Question -
Answer -
(i)┬аcos 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)┬аcos 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)┬аsin 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 тАУ 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)┬аsin 3A + sin 2A тАУ sin A = 4 sin A cos A/2 cos3A/2
Let us consider LHS:
sin 3A + sin 2A тАУ sin A
So now,
(sin 3A тАУ sin A) + sin 2A
By using the formula,
sin A тАУ sin B = 2 cos (A+B)/2 sin (A-B)/2
(sin 3A тАУ sin A) + sin 2A = 2 cos (3A + A)/2 sin (3A тАУA)/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)┬аcos 20o┬аcos 100o┬а+ cos100o┬аcos 140o┬атАУ cos 140o┬аcos200o┬а= тАУ 3/4
Let us consider LHS:
cos 20o┬аcos 100o┬а+ cos100o┬аcos 140o┬атАУ cos 140o┬аcos200o┬а=
We shall multiply and divide by 2 we get,
= 1/2 [2 cos 100o┬аcos 20o┬а+2 cos 140o┬аcos 100o┬атАУ 2 cos 200o┬аcos140o]
We know that 2 cos A cos B = cos (A+B) + cos (A-B)
So,
= 1/2 [cos (100o┬а+ 20o) +cos (100o┬атАУ 20o) + cos (140o┬а+ 100o)+ cos (140o┬атАУ 100o) тАУ cos (200o┬а+140o) тАУ cos (200o┬атАУ 140o)]]
= 1/2 [cos 120o┬а+ cos 80o┬а+cos 240o┬а+ cos 40o┬атАУ cos 340o┬атАУcos 60o]
= 1/2 [cos (90o┬а+ 30o) +cos 80o┬а+ cos (180o┬а+ 60o) + cos 40o┬атАУcos (360o┬атАУ 20o) тАУ cos 60o]
We know, cos (180o┬а+ A) = тАУ cos A, cos(90o┬а+ A) = тАУ sin A, cos (360o┬атАУ A) = cos A
So,
= 1/2 [- sin 30o┬а+ cos 80o┬атАУcos 60o┬а+ cos 40o┬атАУ cos 20o┬атАУcos 60o]
= 1/2 [- sin 30o┬а+ cos 80o┬а+cos 40o┬атАУ cos 20o┬атАУ 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┬а+ 2 cos (80o+40o)/2cos (80o-40o)/2 тАУ cos 20o┬атАУ 2 ├Ч 1/2]
= 1/2 [- sin 30o┬а+ 2 cos 120o/2cos 40o/2 тАУ cos 20o┬атАУ 1]
= 1/2 [- sin 30o┬а+ 2 cos 60o┬аcos20o┬атАУ cos 20o┬атАУ 1]
= 1/2 [- 1/2 + 2├Ч1/2├Чcos 20o┬атАУ cos 20o┬атАУ1]
= 1/2 [-1/2 + cos 20o┬атАУ cos 20o┬атАУ1]
= 1/2 [-1/2 -1]
= 1/2 [-3/2]
= -3/4
= RHS
Hence proved.
(vi)┬аsin 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) тАУ cos (A+B)
So,
= 1/2 [cos (7x/2 тАУ x/2) тАУ cos (7x/2 + x/2) + cos(11x/2 тАУ 3x/2) тАУ cos (11x/2 + 3x/2)]
= 1/2 [cos (7x-x)/2 тАУ cos (7x+x)/2 + cos (11x-3x)/2 тАУcos (11x+3x)/2]
= 1/2 [cos 6x/2 тАУ cos 8x/2 + cos 8x/2 тАУ cos 14x/2]
= 1/2 [cos 3x тАУ cos 7x]
= тАУ 1/2 [cos 7x тАУ cos 3x]
Again by using the formula,
cos A тАУ 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)┬аcos x cos x/2 тАУ cos 3x cos 9x/2 = sin 4x sin 7x/2
Let us consider LHS:
cos x cos x/2 тАУ cos 3x cos 9x/2 =
We shall multiply and divide by 2 we get,
= 1/2 [2 cos x cos x/2 тАУ 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 тАУ x/2) тАУ cos (9x/2 + 3x)тАУ cos (9x/2 тАУ 3x)]
= 1/2 [cos (2x+x)/2 + cos (2x-x)/2 тАУ cos (9x+6x)/2 тАУcos (9x-6x)/2]
= 1/2 [cos 3x/2 + cos x/2 тАУ cos 15x/2 тАУ cos 3x/2]
= 1/2 [cos x/2 тАУ cos 15x/2]
= тАУ 1/2 [cos 15x/2 тАУ cos x/2]
Again by using the formula,
cos A тАУ cos B = -2 sin (A+B)/2 sin (A-B)/2
= тАУ 1/2 [-2 sin (15x/2 + x/2)/2 sin (15x/2 тАУ x/2)/2]
= -1/2 [-2 sin (16x/2)/2 sin (14x/2)/2]
= -1/2 [-2 sin 16x/4 sin 7x/2]
= тАУ 1/2 [-2 sin 4x sin 7x/2]
= -2/-2 [sin 4x sin 7x/2]
= sin 4x sin 7x/2
= RHS
Hence proved.