Question 1128737
In the expression "sin(cos-1(3 - x))"
the "cos-1" or {{{"cos"^(-1)}}} part refers to the inverse cosine function,
so {{{cos^(-1)(3-x)}}} is an angle {{{A}}} , between {{{0}}} and {{{pi}}}
(or between {{{0^o}}} and {{{180^o}}} if you want to visualize angles in degrees)
whose cosine is {{{3-x}}} .
Those angles would have a non-negative sine, so {{{sin(cos^(-1)(3-x))>=0}}} .
Using the trigonometric identity {{{sin^2(A)+cos^2(A)=1}}} ,
{{{sin^2(cos^(-1)(3-x))+cos^2(A)=1}}} ,
{{{sin^2(cos^(-1)(3-x))=1-(3-x)^2}}} ,
{{{sin^2(cos^(-1)(3-x))=1-(9-6x+x^2)}}} ,
{{{sin^2(cos^(-1)(3-x))=1-9+6x-x^2}}} , and
{{{sin^2(cos^(-1)(3-x))=-8+6x-x^2}}}
We are looking for a non-negative result, so
{{{sin(cos^(-1)(3-x))=highlight(sqrt(-8+6x-x^2))}}}


NOTE:
Of course {{{cos^(-1)(3-x)}}} is not defined as a real number for all values of {{{x}}} .
It must be {{{-1<=3-x<=1}}} or {{{2<=x<=4}}} .
That is the domain of the function,
but that is probably not expected as part of the answer.