<s>[INST] How can I understand and prove the "sum and difference formulas" in trigonometry? The "sum and difference" formulas often come in handy, but it's not immediately obvious that they would be true.
\begin{align}
\sin(\alpha \pm \beta) &= \sin \alpha \cos \beta \pm \cos \alpha \sin \beta \\
\cos(\alpha \pm \beta) &= \cos \alpha \cos \beta \mp \sin \alpha \sin \beta
\end{align}
So what I want to know is, 


*

*How can I prove that these formulas are correct?

*More importantly, how can I understand these formulas intuitively?


Ideally, I'm looking for answers that make no reference to Calculus, or to Euler's formula, although such answers are still encouraged, for completeness.
 [/INST] Consider a unit circle with $O$ as the centre. Let $P_{1}$, $P_2$ and $P_{3}$ be points on the circle making angles $A$, $B$ and $A−B$, respectively, with the positive direction of the X-axis.

We know that if two chords subtend equal angle at the centre, then the chords are equal and chords $P_{3}P_{0}$ and $P_1P_2$ subtend equal angles at $O$. Therefore, $\overline{P_{3}P_{0}}=\overline{P_1P_2}$. By distance formula, the distance between the points $P_{0}(1,0)$ and $P_{3}(\cos(A−B),\sin(A−B))$ is $$P_{3}P_{0}=\sqrt{\left(\cos \left( A-B \right) - 1\right) ^{2} + \left(\sin \left( A-B \right) - 0\right) ^{2}}$$ Similarly, the distance between the points $P_{1}(\cos A,\sin A)$ and $P_{2}(\cos B,\sin B)$ is $$P_1P_2= \sqrt{\left( \cos B-\cos A\right) ^{2}+\left( \sin B - \sin A \right) ^{2}}$$
On squaring both sides, we get
$$
\begin{array}{ll}
\Rightarrow & \left\{\cos (A-B)-\left.1\right|^{2}+\sin ^{2}(A-B)=(\cos B-\cos A)^{2}+(\sin B-\sin A)^{2}\right. \\
\Rightarrow & \cos ^{2}(A-B)-2 \cos (A-B)+1+\sin ^{2}(A-B)=\cos ^{2} B+\cos ^{2} A-2 \cos A \cos B \\
& +\sin ^{2} B+\sin ^{2} A-2 \sin A \sin B \\
\Rightarrow & 2-2 \cos (A-B)=2-2 \cos A \cos B-2 \sin A \sin B \\
\Rightarrow & \cos (A-B)=\cos A \cos B+\sin A \sin B \\
\text { Hence, } & \cos (A-B)=\cos A \cos B+\sin A \sin B
\end{array}
$$
</s>