PlanetMath: proof of angle sum identities

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

such that $\angle CAB = x$ and $\angle ABF = y$ :

$\displaystyle \begin{xy} ,(0,0) ;(70,0)**@{-} ;(40,30)**@{-} ;(0,0)**@{-} ,(-2,-2)*{A} ,(72,-2)*{B} ,(40,32)*{C} \end{xy}$

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 from to :

$\displaystyle \begin{xy} ,(0,0) ;(70,0)**@{-} ;(40,30)**@{-} ;(0,0)**@{-} ,(40,30) ;(40,0)**@{-} ,(-2,-2)*{A} ,(72,-2)*{B} ,(40,32)*{C} ,(40,-2)*{D} \end{xy}$

Since

and

are right triangles we have, by definition,

$\displaystyle \cot (\angle CAB) = \overline{AD} / \overline{CD} \qquad \cot (\a... ...BD} / \overline{CD} \qquad \sin (\angle CAB) = \overline{CD} / \overline{AC} .$

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

$\displaystyle \begin{xy} ,(0,0) ;(70,0)**@{-} ;(30,40)**@{-} ;(0,0)**@{-} ,(0,0) ;(35,35)**@{-} ,(-2,-2)*{A} ,(72,-2)*{B} ,(30,42)*{C} ,(36,36)*{E} \end{xy}$

$\displaystyle \begin{xy} ,(0,0) ;(70,0)**@{-} ;(35,35)**@{-} ;(0,0)**@{-} ,(0,0) ;(40,30)**@{-} ,(-2,-2)*{A} ,(72,-2)*{B} ,(41,31)*{C} ,(35,37)*{E} \end{xy}$

Either way,

and

are right triangles, and we have, by definition,

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

Combining these ratios, we find that

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

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 and , we use the identities for complements and negatives of angles.

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