = - cos (60°); cosine is -ive in the II Q

sin (150°) = sin (180° - 30°) = sin (30°) = sine is +ive in II Q

Given expression

sin (600°) cos (330°) + cos (120°) sin (150°)

= - sin (60°) cos (60°) - cos (60°) sin (30°)

= - (√3/2)(½) - ½(½)

= - √3/4 - 1/4

= - ¼(√3+1)