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(i) sin 50° cos 85º = 1-√2sin35°/2√2 |
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Answer» using the formula, 2 sin A cos B = sin (A + B) + sin (A – B) sin A cos B = [sin (A + B) + sin (A – B)]/2 sin 50° cos 85° = [sin(50° + 85°) + sin (50° – 85°)]/ = [sin (135°) + sin (-35°)]/2 = [sin (135°) – sin (35°)]/2 (SINCE, sin (-x) = -sin x) = [sin (180° – 45°) – sin 35°]/2 = [sin 45° – sin 35°]/2 = [(1/√2) – sin 35°]/2 = [(1 – sin 35°)/√2]/2 = (1 – sin 35°)/2√2 Thus PROVED. (ii) sin 25° cos 115° = 1/2 {sin 140° – 1} On using the formula, 2 sin A cos B = sin (A + B) + sin (A – B) sin A cos B = [sin (A + B) + sin (A – B)]/2 sin 20° cos 115° = [sin(25° + 115°) + sin (25° – 115°)]/2 = [sin (140°) + sin (-90°)]/2 = [sin (140°) – sin (90°)]/2 (since, sin (-x) = -sin x) = 1/2 {sin 140° – 1} Thus proved. |
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