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Question -

showthat:

(i) sin 50o┬аcos 85o┬а=(1 тАУ тИЪ2sin 35o)/2тИЪ2

(ii) sin25o┬аcos 115o┬а= 1/2 {sin 140o┬атАУ 1}



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.

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