(i) Given as sin 50° + sin 10° = cos 20°
Let us consider the LHS
sin 50° + sin 10°
On using the formula,
sin A + sin B = 2 sin (A + B)/2 cos (A - B)/2
sin 50° + sin 10° = 2 sin (50° + 10°)/2 cos (50° – 10°)/2
= 2 sin 60°/2 cos 40°/2
= 2 sin 30° cos 20°
= 2 × 1/2 × cos 20°
= cos 20°
= RHS
Thus proved.
(ii) sin 23° + sin 37° = cos 7°
Let us consider the LHS
sin 23° + sin 37°
On using the formula,
sin A + sin B = 2 sin (A + B)/2 cos (A - B)/2
sin 23° + sin 37° = 2 sin (23° + 37°)/2 cos (23° – 37°)/2
= 2 sin 60°/2 cos -14°/2
= 2 sin 30° cos -7°
= 2 × 1/2 × cos -7°
= cos 7° (since, {cos (-A) = cos A})
= RHS
Thus proved.
(iii) sin 105° + cos 105° = cos 45°
Let us consider the LHS
sin 105° + cos 105° = sin 105° + sin (90° – 105°) [since, {sin (90° – A) = cos A}]
= sin 105° + sin (-15°)
= sin 105° – sin 15° [{sin(-A) = – sin A}]
On using the formula,
Sin A – sin B = 2 cos (A + B)/2 sin (A - B)/2
sin 105° – sin 15° = 2 cos (105° + 15°)/2 sin (105° – 15°)/2
= 2 cos 120°/2 sin 90°/2
= 2 cos 60° sin 45°
= 2 × 1/2 × 1/√2
= 1/√2
= cos 45°
= RHS
Thus proved.
(iv) Given as sin 40° + sin 20° = cos 10°
Let us consider the LHS:
sin 40° + sin 20°
On using the formula,
sin A + sin B = 2 sin (A + B)/2 cos (A - B)/2
sin 40° + sin 20° = 2 sin (40° + 20°)/2 cos (40° – 20°)/2
= 2 sin 60°/2 cos 20°/2
= 2 sin 30° cos 10°
= 2 × 1/2 × cos 10°
= cos 10°
= RHS
Thus proved.
