sin . x/2 , cos . x/2 तथा tan . x/2 ज्ञात कीजिएः` sin x = 1/4 ,x` द्वितीय चतुर्थाश में है ।

Explanation

` sin x = 1/4 " " sin^(2) x + cos^(2) x = 1`
` rArr (1/4)^(2) + cos^(2) x = 1`
` rArr cos^(2) x = 1 - 1/16 = 15/16`
` rArr cos x = -(sqrt(15))/4` (क्योकि x दूसरे चतुर्थाश में है । )
` :' cos x = - sqrt(15)/4 rArr 2 cos^(2) .x/2 -1 = -(sqrt(15))/4`
` rArr 2 cos^(2) . x/2 = 1- (sqrt(15)) = (4- sqrt(15))/4`
` rArr cos^(2) .x/2 = (4- sqrt(15))/8`
` rArr cos.x/2 = sqrt((4-sqrt(15))/8)=sqrt((8-2sqrt(15))/16)`
` :' x " द्वितीय चतुर्थाश में है " rArrpi/2 lt c lt pi`
तब `pi/2` पहले चतुर्थाश में गोगा जिसे सभी अनुपात सदैव धनात्मक होंगे ।
` :' cos.x/2 = sqrt(8-2sqrt(15))/4`
` :. cos ^(2) . x/2 = ( 8 - 2sqrt(15))/16 "तब " sin^(2).x/2 = 1- cos^(2) .x/2 `
या ` sin^(2). x/2 = 1- (8-2sqrt(15))/16`
या ` sin^(2) .x/2 = (8 +2sqrt(15))/16`
या ` sin.x/2 = sqrt((8+2sqrt(15))/16)`
या ` sin .x/2 = sqrt(8+2sqrt(15))/4`
अब ` tan.x/2 = (sin.x/2)/(cos.x/2)= (sqrt(8+2sqrt(15))/4)/(sqrt(8-2sqrt(15))/4)`
` = sqrt(8+2sqrt(15))/(sqrt(8-2sqrt(15)))xx (sqrt(8 + 2 sqrt(15)))/(sqrt(8+2sqrt(15)))`
` = (sqrt(8+2(sqrt(15)))^(2))/(sqrt(8)^(2)-(2sqrt(15))^(2))= (8+2sqrt(15))/(sqrt(64-60))`
` = ( 8 + 2 sqrt(15 ))/2 -= 4 + sqrt(15)`
अतः ` sin.x/2 = (sqrt8+2sqrt(15))/4 ,cos. x/2 = (sqrt(8-2sqrt(15)))/4`
तथा ` tan. x/2 = 4 + sqrt(15)`