Question -
Answer -
(i)┬аsin 50o┬аcos85o┬а= (1 тАУ тИЪ2sin 35o)/2тИЪ2
By using the formula,
2 sin A cos B = sin (A +B) + sin (A тАУ B)
sin A cos B = [sin (A +B) + sin (A тАУ B)] / 2
sin 50o┬аcos 85o┬а=[sin(50o┬а+ 85o) + sin (50o┬атАУ85o)] / 2
= [sin (135o) + sin (-35o)]/ 2
= [sin (135o) тАУ sin (35o)]/ 2 (since, sin (-x) = -sin x)
= [sin (180o┬атАУ 45o)тАУ sin 35o] / 2
= [sin 45o┬атАУ sin 35o]/ 2
= [(1/тИЪ2) тАУ sin 35o] / 2
= [(1 тАУ sin 35o)/тИЪ2] / 2
= (1 тАУ sin 35o) / 2тИЪ2
Hence proved.
(ii)┬аsin 25o┬аcos115o┬а= 1/2 {sin 140o┬атАУ 1}
By using the formula,
2 sin A cos B = sin (A +B) + sin (A тАУ B)
sin A cos B = [sin (A +B) + sin (A тАУ B)] / 2
sin 20o┬аcos 115o┬а=[sin(25o┬а+ 115o) + sin (25o┬атАУ115o)] / 2
= [sin (140o) + sin (-90o)]/ 2
= [sin (140o) тАУ sin (90o)]/ 2 (since, sin (-x) = -sin x)
= 1/2 {sin 140o┬атАУ 1}
Hence proved.