sin(x+y) does not equal sin(x)+sin(y)

Following on from my physics talk about vector quantities, it reminded me of what my Real Analysis (second year math subject) lecturer told our class during a lecture. I already knew it, but it does seem to be a mistake some people make. I paraphrase slightly:

f(x+y) does not always equal f(x)+f(y). An example is, sin(x+y) does not equal sin(x)+sin(y). Some people sometimes write it on exams, and it's not correct. Actually, sin(x+y)=sin(x)cos(y)+sin(y)cos(x).

[note: the angle expansion formula is not required for the GAMSAT, so don't worry about memorizing it if you aren't doing math subjects or subjects which require significant math, but you should know that sin(x)+sin(y) doesn't equal sin(x+y), in case you have to solve some algebra and think you've found a neat shortcut which doesn't exist]

I think some people get confused because a(b+c)=ab+ac, so they think sin(x+y)=sin(x)+sin(y), but it's not like that. a(b+c) means a times (b+c), but sin(x+y) doesn't mean sin times (x+y).

How does this relate to the GAMSAT? I think sometimes you might have to manipulate logs in certain questions. No, I don't mean logs you find from trees, even if I say "natural log". I mean logarithms. In radioactive decay or other types of decay with half-lives, you might need logarithms to find out stuff. The "log" also pops up in the decibel formula. There's a whole set of log laws, which can be thought of as the reverse of exponent laws.

In particular, loga(x)+loga(y)=loga(x*y), provided that x and y are positive, and a is positive (which is required for the logarithm to exist). Don't think that loga(x)+loga(y)=loga(x+y), because it's not.

Also, loga(x^n)=n*loga(x) is a law you might need.

Anyway, perhaps I'll finish with an example. Simple one, but one with bad numbers, so you can't guess whole numbers easily.


"A radioactive compound initially is 1000 grams. After 60 hours, there is 100 g of the original compound remaining. What is the half life?"

Actually, before I'll begin, I'll mention that there is some weird formula for calculating this type of stuff with a e^(logex) type thing in it which I haven't bothered to memorize. The advantage is that you don't have to remember your log laws when using it, but the disadvantage is that it's not that easy to remember. Personally, I like to remember the formula more logically, even if it means I have to remember the log laws to solve something on the calculator.

Basically, after every half life, the amount of original compound remaining is halved. So after one half life there's half remaining, and after the second half life there's 1/2 * 1/2=(1/2)^2=1/4 remaining; after the next one it's 1/4*1/2=(1/2)^3 remaining. See the pattern? But in the question, we have 1/10 remaining, and 10 is not a multiple of 2, so we can't exactly just work it out the easy way.

Anyway, from the discussion above (there's more rigorous ways to prove it though, but for our purposes an illustration is OK), the proportion remaining of the original compound is (1/2)^n, where n=number of half lives.

Then to convert that into an actual amount, multiply by the original amount.

So, N=N0(1/2)^n, where N=amount remaining, N0=initial amount of compound. To me, knowing this formula is simpler than memorizing a whole formula with e^loge etc. However, to use it backwards you need to know your log laws, unlike the other way where you don't need to know so much.

Anyway, back to the question.

N=1000, N0=100.

100=1000(1/2)^n
1/10=(1/2)^n

now if you use this formula, this is where we get logs into play. Take the log of both sides (it doesn't matter which base, but your calculator has two logs; base 10 (the button "log") or base e (the button "ln"), so it has to be one of these. It would be simpler if there was a log base 2 for this particular question, but it doesn't exist on your calculator. So if we don't have a log base 2 on the calculator, we need a work-around.

log (1/10)=log (1/2)^n

Now use the log law: loga(x)^n=n*loga(x) on the RHS of the equation:

log (1/10)=n*log (1/2)
n=log(1/10)/log(1/2) (put this onto your calculator)
=3.3219

Now this is the number of half lives. We wanted to know the half life. Now, we know from the question that the time taken was 60 hours, so:

60 hours=3.3219 half lives

What's one half life? Divide 60 by 3.3219.

Answer is 18.06

So half life is 18 hours. That is the answer.

As I said before, there's another formula which is more complex to remember that some people use for it, but which means you don't need to know your log laws. Personally, because I do remember my log laws, I find this way simpler because I don't really have to remember a more complex formula for exponential decay involving half lives.

Hope this helps. I don't really plan to turn this blog into a GAMSAT blog, but I might still mention a few things here and there.