Question -
Answer -
(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.