(i) Given as cos 55° + cos 65° + cos 175° = 0
Let us consider the LHS
cos 55° + cos 65° + cos 175°
On using the formula,
cos A + cos B = 2 cos (A + B)/2 cos (A - B)/2
cos 55° + cos 65° + cos 175° = 2 cos (55° + 65°)/2 cos (55° – 65°) + cos (180° – 5°)
= 2 cos 120°/2 cos (-10°)/2 – cos 5° (since, {cos (180° – A) = – cos A})
= 2 cos 60° cos (-5°) – cos 5° (since, {cos (-A) = cos A})
= 2 × 1/2 × cos 5° – cos 5°
= 0
= RHS
Thus proved.
(ii) sin 50° – sin 70° + sin 10° = 0
Let us consider the LHS
sin 50° – sin 70° + sin 10°
On using the formula,
sin A – sin B = 2 cos (A + B)/2 sin (A - B)/2
sin 50° – sin 70° + sin 10° = 2 cos (50° + 70°)/2 sin (50° – 70°) + sin 10°
= 2 cos 120°/2 sin (-20°)/2 + sin 10°
= 2 cos 60° (- sin 10°) + sin 10° [since,{sin (-A) = -sin (A)}]
= 2 × 1/2 × – sin 10° + sin 10°
= 0
= RHS
Thus proved.
(iii) cos 80° + cos 40° – cos 20° = 0
Let us consider the LHS
cos 80° + cos 40° – cos 20°
On using the formula,
cos A + cos B = 2 cos (A + B)/2 cos (A - B)/2
cos 80° + cos 40° – cos 20° = 2 cos (80° + 40°)/2 cos (80° – 40°) – cos 20°
= 2 cos 120°/2 cos 40°/2 – cos 20°
= 2 cos 60° cos 20° – cos 20°
= 2 × 1/2 × cos 20° – cos 20°
= 0
= RHS
Thus proved.
(iv) cos 20° + cos 100° + cos 140° = 0
Let us consider the LHS
cos 20° + cos 100° + cos 140°
On using the formula,
cos A + cos B = 2 cos (A + B)/2 cos (A - B)/2
cos 20° + cos 100° + cos 140° = 2 cos (20° + 100°)/2 cos (20° – 100°) + cos (180° – 40°)
= 2 cos 120°/2 cos (-80°)/2 – cos 40° (since, {cos (180° – A) = – cos A})
= 2 cos 60° cos (-40°) – cos 40° (since, {cos (-A) = cos A})
= 2 × 1/2 × cos 40° – cos 40°
= 0
= RHS
Thus proved.
