Question

Find `sin  x/2, cos  x/2 and tan  x/2` of the following:

`sin x =  1/4`, x in quadrant II

Answer 1

x, is in the II quadrant.

⇒ 90° < `x/2` < 180°

Dividing by 2 gives 45° < `x/2` < 90°

⇒ `x/2` is in the first quadrant

∴ `sin  x/2, cos  x/2, tan  x/2 "All three are positive."`

`sinx = 1/4 , cos x = - sqrt1 - sin^2 x = - sqrt(1 - 1/16) = - sqrt15/4` [x is in the second quadrant]

`sin  x/2 = +  sqrt((1 - cos x)/2 = sqrt((1 + sqrt15/4)/2)`

= `sqrt(15+ 4/8)  = sqrt(((15+4)2)/16`

= `sqrt((2sqrt15 + 8)/4`

= `cos  x/2 = + sqrt((1 + cosx)/2`

= `sqrt((1- sqrt15/4)/2   = sqrt(4- sqrt15)/8`

= tan `x /2 = sqrt((1-cosx)/(1 + cosx ` = `sqrt((1 +sqrt15/4)/(1 - sqrt15/4`

= `sqrt(4+sqrt15/4)/((4 - sqrt15)/4`

= `sqrt((4 +sqrt15)/(4- sqrt15`

= `sqrt((4 +sqrt15 xx sqrt4 + sqrt15)/(4 - sqrt15 xx sqrt4 + sqrt15)`

= `sqrt((4 + sqrt15)^2)/(16 - 15)`

= 4 + `sqrt15`

Hence sin `x/2 = sqrt((2sqrt15+8)/4, cos  x/2 = sqrt(8-2sqrt15)/4 and tan  x/2 = 4 + sqrt15`