Let $△AFD$ be a right triangle with hypotenuse $1$ and an acute angle with measure $x+y.$

By definition, the sine of an angle is the quotient between the length of the opposite side and the hypotenuse. $$\begin{array}{c}\sin (x+y)=\frac{DF}{1}\\ \Downarrow \\ DF=\sin (x+y)\end{array}$$ Next, by drawing a ray, $\angle A$ can be divided into two angles with measures $x$ and $y.$ Let $C$ be a point on this ray such that $△ACD$ and $△ABC$ are right triangles.

Consider $△ACD.$ By calculating the sine and cosine of $x,$ the legs of this triangle can be rewritten. $$\begin{array}{c}\sin x=\frac{DC}{1}\Rightarrow DC=\sin x\\ \cos x=\frac{AC}{1}\Rightarrow AC=\cos x\end{array}$$ Consider $△ABC.$ Knowing that $AC=\cos x,$ the sine of $y$ can be used to write $BC$ in terms of $x$ and $y.$ $$\begin{array}{rl}\sin y& =\frac{BC}{AC}\\ \sin y& =\frac{BC}{\cos x}\Rightarrow BC=\cos x\sin y\end{array}$$ Let $G$ be the point of intersection between $\overline{FD}$ and $\overline{AC}.$ Notice that $\angle AGF\cong \angle DGC$ by the Vertical Angles Theorem.

By the Third Angle Theorem, it is known that $\angle GAF\cong \angle GDC.$ Therefore, $m\angle GDC=y.$

Now, let $E$ be a point on $\overline{FD}$ such that $\overline{EC}\parallel \overline{AB}.$ This implies that $EF=\cos x\sin y.$ In addition, $△CED$ is a right triangle and then $\cos y$ can be calculated using $△CED.$

Finally, by the Segment Addition Postulate, $DF$ is equal to the sum of $DE$ and $EF.$ All these lengths have been rewritten in terms of the sine and cosine of $x$ and $y.$