Dealing with the R sin alpha nonsense that comes up in C3 is the kind of thing our arch-nemesis, the mathematical pirate, would grumble about constantly. They’d claim it wasn’t really trigonometry, or just that it was too hard, or a complete misuse of the word ‘arr’.

The mathematical ninja, on the other hand, raises an arch eyebrow and says what R and alpha are in the blink of an eye.

## How to solve an $R \sin (x + \alpha)$ sort of problem

They typically ask something like “Express $7 \cos(x) + 24 \sin(x)$ in the form $\sin(x + \alpha)$.” There’ll be conditions on the $R$ ($R > 0$) and on the $\alpha$, normally $0 \le \alpha \lt \frac{\pi}{2}$ or something similar in bad degrees.

You can always get $R$ using Pythagoras. It’s always the same deal: you square each of the numbers, add them up, and find the square root of the answer. For this one, you’d instantly say “49 and 576 is 625, which is 25 squared, so $R = 25$.”

The $\alpha$ wants a little bit more work. There’s going to be an $\arctan$ in there, but the question is, $arctan$ of what? It all comes down to the thing you’re trying to get to. If it’s a $\cos$ you’re after, you divide the number in front of the $\sin (x)$ by the number in front of $\cos(x)$. If it’s a $\sin$, it’s the other way around. In either case, ignore any minus signs.

Here, we’re looking for a $\sin$, so we do $\arctan(\frac{24}{7})$. Looking back to the stuff on small-angle trigonometry [link], which is 1.3 or so.

## Wait, wait, wait…

OK, ok. I looked at $\frac{24}{7}$ and thought: that’s a big number, about 3.4. But if I can figure out $\arctan$ of the reciprocal and take it off of $\frac{\pi}{2}$, that’ll give me the same answer. $\frac{7}{24}$ is a much smaller number, midway between a quarter and a third, so a bit less than 0.3. Meanwhile, $\frac{\pi}{2} \approx 1.57$, so if I take away the 0.3 I get about 1.3 - which is good enough for government work.

## But why does that rule work?

A good question, my friend. It comes from the expansions of $\sin(x \pm \alpha)$ and $\cos(x \pm \alpha)$. If you have, for instance, $A \sin(x) + B\cos(x) \equiv R\sin(x + \alpha) \equiv R \sin(x) \cos(\alpha) + R \cos(x) \sin(\alpha)$, it stands to reason that $R\cos(\alpha) = A$ and $R \sin (\alpha) = B$. Dividing them, you get $\tan(\alpha) = \frac{B}{A}$ — meaning you divided the number with the $\cos$ by the number with the $\sin$, as I said before. The same thing works with $\cos$, but I’ll leave that as an exercise.