15-00-GeometricDerivationOfAdditionFormulasForSineAndCosine.tex

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\pmtitle{geometric derivation of addition formulas for sine and cosine}
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\begin{document}
Here is a geometric derivation of the addition laws for sines (and cosines)


\begin{align*}
\sin(\alpha+\beta)&=\sin(\alpha)\cos(\beta)+\cos(\alpha)\sin(\beta)\\
\cos(\alpha+\beta)&=\cos(\alpha)\cos(\beta)-\sin(\alpha)\sin(\beta)
\end{align*}

First note that, by symmetry, it is clear that
\begin{align*}
\sin\left(x+\frac{\pi}{2}\right)&=\cos x\\
\cos\left(x+\frac{\pi}{2}\right)&=-\sin x\\
\sin(-x)&=-\sin x\\
\cos(-x)&=\cos x
\end{align*}
and so we can reduce proving the addition law to proving it in the case where $\alpha$, $\beta$, and $\alpha+\beta$ are all in the first quadrant.

We then have the situation pictured below:

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Now, from the definitions of $\sin$ and $\cos$, and assuming that $OA=1$, we see that
\begin{align*}
AC&=\sin(\beta)\\
OC&=\cos(\beta)
\end{align*}

Now,
\begin{align*}
\sin(\alpha)&=\sin(\angle DOC)=\frac{CD}{OC}=\frac{CD}{\cos(\beta)}\\
\cos(\alpha)&=\cos(\angle DOC)=\frac{OD}{OC}=\frac{OD}{\cos(\beta)}
\end{align*}
so we have that
\begin{align*}
CD&=\sin(\alpha)\cos(\beta)\\
OD&=\cos(\alpha)\cos(\beta)
\end{align*}

But also, it is clear that $\angle BCA=\angle DOC=\alpha$, and therefore we have similarly
\begin{align*}
\sin(\alpha)&=\sin(\angle BCA)=\frac{AB}{AC}=\frac{AB}{\sin(\beta)}\\
\cos(\alpha)&=\cos(\angle BCA)=\frac{BC}{AC}=\frac{BC}{\sin\beta}
\end{align*}
so that
\begin{align*}
AB&=\sin(\alpha)\sin(\beta)\\
BC&=\cos(\alpha)\sin(\beta)
\end{align*}
Thus $\sin(\alpha+\beta)$ is the $y$-coordinate of $A$, which is the $y$-coordinate of $B$, so
\[\sin(\alpha+\beta)=CD+BC=\sin(\alpha)\cos(\beta)+\cos(\alpha)\sin(\beta)\]
and $\cos(\alpha+\beta)$ is the $x$-coordinate of $A$, which is the $x$-coordinate of $B$ less the difference in $x$-coordinates between $B$ and $A$, so
\[\cos(\alpha+\beta)=OD-AB=\cos(\alpha)\cos(\beta)-\sin(\alpha)\sin(\beta)\]
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