(i) Let us consider the LHS
sin 3A + sin 2A – sin A
Therefore now,
(sin 3A – sin A) + sin 2A
On 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
As 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 on 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
Thus proved.
(ii) Let us consider the LHS
cos 20° cos 100° + cos 100° cos 140° – cos 140° cos 200°
Now, we shall multiply and divide by 2 we get,
= 1/2 [2 cos 100° cos 20° + 2 cos 140° cos 100° – 2 cos 200° cos 140°]
As we know that 2 cos A cos B = cos (A + B) + cos (A - B)
Therefore,
= 1/2 [cos (100° + 20°) + cos (100° – 20°) + cos (140° + 100°) + cos (140° – 100°) – cos (200° + 140°) – cos (200° – 140°)]]
= 1/2 [cos 120° + cos 80° + cos 240° + cos 40° – cos 340° – cos 60°]
= 1/2 [cos (90° + 30°) + cos 80° + cos (180° + 60°) + cos 40° – cos (360° – 20°) – cos 60°]
As we know, cos (180° + A) = – cos A, cos (90° + A) = – sin A, cos (360° – A) = cos A
Therefore,
= 1/2 [- sin 30° + cos 80° – cos 60° + cos 40° – cos 20° – cos 60°]
= 1/2 [- sin 30° + cos 80° + cos 40° – cos 20° – 2 cos 60°]
Again on using the formula,
cos A + cos B = 2 cos (A + B)/2 cos (A - B)/2
= 1/2 [- sin 30° + 2 cos (80° + 40°)/2 cos (80° -40°)/2 – cos 20° – 2 × 1/2]
= 1/2 [- sin 30° + 2 cos 120°/2 cos 40°/2 – cos 20° – 1]
= 1/2 [- sin 30° + 2 cos 60° cos 20° – cos 20° – 1]
= 1/2 [- 1/2 + 2 × 1/2 × cos 20° – cos 20° – 1]
= 1/2 [-1/2 + cos 20° – cos 20° – 1]
= 1/2 [-1/2 - 1]
= 1/2 [-3/2]
= -3/4
= RHS
Thus proved.
(iii) Given as sin x/2 sin 7x/2 + sin 3x/2 sin 11x/2 = sin 2x sin 5x
Let us consider the LHS
sin x/2 sin 7x/2 + sin 3x/2 sin 11x/2 =
Now, 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]
As we know that 2 sin A sin B = cos (A - B) – cos (A + B)
Therefore,
= 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 on 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
Thus proved.
(iv) cos x cos x/2 – cos 3x cos 9x/2 = sin 4x sin 7x/2
Let us consider the LHS
cos x cos x/2 – cos 3x cos 9x/2 =
Now, we shall multiply and divide by 2 we get,
= 1/2 [2 cos x cos x/2 – 2 cos 9x/2 cos 3x]
As we know that 2 cos A cos B = cos (A + B) + cos (A - B)
Therefore,
= 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 on 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
Thus proved.
