Question

Question #b6b30

Answer 1

Given

`sinx=1/4`

`and 0< x < pi/2-> x in "1st quadrant"`

So `cosx=sqrt(1-sin^2x)=sqrt(1-1/16)=sqrt15/4`

Now

`sin(x/2)=sqrt(1/2(1-cosx))`

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

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