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
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
(i) cos 3A + cos 5A + cos 7A + cos 15A = 4 cos 4A cos 5A cos 6A
(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.