Question 1024326
<pre>
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<sup>-1</sup>x + cos<sup>-1</sup>y)

Let A = sin<sup>-1</sup>x  then sinA = x
Let B = cos<sup>-1</sup>y  then cosB = y

and B = cos<sup>-1</sup>x

sin(A + B) = sinA&#8729;cosB + cosA&#8729;sinB
           = x&#8729;y + cosA&#8729;sinB
 
There are four cases.

1. sin<sup>-1</sup>x is non-negative and cos<sup>-1</sup>x is non-negative.
2. sin<sup>-1</sup>x is non-negative and cos<sup>-1</sup>x is negative.
3. sin<sup>-1</sup>x is negative and cos<sup>-1</sup>x is non-negative.
4. sin<sup>-1</sup>x is non-negative and cos<sup>-1</sup>x is non-negative.

But in all cases, since 

cosA = ±&#8730;<span style="text-decoration: overline">1-sin²A</span> = ±&#8730;<span style="text-decoration: overline">1-x²±</span>

sinB = ±&#8730;<span style="text-decoration: overline">1-cos²B</span> = ±&#8730;<span style="text-decoration: overline">1-y²</span>
 
sin(A + B) = xy ± &#8730;<span style="text-decoration: overline">(1-x²)(1-y²)</span> 

Edwin</pre>