Entwined angles

By the sum and product formulas, we know that

$\cos(\alpha + \beta) = \cos \alpha \cos \beta - \sin \alpha \sin \beta.$

Hence, we must have $N = 1, M = -1$ , so $N + M = 0$ .

Now, let us show that $(N=1, M=-1$ is the unique solution.
Let $(N,M)$ be a solution. This implies that: $\cos ( \alpha + \beta) = N \cos \alpha \cos \beta + M \sin \alpha \sin \beta$ is true.

Now, let's subtract $``\cos ( \alpha + \beta)"$ from the left hand side and $``\cos \alpha \cos \beta - \sin \alpha \sin \beta"$ from the right hand side of the equation. We can do this because we know that $``\cos ( \alpha + \beta) =\cos \alpha \cos \beta - \sin \alpha \sin \beta "$ , so we are subtracting the same thing from both sides.

Executing the subtraction:
$\cos ( \alpha + \beta) - \cos ( \alpha + \beta) = N\cos \alpha \cos \beta - \cos \alpha \cos \beta + M \sin \alpha \sin \beta - (-1) \sin \alpha \sin \beta$ , which gives us $0 = (N-1)\cos \alpha \cos \beta + (M+1) \sin \alpha \sin \beta = 0.$

Lastly, dividing both sides by $\cos \alpha \cos \beta$ , we get the equation:

$(N-1) + (M+1) \tan \alpha \tan \beta = 0$

The only way to satisfy this formula for all $\alpha$ and $\beta$ is if $(N-1) = 0,$ and $(M+1) = 0$ , which corresponds to our solution above, that $N = 1$ and that $M=-1$ . Therefore, this solution is unique.