1.  For your information: sin^-1(x) = arcsin(x);   cos^-1(y) = arccos(y).


2.  So, you are given  = arcsin(x)  and   = arccos(y), 
    and you are asked to express  in terms of x and y.


3.  Since   = arcsin(x)  and   = arccos(y),  you have
     = x  and   = y.

    It implies that   = +/-  (depending on in which quadrant the angle  is) 
    and   = +/- (depending on where the angle  is).


4. Now use the formula   = . 

   (If you don't know it, then see the lesson Addition and subtraction formulas in this site).


   Substitute what you know (what is given and what is written above) into this formula, and you will get 

    = .

   with different possible combinations of signs.

   It is your answer.

I think the lady above made a careless error and 
also did not take the four cases of signs of sines and
cosines in the various quadrants.

sin(sin-1x + cos-1y)

Let A = sin-1x  then sinA = x
Let B = cos-1y  then cosB = y

and B = cos-1x

sin(A + B) = sinA∙cosB + cosA∙sinB
           = x∙y + cosA∙sinB
 
There are four cases.

1. sin-1x is non-negative and cos-1x is non-negative.
2. sin-1x is non-negative and cos-1x is negative.
3. sin-1x is negative and cos-1x is non-negative.
4. sin-1x is non-negative and cos-1x is non-negative.

But in all cases, since 

cosA = ±√1-sin²A = ±√1-x²±

sinB = ±√1-cos²B = ±√1-y²
 
sin(A + B) = xy ± √(1-x²)(1-y²) 

Edwin