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 $0^{\circ}<x<90^{\circ}$ and $0^{\circ}<y<90^{\circ}$ for the time being. Then we may draw a triangle $A B C$ 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 $A D$ from $C$ to $A B$ :

????????????wwwwwwwwwwwwwwww

$\textstyle{A}$

$\textstyle{B}$

$\textstyle{C}$

$\textstyle{D}$

Since $A C D$ and $B C D$ 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 $A E$ form $A$ to $B C$ . 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, $A B E$ and $A C E$ 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
Canonical name	ProofOfAngleSumIdentities
Date of creation	2013-05-28 14:35:29
Last modified on	2013-05-28 14:35:29
Owner	rspuzio (6075)
Last modified by	rspuzio (6075)
Numerical id	15
Author	rspuzio (6075)
Entry type	Proof
Classification	msc 43-00
Classification	msc 51-00
Classification	msc 42-00
Classification	msc 33B10