> [!NOTE] Theorem (Addition Formulae) > For all $x,y\in\mathbb{R}$, we have $\begin{align} \cos(x+y) &= \cos x \cos y - \sin x\sin y \\ \sin(x+y) &= \sin x \cos y + \cos x \sin y \end{align}$ >*Proof*. Fix $z\in\mathbb{R}$ and let $f(x)=\cos(x)\cos (z-x) - \sin (x) \sin(z-x)$Note that $f'(x) = -\sin x \cos(z-x) + \cos(x)\sin(z-x) - \cos(x)\sin(z-x) + \sin(x) \cos(z-x) =0 $so for all $x\in\mathbb{R}$, $f(x)$ is constant by [[Real Function with Zero Derivative is Constant|MVT]]. >Note that $f(0)=\cos z$ hence $f(x)=\cos z$. >Substituting $z= x+y$ gives the desired result. >Proof. Use [[Exponential Function in Terms of Trigonometric Functions (Euler's Formula)]].