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Prove that :cos 570° sin 510°+ sin (-330°) cos (-390°) = 0 |
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Answer» 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. |
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