# 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 AngleSumIdentity 2013-03-22 12:50:36 2013-03-22 12:50:36 mathcam (2727) mathcam (2727) 14 mathcam (2727) Theorem msc 51-00 ProofOfDeMoivreIdentity DoubleAngleIdentity