Question

cos570°-sin510°-sin330°cos390°​

Answer 1

LHS

= cos 570° sin 510°+ sin (-330°) cos (-390°)

We know that,

sin (-x) = -sin (x) and cos (-x) = +cos (x).

= cos 570° sin 510°+ [sin (-330°)] cos (-390°)

= cos 570° sin 510° - sin (-330°) cos (-390°)

= cos (90° × 6 + 30°) sin (90° × 5 + 60°) – sin (90° × 3 + 60°) cos (90° × 4 + 30°)

We know that cos is negative at 90° + θ i.e. in Q2 and when n is odd, sin → cos and cos → sin.

= -cos 30° cos 60° - [-cos 60°] cos 30°

= -cos 30° cos 60° + cos 60° cos 30° = 0

= RHS

Hence proved.