angle sum identity

It is desired to prove the identities

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

and

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

Consider the figure

where we have

• $\mathrm{△}Aad\equiv \mathrm{△}Ccb$

• $\mathrm{△}Bba\equiv \mathrm{△}Ddc$

• $ad=dc=1$.

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

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

$$\mathrm{\angle }adc=\theta +\varphi$$

We have, therefore, that the area of the pink parallelogram is $\sin (\theta +\varphi )$. 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 $\sin \varphi \cos \theta$ and $\cos \varphi \sin \theta$. Adding,we have

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

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