Question -
Answer -
(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.