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angle sum identity (Theorem)

It is desired to prove the identities

$\displaystyle \sin(\theta+\phi) = \sin\theta\cos\phi + \cos\theta\sin\phi $
and
$\displaystyle \cos(\theta+\phi) = \cos\theta\cos\phi - \sin\theta\sin\phi $

Consider the figure

\includegraphics[width = 5.3cm, height = 5.0cm]{sclaw1.eps}

where we have

$ \circ$
$ \triangle Aad \equiv \triangle Ccb$
$ \circ$
$ \triangle Bba \equiv \triangle Ddc$
$ \circ$
$ ad = dc = 1$.

Also, everything is Euclidean, and in particular, the interior angles of any triangle sum to $ \pi$.

Call $ \angle Aad = \theta$ and $ \angle baB = \phi$. From the triangle sum rule, we have $ \angle Ada = \frac{\pi}{2}-\theta$ and $ \angle Ddc = \frac{\pi}{2} - \phi$, while the degenerate angle $ \angle AdD = \pi$, so that

$\displaystyle \angle adc = \theta + \phi$
We have, therefore, that the area of the pink parallelogram is $ \sin(\theta + \phi)$. On the other hand, we can rearrange things thus:
\includegraphics[width = 5.6cm, height=5.1cm]{sclaw2.eps}

In this figure we see an equal pink area, but it is composed of two pieces, of areas $ \sin \phi \cos\theta$ and $ \cos\phi \sin\theta$. Adding, we have

$\displaystyle \sin(\theta + \phi) = \sin\phi\cos\theta + \cos\phi\sin\theta$
which gives us the first. From definitions, it then also follows that $ \sin(\theta + \pi/2) = \cos(\theta)$, and $ \sin(\theta+\pi) = - \sin(\theta)$. Writing
$\displaystyle \begin{array}{rcl} \cos(\theta + \phi)&=& \sin(\theta+\phi+\pi/2)... ...\theta) \cos(\phi) \ &=& \cos\theta\cos\phi - \sin\theta\sin\phi \end{array} $



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See Also: proof of de Moivre identity, double angle identity

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Cross-references: definitions, parallelogram, area, angle, sum, triangle, interior angles, Euclidean, identities
There are 5 references to this entry.

This is version 11 of angle sum identity, born on 2002-07-17, modified 2005-03-25.
Object id is 3170, canonical name is AngleSumIdentity.
Accessed 5499 times total.

Classification:
AMS MSC51-00 (Geometry :: General reference works )

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