What does it mean when the question asks "Express the height of the spaceship as a function of the angle of elevation (theta)" versus "Express the angle of elevation (theta) as a function of height (h) of the spaceship?" You don't really need to know the specific question (unless you want to see it) but what does it mean "express (X) as a function of (Y)?
you've got one function that's increasing and another that's decreasing. what about their sum? it could do either, or it could stay constant -- but you also know that one of the increases/decreases is always going faster than the other
meaning that if you take a tiny step, the increase is always going to be more than the decrease
well, you don't know anything about their actual values yet - we're just talking about relative behaviour. they could both be massively negative and still have one increase and the other decrease
you have two functions that are being added together. you don't know whether they're positive or negative (well, you do in this particular case, but it's not relevant here)
if you take a tiny step forward, one of them will increase and the other will decrease. and you know that the increasing function goes faster -- the increase will be more than the decrease
you have two columns of stuff (either of which could be negative). every minute, you add some stuff onto the left column and take it away from the right column (potentially going negative). you always add more than you take away. what can you say about your total amount of stuff over time?
the increase is always going faster than the decrease. and since you're adding them together, that means your total will be growing overall (specifically, it will be growing by [increase amount] - [decrease amount])
@user586228: this isn't a room to just ask for personal tutoring, especially if you can't be bothered to even type the problem out. Mostly it's North's friends here.
Hm, it looks like they ask there fairly often anyways
Should I start making this room gallery?
I would rather not
But
Okay, if random people start dropping by more frequently I'd probably have to change it to gallery-only. But I'd like to make that like a last-resort option
If more people start to think this is a tutoring room then I'll make it gallery only
Anyhow, back to my English thing
So I ended up giving the same piece of evidence as the person in class, because again, I couldn't hear them and now I'm not going to get a 100 on this assignment
I can't afford to take any more non-100s in English, I already have a C
I think Deusovi should be banned from Puzzling. https://puzzling.meta.stackexchange.com/questions/6155/should-we-ban-deusovi-from-puzzle-solving/6160#6160
@GarethMcCaughan I'm a bit upset that I wasted 2 hours working with cough underqualified tutors (not Deusovi, I went to a tutoring site my school provided). I was transferred three times (took up 30 minutes of doing nothing) and my last tutor started the conversation with "So, sin^-1 is 1/sin, right?"
I think I'm confused. So you start with arcsin 2x + cos x = pi/6. Then you subtract arcsin 2x from both sides: cos x = pi/6 - arcsin 2x. Then you take cosines: x = cos (pi/6 - arcsin 2x). This all seems very reasonable but x isn't isolated on the LHS -- you still have an x on the right.
(Ouch! In fairness, the sin^... notation is Really Stupid, albeit also Really Convenient, and I have a lot of sympathy with anyone who finds it confusing.)
So, imagine a horizontal line at some value of y (which actually is pi/6, but never mind that for the moment). You want to know whether it crosses over that red line or not.
If you imagine the line moving, say, upwards from y=-oo towards y=+oo, what happens is that for very-negative y you have no solutions, then at some point you start having solutions, and then you stop and for very-positive y there are no solutions again.
Can you say -- in terms of features of that red line that you might be able to work out -- where you switch from "no solution" to "one solution" and back to "no solution" again?
So, to recap: You figured out that f(x) = arcsin 2x + arccos x is only defined on -1/2 <= x <= +1/2, and that it's a strictly increasing function there. That means that for y between f(-1/2) and f(+1/2) inclusive there's exactly one x with f(x) = y, and for y outside that range there are none. You found that f(-1/2) happens to equal pi/6, which is the value of y you wanted, so you're done.
Yes, cos -1/2 = cos 1/2. If it looks wrong, it's probably because you very seldom actually want cos 1/2: you want things like cos pi/2 and cos -pi/6 and so on. Or arccos 1/2 (but probably not arccos pi/3).
Now, I don't know whether Deusovi had already figured out that the "endpoint" -1/2 was actually going to be the answer to the question or not. He might well have. But if not, there was still some point in what he was doing.
Because then it might have turned out that pi/6 was outside the attainable interval from f(-1/2) to f(+1/2), in which case you'd be able to say "no solutions" without doing any further work.
Or it might have turned out that it was inside the interval, in which case you'd know there was just one solution but not yet know what it was.
At that point, though, it might have turned out profitable to guess some convenient values of x and see what they did, just in case the answer was a nice one. If so, then you'd have been able to say "well, x=... is a solution, as we can see by just calculating arcsin 2x and arccos 2x. And it's the only solution because f(x) is strictly increasing."
It's good to have both ways of thinking in your toolbox.
In practice, when there's some equation to solve it's often helpful to have a (mental or sketched or computer-plotted) picture to give you some intuition for roughly what you expect the solution to look like.
