Answer
The product of the diagonals is equal to the sum of the products of the opposite sides. Therefore:
$sin(A+B) = sin~A~cos~B+sin~B~cos~A$
Note that this is the formula for the sine of the sum of two angles.
Work Step by Step
The line through the middle of the circle is the diameter. We know that any triangle constructed with the diameter as one side and with a point on the circle's circumference is a right triangle. Therefore, the four sides around the outside have lengths $sin~A, cos~A, sin~B,$ and $cos~B$ since the hypotenuse has a length of 1.
In Exercise 53, we saw that for any triangle inscribed in a circle, $\frac{d}{sin~D} = 2r$. In this case, let $d$ be the side opposite the angle $A+B$. Then:
$\frac{d}{sin~(A+B)} = 2r = 1$
$d = sin~(A+B)$
Therefore, the side opposite the angle $A+B$ has a length of $sin(A+B)$
The product of the diagonals is $sin~(A+B)(1) = sin(A+B)$
The product of one pair of opposite sides is $sin~A~cos~B$
The product of the other pair of opposite sides is $sin~B~cos~A$
The sum of these two products is $sin~A~cos~B+sin~B~cos~A$
From the information in the question, the product of the diagonals is equal to the sum of the products of the opposite sides. Therefore:
$sin(A+B) = sin~A~cos~B+sin~B~cos~A$
Note that this is the formula for the sine of the sum of two angles.
