Prove that :(i) sin 38° + sin 22° = sin 82°(ii) cos 100° + cos 20�

1.	Prove that :(i) sin 38° + sin 22° = sin 82°(ii) cos 100° + cos 20° = cos 40°
Answer» (i) sin 38° + sin 22° = sin 82° Let us consider the LHS sin 38° + sin 22° On using the formula, sin A + sin B = 2 sin (A + B)/2 cos (A - B)/2 sin 38° + sin 22° = 2 sin (38° + 22°)/2 cos (38° – 22°)/2 = 2 sin 60°/2 cos 16°/2 = 2 sin 30° cos 8° = 2 × 1/2 × cos 8° = cos 8° = cos (90° – 82°) = sin 82° (since, {cos (90° – A) = sin A}) = RHS Thus proved. (ii) cos 100° + cos 20° = cos 40° Let us consider the LHS: cos 100° + cos 20° On using the formula, cos A + cos B = 2 cos (A + B)/2 cos (A - B)/2 cos 100° + cos 20° = 2 cos (100° + 20°)/2 cos (100° – 20°)/2 = 2 cos 120°/2 cos 80°/2 = 2 cos 60° cos 4° = 2 × 1/2 × cos 40° = cos 40° = RHS Thus proved.

Prove that :(i) sin 38° + sin 22° = sin 82°(ii) cos 100° + cos 20° = cos 40°

Answer»

(i) sin 38° + sin 22° = sin 82°

Let us consider the LHS

sin 38° + sin 22°

On using the formula,

sin A + sin B = 2 sin (A + B)/2 cos (A - B)/2

sin 38° + sin 22° = 2 sin (38° + 22°)/2 cos (38° – 22°)/2

= 2 sin 60°/2 cos 16°/2

= 2 sin 30° cos 8°

= 2 × 1/2 × cos 8°

= cos 8°

= cos (90° – 82°)

= sin 82° (since, {cos (90° – A) = sin A})

= RHS

Thus proved.

(ii) cos 100° + cos 20° = cos 40°

Let us consider the LHS:

cos 100° + cos 20°

On using the formula,

cos A + cos B = 2 cos (A + B)/2 cos (A - B)/2

cos 100° + cos 20° = 2 cos (100° + 20°)/2 cos (100° – 20°)/2

= 2 cos 120°/2 cos 80°/2

= 2 cos 60° cos 4°

= 2 × 1/2 × cos 40°

= cos 40°

= RHS

Thus proved.