# Find the value of `sin(x/2)` given `sinx=-1/3` and `pi<x<(3pi)/2`

We will use formula for `sin (x/2)`:

`sin(x/2)=pm sqrt((1-cos x)/2)`

(`pm` means that sign depends on quadrant in which is `x/2` ` ` )

We have `sin x` but in above formula we need `cos x` which we will calculate from the following formula:

`cos x=pm sqrt(1-sin^2x)`

Since `pi leq x leq (3pi)/2` cosine is negative hence

`cos x= -sqrt(1-(-1/3)^2)=-sqrt(8/9)=-(2sqrt2)/3`

Since `pi leq x leq (3pi)/2=>pi/2 leq x/2 leq (3pi)/4` which means that `x/2` is in second quadrant where sine is positive.

`sin(x/2)=sqrt((1-(-(2sqrt2)/3))/2)=sqrt(1/6(3+2sqrt2))`

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