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Prove that:(i) cos 3A + cos 5A + cos 7A + cos 15A = 4 cos 4A cos 5A cos 6A(ii) cos A + cos 3A + cos 5A + cos 7A = 4 cos A cos 2A cos 4A(iii) sin A + sin 2A + sin 4A + sin 5A = 4 cos A/2 cos 3A/2 sin 3A |
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Answer» (i) Given as cos 3A + cos 5A + cos 7A + cos 15A = 4 cos 4A cos 5A cos 6A Let us consider the LHS cos 3A + cos 5A + cos 7A + cos 15A Therefore now, (cos 5A + cos 3A) + (cos 15A + cos 7A) On using the formula, cos A + cos B = 2 cos (A + B)/2 cos (A - B)/2 (cos 5A + cos 3A) + (cos 15A + cos 7A) = [2 cos (5A + 3A)/2 cos (5A - 3A)/2] + [2 cos (15A + 7A)/2 cos (15A - 7A)/2] = [2 cos 8A/2 cos 2A/2] + [2 cos 22A/2 cos 8A/2] = [2 cos 4A cos A] + [2 cos 11A cos 4A] = 2 cos 4A (cos 11A + cos A) Again on using the formula, cos A + cos B = 2 cos (A + B)/2 cos (A - B)/2 2 cos 4A (cos 11A + cos A) = 2 cos 4A [2 cos (11A + A)/2 cos (11A - A)/2] = 2 cos 4A [2 cos 12A/2 cos 10A/2] = 2 cos 4A [2 cos 6A cos 5A] = 4 cos 4A cos 5A cos 6A = RHS Thus proved. (ii) cos A + cos 3A + cos 5A + cos 7A = 4 cos A cos 2A cos 4A Let us consider the LHS cos A + cos 3A + cos 5A + cos 7A Therefore now, (cos 3A + cos A) + (cos 7A + cos 5A) On using the formula, cos A + cos B = 2 cos (A + B)/2 cos (A - B)/2 (cos 3A + cos A) + (cos 7A + cos 5A) = [2 cos (3A + A)/2 cos (3A - A)/2] + [2 cos (7A + 5A)/2 cos (7A - 5A)/2] = [2 cos 4A/2 cos 2A/2] + [2 cos 12A/2 cos 2A/2] = [2 cos 2A cos A] + [2 cos 6A cos A] = 2 cos A (cos 6A + cos 2A) Again on using the formula, cos A + cos B = 2 cos (A + B)/2 cos (A - B)/2 2 cos A (cos 6A + cos 2A) = 2 cos A [2 cos (6A + 2A)/2 cos (6A - 2A)/2] = 2 cos A [2 cos 8A/2 cos 4A/2] = 2 cos A [2 cos 4A cos 2A] = 4 cos A cos 2A cos 4A = RHS Thus proved. (iii) Given as sin A + sin 2A + sin 4A + sin 5A = 4 cos A/2 cos 3A/2 sin 3A Let us consider the LHS sin A + sin 2A + sin 4A + sin 5A Therefore now, (sin 2A + sin A) + (sin 5A + sin 4A) On using the formula, sin A + sin B = 2 sin (A + B)/2 cos (A - B)/2 (sin 2A + sin A) + (sin 5A + sin 4A) = [2 sin (2A + A)/2 cos (2A - A)/2] + [2 sin (5A + 4A)/2 cos (5A - 4A)/2] = [2 sin 3A/2 cos A/2] + [2 sin 9A/2 cos A/2] = 2 cos A/2 (sin 9A/2 + sin 3A/2) Again on using the formula, sin A + sin B = 2 sin (A + B)/2 cos (A - B)/2 2 cos A/2 (sin 9A/2 + sin 3A/2) = 2 cos A/2 [2 sin (9A/2 + 3A/2)/2 cos (9A/2 – 3A/2)/2] = 2 cos A/2 [2 sin ((9A + 3A)/2)/2 cos ((9A - 3A)/2)/2] = 2 cos A/2 [2 sin 12A/4 cos 6A/4] = 2 cos A/2 [2 sin 3A cos 3A/2] = 4 cos A/2 cos 3A/2 sin 3A = RHS Thus proved. |
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