Question -
Answer -
(i) sin 50o cos85o = (1 – √2sin 35o)/2√2
By 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 50o cos 85o =[sin(50o + 85o) + sin (50o –85o)] / 2
= [sin (135o) + sin (-35o)]/ 2
= [sin (135o) – sin (35o)]/ 2 (since, sin (-x) = -sin x)
= [sin (180o – 45o)– sin 35o] / 2
= [sin 45o – sin 35o]/ 2
= [(1/√2) – sin 35o] / 2
= [(1 – sin 35o)/√2] / 2
= (1 – sin 35o) / 2√2
Hence proved.
(ii) sin 25o cos115o = 1/2 {sin 140o – 1}
By 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 20o cos 115o =[sin(25o + 115o) + sin (25o –115o)] / 2
= [sin (140o) + sin (-90o)]/ 2
= [sin (140o) – sin (90o)]/ 2 (since, sin (-x) = -sin x)
= 1/2 {sin 140o – 1}
Hence proved.