.(a) If A, B and C are interior angles of a triangle ABC, then show that sin (B+C/2) = cos A/2
(b) Express sin 67° + cos 75° in terms of trigonometric ratios of angles between 0° and 45°.
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(a)In a triangle, sum of all the interior angles A + B + C = 180° ⇒ B + C = 180° - A ⇒ (B+C)/2 = (180°-A)/2 ⇒ (B+C)/2 = (90°-A/2) ⇒ sin (B+C)/2 = sin (90°-A/2) ⇒ sin (B+C)/2 = cos A/2(b)sin 67° + cos 75° = sin (90° - 23°) + cos (90° - 15°) = cos 23° + sin 15°
(a)In a triangle, sum of all the interior angles A + B + C = 180° ⇒ B + C = 180° - A ⇒ (B+C)/2 = (180°-A)/2 ⇒ (B+C)/2 = (90°-A/2) ⇒ sin (B+C)/2 = sin (90°-A/2) ⇒ sin (B+C)/2 = cos A/2(b)sin 67° + cos 75° = sin (90° - 23°) + cos (90° - 15°) = cos 23° + sin 15°
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