Theorem 1   $$ \sin (x + y) = \sin (x) \cos (y) + \cos (x) \sin (y)$$

Proof. Let us make the restrictions $0^\circ < x < 90^\circ$ and $0^\circ < y < 90^\circ$ for the time being. Then we may draw a triangle $ABC$ such that $\angle CAB = x$ and $\angle ABF = y$ :

Since the angles of a triangle add up to $180^\circ$ , we must have $\angle BCA = 180^\circ - x - y$ , so we have $\sin (\angle BCA) = \sin (180^\circ - x - y) = \sin (x + y)$ .

We now draw perpendiculars two different ways in order to derive ratios. First, we drop a perpendicular $AD$ from $C$ to $AB$ :

Since $ACD$ and $BCD$ are right triangles we have, by definition,$$ \cot (\angle CAB) = \overline{AD} / \overline{CD} \qquad \cot (\angle ABC) = \overline{BD} / \overline{CD} \qquad \sin (\angle CAB) = \overline{CD} / \overline{AC} .$$

Second, we draw a perpendicular $AE$ form $A$ to $BC$ . Depending on whether $x+y < 90^\circ$ or $x+y < 90^\circ$ the point $E$ will or will not lie between $B$ and $C$ , as illustrated below. (There is also the case $x+y = 90^\circ$ , but it is trivial.)

Either way, $ABE$ and $ACE$ are right triangles, and we have, by definition,$$ \sin (\angle BCA) = \overline{AE} / \overline{AC} \qquad \sin (\angle ABC) = \overline{AE} / \overline{AB} .$$ Combining these ratios, we find that$$ \sin (\angle BCA) / \sin (\angle ABC) = \overline{AB} / \overline{AC} .$$

To finish deriving the sum identity, we manipulate the ratios derived above algebraically and use the fact that $\overline{AD} + \overline{BD} = \overline{AB}$ :

To lift the restriction on the range of $x$ and $y$ , we use the identities for complements and negatives of angles.