Question -
Answer -
(i) cos 3A + cos 5A + cos 7A + cos 15A = 4 cos 4A cos 5Acos 6A
Let us consider LHS:
cos 3A + cos 5A + cos 7A + cos 15A
So now,
(cos 5A + cos 3A) + (cos 15A + cos 7A)
By 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)/2cos (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 by 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)/2cos (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
Hence proved.
(ii) cos A + cos 3A + cos 5A + cos 7A = 4 cos A cos2A cos 4A
Let us consider LHS:
cos A + cos 3A + cos 5A + cos 7A
So now,
(cos 3A + cos A) + (cos 7A + cos 5A)
By 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 by 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)/2cos (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
Hence proved.
(iii) sin A + sin 2A + sin 4A + sin 5A = 4 cos A/2 cos3A/2 sin 3A
Let us consider LHS:
sin A + sin 2A + sin 4A + sin 5A
So now,
(sin 2A + sin A) + (sin 5A + sin 4A)
By 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 by 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
Hence proved.
(iv) sin 3A + sin 2A – sin A = 4 sin A cos A/2 cos3A/2
Let us consider LHS:
sin 3A + sin 2A – sin A
So now,
(sin 3A – sin A) + sin 2A
By using the formula,
sin A – sin B = 2 cos (A+B)/2 sin (A-B)/2
(sin 3A – sin A) + sin 2A = 2 cos (3A + A)/2 sin (3A –A)/2 + sin 2A
= 2 cos 4A/2 sin 2A/2 + sin 2A
We know that, sin 2A = 2 sin A cos A
= 2 cos 2A Sin A + 2 sin A cos A
= 2 sin A (cos 2A + cos A)
Again by using the formula,
cos A + cos B = 2 cos (A+B)/2 cos (A-B)/2
2 sin A (cos 2A + cos A) = 2 sin A [2 cos (2A+A)/2 cos(2A-A)/2]
= 2 sin A [2 cos 3A/2 cos A/2]
= 4 sin A cos A/2 cos 3A/2
= RHS
Hence proved.
(v) cos 20o cos 100o + cos100o cos 140o – cos 140o cos200o = – 3/4
Let us consider LHS:
cos 20o cos 100o + cos100o cos 140o – cos 140o cos200o =
We shall multiply and divide by 2 we get,
= 1/2 [2 cos 100o cos 20o +2 cos 140o cos 100o – 2 cos 200o cos140o]
We know that 2 cos A cos B = cos (A+B) + cos (A-B)
So,
= 1/2 [cos (100o + 20o) +cos (100o – 20o) + cos (140o + 100o)+ cos (140o – 100o) – cos (200o +140o) – cos (200o – 140o)]]
= 1/2 [cos 120o + cos 80o +cos 240o + cos 40o – cos 340o –cos 60o]
= 1/2 [cos (90o + 30o) +cos 80o + cos (180o + 60o) + cos 40o –cos (360o – 20o) – cos 60o]
We know, cos (180o + A) = – cos A, cos(90o + A) = – sin A, cos (360o – A) = cos A
So,
= 1/2 [- sin 30o + cos 80o –cos 60o + cos 40o – cos 20o –cos 60o]
= 1/2 [- sin 30o + cos 80o +cos 40o – cos 20o – 2 cos 60o]
Again by using the formula,
cos A + cos B = 2 cos (A+B)/2 cos (A-B)/2
= 1/2 [- sin 30o + 2 cos (80o+40o)/2cos (80o-40o)/2 – cos 20o – 2 × 1/2]
= 1/2 [- sin 30o + 2 cos 120o/2cos 40o/2 – cos 20o – 1]
= 1/2 [- sin 30o + 2 cos 60o cos20o – cos 20o – 1]
= 1/2 [- 1/2 + 2×1/2×cos 20o – cos 20o –1]
= 1/2 [-1/2 + cos 20o – cos 20o –1]
= 1/2 [-1/2 -1]
= 1/2 [-3/2]
= -3/4
= RHS
Hence proved.
(vi) sin x/2 sin 7x/2 + sin 3x/2 sin 11x/2 = sin 2xsin 5x
Let us consider LHS:
sin x/2 sin 7x/2 + sin 3x/2 sin 11x/2 =
We shall multiply and divide by 2 we get,
= 1/2 [2 sin 7x/2 sin x/2 + 2 sin 11x/2 sin 3x/2]
We know that 2 sin A sin B = cos (A-B) – cos (A+B)
So,
= 1/2 [cos (7x/2 – x/2) – cos (7x/2 + x/2) + cos(11x/2 – 3x/2) – cos (11x/2 + 3x/2)]
= 1/2 [cos (7x-x)/2 – cos (7x+x)/2 + cos (11x-3x)/2 –cos (11x+3x)/2]
= 1/2 [cos 6x/2 – cos 8x/2 + cos 8x/2 – cos 14x/2]
= 1/2 [cos 3x – cos 7x]
= – 1/2 [cos 7x – cos 3x]
Again by using the formula,
cos A – cos B = -2 sin (A+B)/2 sin (A-B)/2
= -1/2 [-2 sin (7x+3x)/2 sin (7x-3x)/2]
= -1/2 [-2 sin 10x/2 sin 4x/2]
= -1/2 [-2 sin 5x sin 2x]
= -2/-2 sin 5x sin 2x
= sin 2x sin 5x
= RHS
Hence proved.
(vii) cos x cos x/2 – cos 3x cos 9x/2 = sin 4x sin 7x/2
Let us consider LHS:
cos x cos x/2 – cos 3x cos 9x/2 =
We shall multiply and divide by 2 we get,
= 1/2 [2 cos x cos x/2 – 2 cos 9x/2 cos 3x]
We know that 2 cos A cos B = cos (A+B) + cos (A-B)
So,
= 1/2 [cos (x + x/2) + cos (x – x/2) – cos (9x/2 + 3x)– cos (9x/2 – 3x)]
= 1/2 [cos (2x+x)/2 + cos (2x-x)/2 – cos (9x+6x)/2 –cos (9x-6x)/2]
= 1/2 [cos 3x/2 + cos x/2 – cos 15x/2 – cos 3x/2]
= 1/2 [cos x/2 – cos 15x/2]
= – 1/2 [cos 15x/2 – cos x/2]
Again by using the formula,
cos A – cos B = -2 sin (A+B)/2 sin (A-B)/2
= – 1/2 [-2 sin (15x/2 + x/2)/2 sin (15x/2 – x/2)/2]
= -1/2 [-2 sin (16x/2)/2 sin (14x/2)/2]
= -1/2 [-2 sin 16x/4 sin 7x/2]
= – 1/2 [-2 sin 4x sin 7x/2]
= -2/-2 [sin 4x sin 7x/2]
= sin 4x sin 7x/2
= RHS
Hence proved.