proof of angle sum identities

We will derive the angle sum identities for the various trigonometricfunctions here. We begin by deriving the identity for the sine by meansof a geometric argument and then obtain the remaining identities byalgebraic manipulation.

Theorem 1.

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

Proof.

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

 ????????????wwwwwwwwwwwwwwww$A$$B$$C$

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

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

 ????????????wwwwwwwwwwwwwwww$A$$B$$C$$D$

Since $ACD$ and $BCD$ are right triangles we have, by definition,

$$\cot (\mathrm{\angle }CAB)=\overline{AD}/\overline{CD}\mathit{\hspace{1em}\hspace{1em}}\cot (\mathrm{\angle }ABC)=\overline{BD}/\overline{CD}\mathit{\hspace{1em}\hspace{1em}}\sin (\mathrm{\angle }CAB)=\overline{CD}/\overline{AC}.$$

Second, we draw a perpendicular $AE$ form $A$ to $BC$. Depending on whether or 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¨¨¨¨¨¨¨¨¨$A$$B$$C$$E$

 ??????????????¨¨¨¨¨¨¨wwwwwwwwwwwwwwww$A$$B$$C$$E$

Either way, $ABE$ and $ACE$ are right triangles, and we have, by definition,

$$\sin (\mathrm{\angle }BCA)=\overline{AE}/\overline{AC}\mathit{\hspace{1em}\hspace{1em}}\sin (\mathrm{\angle }ABC)=\overline{AE}/\overline{AB}.$$

Combining these ratios, we find that

$$\sin (\mathrm{\angle }BCA)/\sin (\mathrm{\angle }ABC)=\overline{AB}/\overline{AC}.$$

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

$\sin (x+y)=\sin (\mathrm{\angle }BCA)$$=\overline{AB}\sin (\mathrm{\angle }ABC)/\overline{AC}$

$=(\overline{AD}+\overline{BD})\sin (\mathrm{\angle }ABC)/\overline{AC}$

$=\overline{CD}\left(\cot (\mathrm{\angle }CAB)+\cot (\mathrm{\angle }ABC)\right)/\sin (\mathrm{\angle }ABC)\overline{AC}$

$=\sin (\mathrm{\angle }CAB)\sin (\mathrm{\angle }ABC)\left({\displaystyle \frac{\cos (\mathrm{\angle }CAB)}{\sin (\mathrm{\angle }CAB)}}+{\displaystyle \frac{\cos (\mathrm{\angle }ABC)}{\sin (\mathrm{\angle }ABC)}}\right)$

$=\sin (\mathrm{\angle }CAB)\cos (\mathrm{\angle }ABC)+\cos (\mathrm{\angle }CAB)\sin (\mathrm{\angle }ABC)$

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

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

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