Q. 2 E

# prove that :<

Answer :

LHS = cos 570o sin 510o + sin (-330o) cos (-390o)

We know that sin (-x) = -sin (x) and cos (-x) = +cos (x).

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

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

= 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.

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