angle sum identity

It is desired to prove the identities

$\sin(\theta+\phi)=\sin\theta\cos\phi+\cos\theta\sin\phi$

and

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

Consider the figure

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 , we have $\angle Ada=\frac{\pi}{2}-\theta$ and $\angle Ddc=\frac{\pi}{2}-\phi$ , while the degenerate angle $\angle AdD=\pi$ , so that

$\angle adc=\theta+\phi$

We have, therefore, that the area of the pink parallelogram is $\sin(\theta+\phi)$ . On the other hand, we can rearrange things thus:

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

$\sin(\theta+\phi)=\sin\phi\cos\theta+\cos\phi\sin\theta$

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

$\begin{array}[]{rcl}\cos(\theta+\phi)&=&\sin(\theta+\phi+\pi/2)\\ &=&\sin(\theta)\cos(\phi+\pi/2)+\cos(\theta)\sin(\phi+\pi/2)\\ &=&\sin(\theta)\sin(\phi+\pi)+\cos(\theta)\cos(\phi)\\ &=&\cos\theta\cos\phi-\sin\theta\sin\phi\end{array}$

Title	angle sum identity
Canonical name	AngleSumIdentity
Date of creation	2013-03-22 12:50:36
Last modified on	2013-03-22 12:50:36
Owner	mathcam (2727)
Last modified by	mathcam (2727)
Numerical id	14
Author	mathcam (2727)
Entry type	Theorem
Classification	msc 51-00
Related topic	ProofOfDeMoivreIdentity
Related topic	DoubleAngleIdentity