RD Chapter 8 Transformation Formulae Ex 8.2 Solutions
Question - 1 : - Express each of the following as the product of sines and cosines:
(i) sin 12x + sin 4x
(ii) sin 5x – sin x
(iii) cos 12x + cos 8x
(iv) cos 12x – cos 4x
(v) sin 2x + cos 4x
Answer - 1 : -
(i) sin 12x + sin 4x
By using the formula,
sin A + sin B = 2 sin (A+B)/2 cos (A-B)/2
sin 12x + sin 4x = 2 sin (12x + 4x)/2 cos (12x – 4x)/2
= 2 sin 16x/2 cos 8x/2
= 2 sin 8x cos 4x
(ii) sin 5x – sin x
By using the formula,
sin A – sin B = 2 cos (A+B)/2 sin (A-B)/2
sin 5x – sin x = 2 cos (5x + x)/2 sin (5x – x)/2
= 2 cos 6x/2 sin 4x/2
= 2 cos 3x sin 2x
(iii) cos 12x + cos 8x
By using the formula,
cos A + cos B = 2 cos (A+B)/2 cos (A-B)/2
cos 12x + cos 8x = 2 cos (12x + 8x)/2 cos (12x – 8x)/2
= 2 cos 20x/2 cos 4x/2
= 2 cos 10x cos 2x
(iv) cos 12x – cos 4x
By using the formula,
cos A – cos B = -2 sin (A+B)/2 sin (A-B)/2
cos 12x – cos 4x = -2 sin (12x + 4x)/2 sin (12x –4x)/2
= -2 sin 16x/2 sin 8x/2
= -2 sin 8x sin 4x
(v) sin 2x + cos 4x
sin 2x + cos 4x = sin 2x + sin (90o –4x)
By using the formula,
sin A + sin B = 2 sin (A+B)/2 cos (A-B)/2
sin 2x + sin (90o – 4x) = 2 sin (2x +90o – 4x)/2 cos (2x – 90o + 4x)/2
= 2 sin (90o – 2x)/2 cos (6x – 90o)/2
= 2 sin (45° – x) cos (3x – 45°)
= 2 sin (45° – x) cos {-(45° – 3x)} (since, {cos (-x)= cos x})
= 2 sin (45° – x) cos (45° – 3x)
= 2 sin (π/4 – x) cos (π/4 – 3x)
Question - 2 : - Provethat :
(i) sin 38°+ sin 22° = sin 82°
(ii) cos100° + cos 20° = cos 40°
(iii) sin50° + sin 10° = cos 20°
(iv) sin23° + sin 37° = cos 7°
(v) sin105° + cos 105° = cos 45°
(vi) sin40° + sin 20° = cos 10°
Answer - 2 : -
(i) sin 38° + sin 22° = sin 82°
Let us consider LHS:
sin 38° + sin 22°
By using the formula,
sin A + sin B = 2 sin (A+B)/2 cos (A-B)/2
sin 38° + sin 22° = 2 sin (38o + 22o)/2cos (38o – 22o)/2
= 2 sin 60o/2 cos 16o/2
= 2 sin 30o cos 8o
= 2 × 1/2 × cos 8o
= cos 8o
= cos (90° – 82°)
= sin 82° (since, {cos (90° – A) = sin A})
= RHS
Hence Proved.
(ii) cos 100° + cos 20° = cos 40°
Let us consider LHS:
cos 100° + cos 20°
By using the formula,
cos A + cos B = 2 cos (A+B)/2 cos (A-B)/2
cos 100° + cos 20° = 2 cos (100o + 20o)/2cos (100o – 20o)/2
= 2 cos 120o/2 cos 80o/2
= 2 cos 60o cos 4o
= 2 × 1/2 × cos 40o
= cos 40o
= RHS
Hence Proved.
(iii) sin 50° + sin 10° = cos 20°
Let us consider LHS:
sin 50° + sin 10°
By using the formula,
sin A + sin B = 2 sin (A+B)/2 cos (A-B)/2
sin 50° + sin 10° = 2 sin (50o + 10o)/2cos (50o – 10o)/2
= 2 sin 60o/2 cos 40o/2
= 2 sin 30o cos 20o
= 2 × 1/2 × cos 20o
= cos 20o
= RHS
Hence Proved.
(iv) sin 23° + sin 37° = cos 7°
Let us consider LHS:
sin 23° + sin 37°
By using the formula,
sin A + sin B = 2 sin (A+B)/2 cos (A-B)/2
sin 23° + sin 37° = 2 sin (23o + 37o)/2cos (23o – 37o)/2
= 2 sin 60o/2 cos -14o/2
= 2 sin 30o cos -7o
= 2 × 1/2 × cos -7o
= cos 7o (since, {cos (-A) = cos A})
= RHS
Hence Proved.
(v) sin 105° + cos 105° = cos 45°
Let us consider LHS: sin 105° + cos 105°
sin 105° + cos 105° = sin 105o + sin(90o – 105o) [since, {sin (90° – A) = cos A}]
= sin 105o + sin (-15o)
= sin 105o – sin 15o [{sin(-A)= – sin A}]
By using the formula,
Sin A – sin B = 2 cos (A+B)/2 sin (A-B)/2
sin 105o – sin 15o = 2cos (105o + 15o)/2 sin (105o – 15o)/2
= 2 cos 120o/2 sin 90o/2
= 2 cos 60o sin 45o
= 2 × 1/2 × 1/√2
= 1/√2
= cos 45o
= RHS
Hence proved.
(vi) sin 40° + sin 20° = cos 10°
Let us consider LHS:
sin 40° + sin 20°
By using the formula,
sin A + sin B = 2 sin (A+B)/2 cos (A-B)/2
sin 40° + sin 20° = 2 sin (40o + 20o)/2cos (40o – 20o)/2
= 2 sin 60o/2 cos 20o/2
= 2 sin 30o cos 10o
= 2 × 1/2 × cos 10o
= cos 10o
= RHS
Hence Proved.
Question - 3 : - Provethat:
(i) cos55° + cos 65° + cos 175° = 0
(ii) sin50° – sin 70° + sin 10° = 0
(iii) cos80° + cos 40° – cos 20° = 0
(iv) cos20° + cos 100° + cos 140° = 0
(v) sin5π/18 – cos 4π/9 = √3 sin π/9
(vi) cosπ/12 – sin π/12 = 1/√2
(vii) sin80° – cos 70° = cos 50°
(viii)sin 51° + cos 81° = cos 21°
Answer - 3 : -
(i) cos 55° + cos 65° + cos 175° = 0
Let us consider LHS:
cos 55° + cos 65° + cos 175°
By using the formula,
cos A + cos B = 2 cos (A+B)/2 cos (A-B)/2
cos 55° + cos 65° + cos 175° = 2 cos (55o +65o)/2 cos (55o – 65o) + cos (180o –5o)
= 2 cos 120o/2 cos (-10o)/2 –cos 5o (since, {cos (180° – A) = – cos A})
= 2 cos 60° cos (-5°) – cos 5° (since, {cos (-A) = cosA})
= 2 × 1/2 × cos 5o – cos 5o
= 0
= RHS
Hence Proved.
(ii) sin 50° – sin 70° + sin 10° = 0
Let us consider LHS:
sin 50° – sin 70° + sin 10°
By using the formula,
sin A – sin B = 2 cos (A+B)/2 sin (A-B)/2
sin 50° – sin 70° + sin 10° = 2 cos (50o +70o)/2 sin (50o – 70o) + sin 10o
= 2 cos 120o/2 sin (-20o)/2 +sin 10o
= 2 cos 60o (- sin 10o) +sin 10o [since,{sin (-A) = -sin (A)}]
= 2 × 1/2 × – sin 10o + sin 10o
= 0
= RHS
Hence proved.
(iii) cos 80° + cos 40° – cos 20° = 0
Let us consider LHS:
cos 80° + cos 40° – cos 20°
By using the formula,
cos A + cos B = 2 cos (A+B)/2 cos (A-B)/2
cos 80° + cos 40° – cos 20° = 2 cos (80o +40o)/2 cos (80o – 40o) – cos 20o
= 2 cos 120o/2 cos 40o/2 – cos20o
= 2 cos 60° cos 20o – cos 20°
= 2 × 1/2 × cos 20o – cos 20o
= 0
= RHS
Hence Proved.
(iv) cos 20° + cos 100° + cos 140° = 0
Let us consider LHS:
cos 20° + cos 100° + cos 140°
By using the formula,
cos A + cos B = 2 cos (A+B)/2 cos (A-B)/2
cos 20° + cos 100° + cos 140° = 2 cos (20o +100o)/2 cos (20o – 100o) + cos (180o –40o)
= 2 cos 120o/2 cos (-80o)/2 –cos 40o (since, {cos (180° – A) = – cos A})
= 2 cos 60° cos (-40°) – cos 40° (since, {cos (-A) =cos A})
= 2 × 1/2 × cos 40o – cos 40o
= 0
= RHS
Hence Proved.
(v) sin 5π/18 – cos 4π/9 = √3 sin π/9
Let us consider LHS:
sin 5π/18 – cos 4π/9 = sin 5π/18 – sin (π/2 – 4π/9)(since, cos A = sin (90o – A))
= sin 5π/18 – sin (9π – 8π)/18
= sin 5π/18 – sin π/18
By using the formula,
sin A – sin B = 2 cos (A+B)/2 sin (A-B)/2
= 2 cos (6π/36) sin (4π/36)
= 2 cos π/6 sin π/9
= 2 cos 30o sin π/9
= 2 × √3/2 × sin π/9
= √3 sin π/9
= RHS
Hence proved.
(vi) cos π/12 – sin π/12 = 1/√2
Let us consider LHS:
cos π/12 – sin π/12 = sin (π/2 – π/12) – sin π/12(since, cos A = sin(90o – A))
= sin (6π – 5π)/12 – sin π/12
= sin 5π/12 – sin π/12
By using the formula,
sin A – sin B = 2 cos (A+B)/2 sin (A-B)/2
= 2 cos (6π/24) sin (4π/24)
= 2 cos π/4 sin π/6
= 2 cos 45o sin 30o
= 2 × 1/√2 × 1/2
= 1/√2
= RHS
Hence proved.
(vii) sin 80° – cos 70° = cos 50°
sin 80° = cos 50° + cos 70o
So, now let us consider RHS
cos 50° + cos 70o
By using the formula,
cos A + cos B = 2 cos (A+B)/2 cos (A-B)/2
cos 50° + cos 70o = 2 cos (50o +70o)/2 cos (50o – 70o)/2
= 2 cos 120o/2 cos (-20o)/2
= 2 cos 60o cos (-10o)
= 2 × 1/2 × cos 10o (since, cos (-A) =cos A)
= cos 10o
= cos (90° – 80°)
= sin 80° (since, cos (90° – A) = sin A)
= LHS
Hence Proved.
(viii) sin 51° + cos 81° = cos 21°
Let us consider LHS:
sin 51° + cos 81° = sin 51o + sin (90o –81o)
= sin 51o + sin 9o (since,sin (90° – A) = cos A)
By using the formula,
sin A + sin B = 2 sin (A+B)/2 cos (A-B)/2
sin 51o + sin 9o = 2sin (51o + 9o)/2 cos (51o – 9o)/2
= 2 sin 60o/2 cos 42o/2
= 2 sin 30o cos 21o
= 2 × 1/2 × cos 21o
= cos 21o
= RHS
Hence proved.
Question - 4 : - Prove that:
(i) cos (3π/4 + x) – cos (3π/4 – x) = -√2 sin x
(ii) cos (π/4 + x) + cos (π/4 – x) = √2 cos x
Answer - 4 : -
(i) cos (3π/4 + x) – cos (3π/4 – x) = -√2 sin x
Let us consider LHS:
cos (3π/4 + x) – cos (3π/4 – x)
By using the formula,
cos A – cos B = -2 sin (A+B)/2 sin (A-B)/2
cos (3π/4 + x) – cos (3π/4 – x) = -2 sin (3π/4 + x + 3π/4 – x)/2 sin (3π/4 + x – 3π/4 + x)/2
= -2 sin (6π/4)/2 sin 2x/2
= -2 sin 6π/8 sin x
= -2 sin 3π/4 sin x
= -2 sin (π – π/4) sin x
= -2 sin π/4 sin x (since, (π-A) = sin A)
= -2 × 1/√2 × sin x
= -√2 sin x
= RHS
Hence proved.
(ii) cos (π/4 + x) + cos (π/4 – x) = √2 cos x
Let us consider LHS:
cos (π/4 + x) + cos (π/4 – x)
By using the formula,
cos A + cos B = 2 cos (A+B)/2 cos (A-B)/2
cos (π/4 + x) + cos (π/4 – x) = 2 cos (π/4 + x + π/4 – x)/2 cos (π/4 + x – π/4 + x)/2
= 2 cos (2π/4)/2 cos 2x/2
= 2 cos 2π/8 cos x
= 2 sin π/4 cos x
= 2 × 1/√2 × cos x
= √2 cos x
= RHS
Hence proved.
Question - 5 : - Provethat:
(i) sin65o + cos 65o = √2 cos 20o
(ii) sin47o + cos 77o = cos 17o
Answer - 5 : -
(i) sin 65o + cos 65o = √2cos 20o
Let us consider LHS:
sin 65o + cos 65o =sin 65o + sin (90o – 65o)
= sin 65o + sin 25o
By using the formula,
sin A + sin B = 2 sin (A+B)/2 cos (A-B)/2
sin 65o + sin 25o = 2sin (65o + 25o)/2 cos (65o – 25o)/2
= 2 sin 90o/2 cos 40o/2
= 2 sin 45o cos 20o
= 2 × 1/√2 × cos 20o
= √2 cos 20o
= RHS
Hence proved.
(ii) sin 47o + cos 77o =cos 17o
Let us consider LHS:
sin 47o + cos 77o =sin 47o + sin (90o – 77o)
= sin 47o + sin 13o
By using the formula,
sin A + sin B = 2 sin (A+B)/2 cos (A-B)/2
sin 47o + sin 13o = 2sin (47o + 13o)/2 cos (47o – 13o)/2
= 2 sin 60o/2 cos 34o/2
= 2 sin 30o cos 17o
= 2 × 1/2 × cos 17o
= cos 17o
= RHS
Hence proved.
Question - 6 : - Provethat:
(i) cos 3A +cos 5A + cos 7A + cos 15A = 4 cos 4A cos 5A cos 6A
(ii) cosA + cos 3A + cos 5A + cos 7A = 4 cos A cos 2A cos 4A
(iii) sinA + sin 2A + sin 4A + sin 5A = 4 cos A/2 cos 3A/2 sin 3A
(iv) sin3A + sin 2A – sin A = 4 sin A cos A/2 cos 3A/2
(v) cos20o cos 100o + cos 100o cos 140o –cos 140o cos 200o = – 3/4
(vi) sinx/2 sin 7x/2 + sin 3x/2 sin 11x/2 = sin 2x sin 5x
(vii) cosx cos x/2 – cos 3x cos 9x/2 = sin 4x sin 7x/2
Answer - 6 : -
(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.
Question - 7 : - Provethat:
Answer - 7 : -
Question - 8 : - Prove that:
Answer - 8 : -
= cot 6A
= RHS
Hence proved.
By using the formulas,
sin A – sin B = 2 cos (A+B)/2 sin (A-B)/2
cos A – cos B = -2 sin (A+B)/2 sin (A-B)/2
= RHS
Hence proved.
Prove that:
(i) sin α + sin β + sin γ – sin (α + β + γ) = 4 sin (α + β)/2 sin (β + γ)/2 sin (α + γ)/2
(ii) cos (A + B + C) + cos (A – B + C) + cos (A + B – C) + cos (-A + B + C) = 4 cos A cos B cos C
Answer - 9 : -
(i) sin α + sin β + sin γ – sin (α + β + γ) = 4 sin (α + β)/2 sin (β + γ)/2 sin (α + γ)/2
Let us consider LHS:
sin α + sin β + sin γ – sin (α + β + γ)
By using the formulas,
Sin A + sin B = 2 sin (A+B)/2 cos (A-B)/2
Sin A – sin B = 2 cos (A+B)/2 sin (A-B)/2
= RHS
Hence proved.
(ii) cos (A + B + C) + cos (A – B + C) + cos (A + B – C) + cos (-A + B + C) = 4 cos A cos B cos C
Let us consider LHS:
cos (A + B + C) + cos (A – B + C) + cos (A + B – C) + cos (-A + B + C)
so,
cos (A + B + C) + cos (A – B + C) + cos (A + B – C) + cos (-A + B + C) =
={cos (A + B + C) + cos (A – B + C)} + {cos (A + B – C) + cos (-A + B + C)}
By using the formula,
Cos A + cos B = 2 cos (A+B)/2 cos (A-B)/2
= 4 cos A cos B cos C
= RHS
Hence proved.