proof of angle sum identities
ProofOfAngleSumIdentities
2007-07-26 18:25:12
2007-07-26 22:49:49
Proof
trigonometry
0
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\newtheorem{thm}{Theorem}
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.
\begin{thm}
\[
\sin (x + y) = \sin (x) \cos (y) + \cos (x) \sin (y)
\]
\end{thm}
\begin{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 $\angle ABF = y$:
\[
\begin{xy}
,(0,0)
;(70,0)**@{-}
;(40,30)**@{-}
;(0,0)**@{-}
,(-2,-2)*{A}
,(72,-2)*{B}
,(40,32)*{C}
\end{xy}
\]
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 $AD$ from $C$ to $AB$:
\[
\begin{xy}
,(0,0)
;(70,0)**@{-}
;(40,30)**@{-}
;(0,0)**@{-}
,(40,30)
;(40,0)**@{-}
,(-2,-2)*{A}
,(72,-2)*{B}
,(40,32)*{C}
,(40,-2)*{D}
\end{xy}
\]
Since $ACD$ 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.)
\[
\begin{xy}
,(0,0)
;(70,0)**@{-}
;(30,40)**@{-}
;(0,0)**@{-}
,(0,0)
;(35,35)**@{-}
,(-2,-2)*{A}
,(72,-2)*{B}
,(30,42)*{C}
,(36,36)*{E}
\end{xy}
\]
\[
\begin{xy}
,(0,0)
;(70,0)**@{-}
;(35,35)**@{-}
;(0,0)**@{-}
,(0,0)
;(40,30)**@{-}
,(-2,-2)*{A}
,(72,-2)*{B}
,(41,31)*{C}
,(35,37)*{E}
\end{xy}
\]
Either way, $ABE$ 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 $\overline{AD} + \overline{BD} =
\overline{AB}$:
\begin{align*}
\sin (x + y) = \sin (\angle BCA) &=
\overline{AB} \, \sin (\angle ABC) / \overline{AC} \\ &=
(\overline{AD} + \overline{BD}) \sin (\angle ABC) / \overline{AC} \\ &=
\overline{CD} \left( \cot (\angle CAB) + \cot (\angle ABC) \right) /
\sin (\angle ABC) \overline{AC} \\ &=
\sin (\angle CAB) \sin (\angle ABC)
\left( {\cos (\angle CAB) \over \sin (\angle CAB)} +
{\cos (\angle ABC) \over \sin (\angle ABC)} \right) \\ &=
\sin (\angle CAB) \cos (\angle ABC) +
\cos (\angle CAB) \sin (\angle ABC) \\ &=
\sin (x) \cos (y) + \cos (x) \sin (y)
\end{align*}
To lift the restriction on the range of $x$ and $y$, we use the identities
for complements and negatives of angles.
\end{proof}
\textbf{Entry under construction}