# A dreadful question

I state this precisely as it was stated to me:

$\sin(a+7) = \cos(b-10)$ In the equation above, $A$ and $B$ are constants. What is the value of $A+B$?

I mean, let’s leave aside for a moment the capitalisation – although it’s a signal that this *might* not be the best-thought-out question – and that presumably the angles should be marked in degrees.

However, even if we generously interpret the first line as $\sin((A+7)^o) = \cos((B-10)^o)$, the answer is still “any real number”.

What I *presume* the answer they want is $A+B=93$. Where does that come from? Well, $\cos(x^o) \equiv \sin((90-x)^o)$, so we have $\sin((A+7)^o) = \sin((100-B)^o)$, and inverting both of those sines gives $A+7 = 80-B$, so $A+B=93$.

But hold on. The sine function isn’t one-to-one, so inverting it like that is super-dicey. There are other important identities at play here as well:

- $\cos(x^o) \equiv \cos((x+360)^o)$
- $\sin(x^o) \equiv \sin((x+360)^o)$
- $\cos(x^o) \equiv \sin((90+x)^o)$

The first two are fairly boring – we can add or subtract any multiple of 360 degrees to either $A$, $B$ or both, without changing the value of their sine or cosine. So *those* are also solutions. Does that make the solution $A+B=93 + 360n$ for integer $n$? Well, no.

The third solution leads to $\sin((A+7)^o) = \sin((80+B)^o)$, and removing the sines gives $A+7=80+B$ - or $A-B = 73$, plus any integer multiple of 360.

But in particular, that doesn’t tell us anything about the value of $A+B$. If I want $A+B$ to be 83 (say), I can set $A=78$ and $B=5$ and get a perfectly valid solution to the original equation.

In short, it’s a really sloppy question, both in terms of typesetting and construction. Hopefully it’s an interesting investigation all the same.