Question

(i) sin 50° cos 85º = 1-√2sin35°/2√2

Answer 1

(i) On using the formula,

2 sin A cos B = sin (A + B) + sin (A – B)

sin A cos B = [sin (A + B) + sin (A – B)]/2

sin 50° cos 85° = [sin(50° + 85°) + sin (50° – 85°)]/

= [sin (135°) + sin (-35°)]/2

= [sin (135°) – sin (35°)]/2 (since, sin (-x)

= -sin x) = [sin (180° – 45°) – sin 35°]/2 = [sin 45° – sin 35°]/2

= [(1/√2) – sin 35°]/2

= [(1 – sin 35°)/√2]/2

= (1 – sin 35°)/2√2 Thus proved.

(ii) sin 25° cos 115°

= 1/2 {sin 140° – 1} On using the formula, 2 sin A cos B

= sin (A + B) + sin (A – B) sin A cos B

= [sin (A + B) + sin (A – B)]/2 sin 20° cos 115° = [sin(25° + 115°) + sin (25° – 115°)]/2

= [sin (140°) + sin (-90°)]/2

= [sin (140°) – sin (90°)]/2 (since, sin (-x)

= -sin x) = 1/2 {sin 140° – 1}

Thus proved.