Trigonometry addition formula cos(a+b)=cos a cos b - sin a sin b

We are going to show that for any angles a, b the trigonometry formula cos (a+b)=cos a cos b - sin a sin b

Let $(O;\overrightarrow{i},\overrightarrow{j})$ be an orthonormal reference frame, $a$ and $b$ two real numbers defined as follows:

$$\begin{array}{rl}a& =(\overrightarrow{i},\overrightarrow{OA})\\ b& =(\overrightarrow{OA},\overrightarrow{OB})\end{array}$$

Where $A$ and $B$ are the points defined on the trigonometric circle relative to the angles $a$ and $b$.

We then have:

$$a+\frac{\pi }{2}=(\overrightarrow{i},\overrightarrow{O{A}^{\prime }})$$

Where $A^{\prime }$ is the point defined on the trigonometric circle relative to the angle ${\displaystyle a+\frac{\pi }{2}}$ with $(\overrightarrow{OA},\overrightarrow{O{A}^{\prime }})={\displaystyle \frac{\pi }{2}}$.

By definition, $\overrightarrow{OA}$ is defined as:

$$\overrightarrow{OA}=\cos a\times \overrightarrow{i}+\sin a\times \overrightarrow{j}$$

$\overrightarrow{O{A}^{\prime }}$ is defined as:

$$\overrightarrow{O{A}^{\prime }}=\cos (a+\frac{\pi }{2})\times \overrightarrow{i}+\sin (a+\frac{\pi }{2})\times \overrightarrow{j}=-\sin a\times \overrightarrow{i}+\cos a\times \overrightarrow{j}$$

$\overrightarrow{OB}$ is defined by:

$$\overrightarrow{OB}=\cos (a+b)\times \overrightarrow{i}+\sin (a+b)\times \overrightarrow{j}$$

Consider the orthonormal reference frame $(O;\overrightarrow{OA},\overrightarrow{O{A}^{\prime }})$. The vector $\overrightarrow{OB}$ in this reference frame is defined as:

$$\begin{array}{rl}\overrightarrow{OB}& =\cos b\times \overrightarrow{OA}+\sin b\times \overrightarrow{O{A}^{\prime }}\\ & =\cos b\times (\cos a\times \overrightarrow{i}+\sin a\times \overrightarrow{j})+\sin b\times (-\sin a\times \overrightarrow{i}+\cos a\times \overrightarrow{j})\\ & =(\cos a\times \cos b-\sin a\times \sin b)\times \overrightarrow{i}+(\sin a\times \cos b+\cos a\times \sin b)\times \overrightarrow{j}\end{array}$$

But we have shown that

$$\overrightarrow{OB}=\cos (a+b)\times \overrightarrow{i}+\sin (a+b)\times \overrightarrow{j}$$

We then obtain by identification:

$$\begin{array}{rl}\cos (a+b)& =\cos a\times \cos b-\sin a\times \sin b\\ \sin (a+b)& =\sin a\times \cos b+\cos a\times \sin b\end{array}$$

We have thus demonstrated: $\mathrm{\forall }a,b\in \mathbb{R},\cos (a+b)=\cos a\cos b-\sin a\sin b$