Question

If cos 3A + cos 7A + cos 15A + cos 25A = 4 cos 5A cos 9A cos kA then k equals​

Answer 1

Answer:

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

Answer 2

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