# proof of angle sum identities

We will derive the angle sum identities for the various trigonometric functions here. We begin by deriving the identity for the sine by means of a geometric argument and then obtain the remaining identities by algebraic manipulation.

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

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

 ????????????wwwwwwwwwwwwwwww$\textstyle{A}$$\textstyle{B}$$\textstyle{C}$

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 $AB$:

 ????????????wwwwwwwwwwwwwwww$\textstyle{A}$$\textstyle{B}$$\textstyle{C}$$\textstyle{D}$

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.)

 ???????????????? ΥΥΥΥΥΥΥ ¨¨¨¨¨¨¨¨¨ $\textstyle{A}$$\textstyle{B}$$\textstyle{C}$$\textstyle{E}$
 ?????????????? ¨¨¨¨¨¨¨ wwwwwwwwwwwwwwww$\textstyle{A}$$\textstyle{B}$$\textstyle{C}$$\textstyle{E}$

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 $\overline{AD}+\overline{BD}=\overline{AB}$:

 $\displaystyle\sin(x+y)=\sin(\angle BCA)$ $\displaystyle=\overline{AB}\,\sin(\angle ABC)/\overline{AC}$ $\displaystyle=(\overline{AD}+\overline{BD})\sin(\angle ABC)/\overline{AC}$ $\displaystyle=\overline{CD}\left(\cot(\angle CAB)+\cot(\angle ABC)\right)/\sin% (\angle ABC)\overline{AC}$ $\displaystyle=\sin(\angle CAB)\sin(\angle ABC)\left({\cos(\angle CAB)\over\sin% (\angle CAB)}+{\cos(\angle ABC)\over\sin(\angle ABC)}\right)$ $\displaystyle=\sin(\angle CAB)\cos(\angle ABC)+\cos(\angle CAB)\sin(\angle ABC)$ $\displaystyle=\sin(x)\cos(y)+\cos(x)\sin(y)$

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

Entry under construction

Title proof of angle sum identities ProofOfAngleSumIdentities 2013-05-28 14:35:29 2013-05-28 14:35:29 rspuzio (6075) rspuzio (6075) 15 rspuzio (6075) Proof msc 43-00 msc 51-00 msc 42-00 msc 33B10