The $\cos(\alpha-\beta)$ formula always need $\alpha > \beta$ or not? I'm a beginner student study the proof of sum and difference trigonometry formula.
There is a formula that:
$$\cos(\alpha-\beta) = \cos(\alpha)\cos(\beta)+\sin(\alpha)\sin(\beta)$$
In the tutorial it shows only the case when $\alpha > \beta$
Can this formula be used when $\alpha < \beta$ ? and is the answer mathematically legitimate? and why it is? or why it is not?
please give me very detail and basic step by step explanation. I'm newbie ^^"
 A: No, because $\cos x = \cos (-x)$ which can be proven using the unit circle.
A: Yes, it is most definitely valid for $\alpha < \beta$.  This is simple to show.
Assume it's correct for $\alpha > \beta$:
$$
\cos(\alpha - \beta) = \cos(\alpha)\cos(\beta) + \sin(\alpha)\sin(\beta)
$$
Consider the following, $\cos(\beta - \alpha) = \cos(-(\alpha - \beta))$.  In this case we have $\alpha > \beta \rightarrow \beta < \alpha$.  But, because the cosine is an even function, i.e. $\cos(-x) = \cos(x)$, we have that:
\begin{align}
\cos(\beta - \alpha) =&\ \cos(-(\beta - \alpha)\\
 =&\ \cos(\alpha - \beta)\\
 =&\ \cos(\alpha)\cos(\beta) + \sin(\alpha)\sin(\beta)
\end{align}
It's even easier to see through symmetry via "change of indexes":
$$
\cos(\alpha - \beta) = \cos(\alpha)\cos(\beta) + \sin(\alpha)\sin(\beta)\\
\cos(\beta - \alpha) = \cos(\beta)\cos(\alpha) + \sin(\beta)\sin(\alpha)
$$
These two are clearly the same via commutative property of multiplication.  That is, if it's true for $\alpha < \beta$ then it's also true for $\beta < \alpha \rightarrow \alpha > \beta$.
I want to point out the importance of the symmetry argument because this would seemingly not hold for a sine function, consider the following, proved for $\alpha > \beta$:
$$
\sin(\alpha - \beta) = \cos(\beta)\sin(\alpha)-\cos(\alpha)\sin(\beta)
$$
Does it hold when $\alpha < \beta$?  Consider the following when $\beta > \alpha$:
$$
\sin(\beta - \alpha) = \cos(\alpha)\sin(\beta) - \cos(\beta)\sin(\alpha)
$$
This is where the fact that $\cos(x) = \cos(-x)$ and $\sin(x) = -\sin(-x)$ is important.  Indeed, if you look at the above $\sin(\alpha - \beta) = -\sin(\beta - \alpha)$ therefore the above "rule" works whether $\alpha > \beta$ or if $\beta > \alpha \rightarrow \alpha < \beta$.
A: The formula holds for all $\alpha,\beta$. If you only know it for the case that $\alpha-\beta>0$ note that for $\alpha-\beta<0$,
$$ \cos(\alpha-\beta)=\cos(\beta-\alpha)=\cos(\beta)\cos(\alpha)+\sin(\beta)\sin(\alpha)$$
(and of course for $\alpha=\beta$, $1=\cos 0=\cos^2(\alpha)+\sin^2(\alpha)$
A: Swap $ \alpha , \beta$ . Has anything happened to the result? No. So cos function's argument can be either >0 or <0, cos is an even function, $\cos(-x)=\cos(x),$ or,
$$ \cos (\alpha-\beta)  = \cos ( \beta -\alpha ) $$
