Why is $\cos(\alpha+\beta) = \cos(\alpha)\cos(\beta)−\sin(\alpha)\sin(\beta)$?

Solution 1:

1. The rotation of $\mathbb R^2$ through angle $\alpha$ is a linear transformation with matrix $\left( \begin{matrix} \cos\alpha & -\sin\alpha\\ \sin\alpha &\cos\alpha \end{matrix}\right)$

2. The rotation of $\mathbb R^2$ through angle $\beta$ is a linear transformation with matrix $\left( \begin{matrix} \cos\beta & -\sin\beta\\ \sin\beta &\cos\beta \end{matrix}\right)$

3. The rotation of $\mathbb R^2$ through angle $\alpha+\beta$ is a linear transformation with matrix $\left( \begin{matrix} \cos(\alpha+\beta) & -\sin(\alpha+\beta)\\ \sin(\alpha+\beta) &\cos(\alpha+\beta) \end{matrix}\right)$

4. The rotation of $\mathbb R^2$ through angle $\alpha+\beta$ is the composition of the rotation of $\mathbb R^2$ through angle $\alpha$ and the rotation of $\mathbb R^2$ through angle $\beta$.

5. The matrix of the composition of two linear transformations is the product of matrices of these transformations.

6. So $$ \left( \begin{matrix} \cos\alpha & -\sin\alpha\\ \sin\alpha &\cos\alpha \end{matrix}\right)\cdot \left( \begin{matrix} \cos\beta & -\sin\beta\\ \sin\beta &\cos\beta \end{matrix}\right) = \left( \begin{matrix} \cos(\alpha+\beta) & -\sin(\alpha+\beta)\\ \sin(\alpha+\beta) &\cos(\alpha+\beta) \end{matrix}\right)$$

7. So, in particular, $$\cos(\alpha+\beta)=\cos\alpha\cos\beta-\sin\alpha\sin\beta.$$

Solution 2:

You can also simply prove it using complex numbers : $$ e^{i(\alpha + \beta)} = e^{i\alpha} \times e^{i\beta} \Leftrightarrow \cos (a+b)+i \sin (a+b)=(\cos a+i \sin a) \times(\cos b+i \sin b) $$ Finally we obtain, after distributing :

$$ \cos (a+b)+i \sin (a+b) =\cos a \cos b-\sin a \sin b+i(\sin a \cos b+\cos a \sin b) $$

By identifying the real and imaginary parts we get

$$ \cos (a+b)=\cos a \cos b-\sin a \sin b $$ $$ \sin (a+b)=\sin a \cos b+\cos a \sin b $$

Of course you need know basics about complex numbers but if you already know all of this it is usually really quick to proove most of this cumbersome trigonometric formulas with complex numbers like what I did here.

Solution 3:

You can also prove it by using a simple lemma:

Let $f:{\mathbb R}\to {\mathbb R}$ be twice differentiable, such that $f''=-f$. Then $f(x) = f(0)\cos x + f'(0) \sin x$.

Indeed, if $g(x)= f(x) - f(0)\cos x - f'(0)\sin x$, then $g''=-g$ and $g(0)=g'(0)=0$. But $(g'^2+g^2)' = 2 g'(g''+g) = 0$, hence $g'^2+g^2$ is a constant, which is $0$ because of its value at $x=0$. Hence $g(x)=0$ everywhere.

Applying this to $f(x) = \cos(x+\beta)$, we have $f(0) = \cos\beta$ and $f'(0) = -\sin\beta$, hence \begin{equation} \cos(x+\beta) = f(x) = \cos x\cos\beta - \sin x\sin\beta \end{equation}

In the case of $h(x) = \sin(x+\beta)$, we have $h(0) = \sin\beta$ and $h'(0) = \cos\beta$, hence \begin{equation} \sin(x+\beta) = h(x) = \cos x\sin\beta + \sin x\cos\beta \end{equation}

Solution 4:

By rotating the unit circle by an angle of $b$ we can see that the point $(cos(a-b),sin(a-b))$ maps to $(cos(a),sin(a))$ and the point $(1,0)$ maps to $(cos(b),sin(b))$.

Since rotations preserve distances,the distance between the points $(cos(a-b),sin(a-b))$ and $(1,0)$ is equal to the distance between points $(cos(a),sin(a))$ and $(cos(b),sin(b))$. The distance formula then gives:

$\sqrt{(cos(a-b)-1)^2+(sin(a-b)-0)^2}=\sqrt{(cos(a)-cos(b))^2+(sin(a)-sin(b))^2}$

With a lot less effort than you'd think (and using $sin(x)^2+cos(x)^2=1$ three times), you arrive at the desired identity