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00:55
@bobble Are you free for tutoring?
here
Okay so this is for physics
And it was on Khan Academy
It was asking about velocity at a given point, (i.e. instantaneous rate of change) but I'm not sure how I was supposed to solve that without you know, calculus or just guessing
what information do you have?
acceleration, distance, time, etc.
Hold on
I had a graph
A velocity vs time for an oscilating grpah
Let me see if I can pull it up again
oscillating? eesh. so not constant acceleration
00:59
Yeah....
Basically tagent lines in sine/cosine graph. I know it's zero at the "peaks" but like
wait if it's velocity vs time, can't you just look up your specific time?
if you have a graph of velocity vs time, and you need to find the velocity at t=2, you just look at the y value when x=2
Oh, it just works like that?
Wait hold on
Back up
to where?
01:01
Let me find the question again and you can walk me through it (might take a couple minutes)
Sorry, it was a displacement vs time graph
okay so that's harder, yeah
ah, so that's another Special Point for sin/cos graphs
where they cross the midline
yeah
01:04
I only remember the max point being=0
I know the answer btw is -4 because it's the only negative choice
But is there a reason for that?
okay, I need to remember how you do this without calc
Should we ping deus?
It's the slope of the graph at any point... isn't it?
(as an aside: my physics class has calc as a prereq)
and hi >.> (interrupting)
01:06
@Graylocke Is it? Khan academy seemed to have picked two random values smaller than 2 and greater than 2 for finding IRC
Which I know you can estimate IRC by doing that but Bobble said this is a special case
@bobble Really? Ours is only "you need to be enrolled in Algebra 2/Trig"
okay so far my brain has remembered that when sin(x) or cos(x) crosses midline the slope is (+/-)1
So velocity is your change in displacement over time... so the rate of change at a point is the slope of the displacement graph there...
Isn't that average rate of change?
Okay I'm going to ping deus, if you don't mind
we can't use calc, Gray
no I don't mind
@Deusovi Can you help?
01:09
hello!
Hi! So urm physics help
trying to find slope when sin/cos graph crosses the midline without calc
well the slope is changing over time >.>; and I would use that fact plus the one bobble just said (which is true for a sin (x) graph... but this is a bit taller than a normal sin :) )
yeah a * sin(x) has slope +/-a when crossing midline
the problem is this isn't just x in the middle
yeah uh i don't see anything except for calculus or estimation
01:10
Khan Academy seems to have more or less "estimated" the IRC by plugging in numbers close to 2
Which I did learn in pre-calc
Oh, I thought there was a cool trick
Is it a coincidence that the IRC is -4 and the midline is 4?
I coulda sworn there was a special case, but my mind is infected with calc at this point
yes, coincidence - shift up/down doesn't affect slope
It looks like 3.sin(x)+4 ? >.>;
yes, it's a coincidence; agreed that that graph is 3sin(t)+4
? it has a period of 4
Oh, I thought there was a really special property where the IRC is equal to the midline
01:12
correction: 3sin(kt)+4, where k is some constant
well there is a conversion on t
yeah ;p
that i'm too lazy to calculate
Lol don't worry about the graph, it's not as important for this specific exercise
t.pi/2 I guess
I believe it's 3sin(pi/2*t)+4?
01:14
oh yeah lol wavelength of 5
duh
wait
nah wavelength is 4
it is 4 >.>;
4, made a mistake :P
I revert to my original answer ;p
So the correct way was just to estimate by using numbers close to it. Boring :/
Urm, what's the calculus way? Or is that really really complicated?
01:16
calculus way is find the function, take the derivative, plug in 2
What's a derivative?
We found the function
literally a special way of getting the slope ;p
Oh drat
grumble grumble Isaac Newton grumble grumble
in the case of sin and cos they magically loop around each other because well... that is how circles work.
<glosses over things> >.>;
if you have f(x), then the derivative of f(x) (we call it f'(x)) has the property that f'(a) is the slope of the tangent line at (a, f(a))
(tangent line of f(x), to be clear)
bobble said it in a much cleverer way ;p
Ok
So how do you find that?
well there is the boring long limit way, or we just apply special rules
Neat patterns you can apply to functions :)
calculus things that you'll learn later
i don't think it's worth going into now
01:19
Mm okay
Yeah, I still need to understand the half-angle shenanigans in pre-calc
Speaking of which
I need help with that, but the problem is, I don't know why I'm not understanding it
i have like 40 minutes, can help walk you through stuff?
I would love help if I could figure out the reason why I can't remember the formulas
first off, what have you tried?
Like I know how to derive it and everything, but just something about the formulas aren't sticking
making up silly songs, flashcards, etc.
01:22
It's probably because I haven't practiced enough, tbh
And also because my math teacher's handwriting is horrible geez
<interested in how other ppl get taught things>
Okay let me pull up the equation reductions for ones where I couldn't understand them
okay, so what do you remember about them off the top of your head
I know how to get to the double angle formula, and I can use those to find the half-angles?
hold on while i process my notes
Oh, right, I was confused on how the stuff got derived for half angle formula of cosine
So
cos(2x)=2cos(x)-1. All good there
you want me to help with deriving them?
01:27
Yeah
Hold on I need to take the trash out
I'll just finish the derivation to make sure I know how
so you want to do the derivation?
Sure
2cos^2(x)-1=cos(2x)
cos^2(x)=[cos(2x)+1]/2
Ohhh that makes sense
good!
01:33
I kept thinking the half-angle formula was dividing the two out of the x in cos(2x) which was why it was confusing me
cos(x)=+/- sqrt([cos(2x)+1]/2)
And when you get half of an angle like 22.5, you can plug it into there
yep!
Okay, can you help me with the tan derivation?
okay
(As in, just sit and watch and point out if I made any errors)
tan2x=(2tanx)/(1-tan^2(x))
you're missing some parentheses there
better
01:38
Hm, why is it tan(2x)=(sin(2x)/cos(2x))? Do I need to simplify the right side or is it because of the trig identities, i.e., tan(x)=sin(x)/cos(x)?
see i never bothered with the tan formulas myself
I just calculated the sin and cos and divided
So is tan(2x) just sin(2x)/cos(2x)?
(me too ;p less to remember)
simplified some, yes I think
tan = sin/cos always. tan(elephant) = sin(elephant)/cos(elephant)
4
tan(2x) = sin(2x)/cos(2x) for exactly the same reason as tan(x) = sin(x)/cos(x)
01:40
Ah, good to know
... how do you find the sin and cos of elephants
though tan elephants are rarer than grey
no need for any identities or anything
ok gareth, time to explain quantum physics with elephants
if this isn't obvious to you, that suggests there's something suboptimal in your understanding of functions somewhere around here that may cause trouble in the future. I don't mean that there's some fact you don't know, just that it seems like in some way you may not be thinking about things right. I am aware that this is a super-unhelpful thing to say.
("this" = "tan 2x = sin 2x / cos 2x, given what you know about the relationship between sin, cos, tan")
01:42
So tan((x+23)-(pi/e))=(sin((x+23)-(pi/e)))/(cos((x+23)-(pi/e)))
yep
I haven't checked you copied everything correctly, but yes
(mmm, pie)
well assuming everything exists
I knew tanx=sinx/cosx, I just didn't know it applied to everything within the tan as well
well set x=elephant
01:43
well X can literally be anything >.>;
so when they say "tan x = sin x / cos x", this isn't a fact about some particular x
(I mean, sometimes people will write down something that applies to a particular x, but that isn't what's happening here)
you should be thinking of that as meaning "tan [anything] = sin [same thing] / cos [same thing]"
Write, because
your 2x is just two of a different x
01:43
44 secs ago, by Graylocke
well X can literally be anything >.>;
Okay, that makes more sense
(same for sin = 1/csc and cos = 1/sec and cot = cos/sin and all that)
that's the thing that should feel obvious and if it doesn't it may be worth staring at it for a while and thinking to see whether it becomes more obvious
I was thinking of x being a specific value, rather than "any value"
yes :) that is the key there
x can be anything. even elephants
01:45
sometimes you do need to think of it as meaning a specific value. If someone writes down x^3-x^2+x-1=0 and says "solve for x", they're asking you what particular values of x make the thing true.
That's where it kind of threw me off, because I've only really dealt with stuff outside the parenthesis
But it's also common practice to write down something with free variables in it and implicitly mean "this is true no matter what x is".
Dec 13 at 19:45, by bobble
a trick that might help: these equations must be valid for any value of x
Dec 13 at 19:45, by bobble
therefore, you can set x to any value you please
once it has a meaning in the universe, you have to stick to that meaning. when you are using it as a pattern, you can choose what it represents...
that's also why this trick I showed your earlier works
01:46
When you want to be explicit about it, there's some useful notation -- an upside-down "A" that means "for all".
So you could write ∀x tan x = sin x / cos x
Oh, that's why tan(x+y)=sin(x+y)/cos(x+y) and then you can just reduce it down
and you might (or might not) find it helpful to think about identities as having "for all x" or "for all x,y" or whatever in front of them
when you know ∀x [stuff], that's your licence to replace [stuff] with anything you like, so long as you use the same anything each time.
That was a very helpful reality check
Ok, I should be good now
01:48
It's an unfortunate fact that the underlying idea of algebra -- you can give names to things -- turns out to have two quite fundamentally different uses (1. expressing that something is always true by using a variable to stand for "anything"; 2. a notation for when you want to say something about particular things without yet saying exactly what they are).
I suspect a lot of people are at least a bit confused by this.
I agree - it's one reason that I think that they should be given separate letters when being set up... then you can think of them separately until that becomes second nature :/
Ah
I think I confused the 1 & 2 up
Used to thinking of it as 2
I don't, for the avoidance of doubt, mean that it's wrong to use the same notation for 1 and 2. But I think it's potentially confusing.
Solve for "x", etc.
It would be an interesting experiment to try presenting this stuff in schools with explicit quantifiers everywhere -- quantifiers are things like the ∀ for "for all", and there's another one with an upside-down "E" meaning "there exists" -- and see whether it ends up being less confusing or more confusing.
01:51
*backwards?
Why do we like x & y so much? X-axis, y-axis, solve for x, x is a variable which means anything, etc.
Why not? :-)
@Graylocke Rotated through 180 degrees.
<leaving for dinner>
It happens that for an A that's the same as reflecting about a horizontal line, and for an E it's the same as reflecting about a vertical one, but I'm pretty sure the right way to think of it is as a half-turn in both cases.
01:53
I have never thought about it before o.o;
I'd personally find it more helpful if we used different variables to mean different things, but inevitably will probably become simplified because people are lazy
like my proofs for example
The variables magically disappear when they ask me to simplify a trig function
(mostly because the variables themselves weren't being manipulated at this point, so I could just drop the "x" or the "u" or whatever. Bad habit though)
One remark about that "gradient of a sine curve where it crosses its midline" thing earlier. This is the same as asking about the gradient of the graph of sin t at t=0, of course. And, provided you express angles in radians which everyone always should, the key fact is that sin t ~= t when t is small. And that's "obvious" because sin t is the y-coordinate of a point on the unit circle that's moved anticlockwise through a distance t, and when t is small [... continues]
... that bit of the unit circle looks a lot like the straight line x=1, and moving a distance t along that line obviously takes you to (1,t) whose y-coord is just t.
I have... zero clue what you just wrote
Also why is radians better than degrees?
Ha. Feel free to ignore it if it was way too confusing.
Radians are better than degrees because (1) the definition is super-simple and has no arbitrary magic numbers in it and (2) lots of facts about trig functions and angles and whatnot are simpler when expressed that way.
But degrees look so much nicer...
01:58
Nope.
:-)
they turn out not to be...
but radians don't require a separate hard-to-type symbol
Well in my level of mathematics it is >:)
in the same way that metric turns out to be easier
97° vs. 97 rad
01:59
Definition of cos and sin: if you start at (1,0) and walk a distance t around the unit circle (anticlockwise, by convention), then you end up at (cos t, sin t). Note that I never used the word "triangle"; trigonometry is about circles, not triangles.
Though the true sin of degrees is that YOU NEED TO CHANGE THE CALCULATOR FUNCTION TO CALCULATE ANYTHING IN DEGREES GAHHHHH
Not "97° vs. 97 rad", "97° vs. 97". The radian is a natural unit and doesn't need any notation.
when you forget to change the mode to degrees smh
and the total distance around a circle (its circumference) is... 2pi
@GarethMcCaughan "natural" meaning...?
02:01
Radians is just distance
@Graylocke tau
or that, sure.
@merrybot Yes, exactly what PNL said.
@PrinceNorthLæraðr i don't think you've realized the pun here
"sin"
Oh I totally did :P
02:02
hm
Btw is pi or tau better to use? I assume pi, but Idk anything about math so
Everyone knows pi, not everyone knows tau, so use pi.
It might have been better if we'd gone with tau instead of pi originally, but we didn't and the world is not going to change.
Some things come out nicer in terms of pi, some things come out nicer in terms of tau, some things would actually be nicer if the thing we had a special symbol for were pi/2.
It doesn't matter all that much which one we choose, we just need to choose one, and it happens that (maybe unfortunately) we chose pi.
02:06
pi/2=hmm (which greek letter isn't taken)
for pi/2 i propose a pi symbol with four legs
since tau/1 = 1 leg
tau/2 = pi = 2 legs
tau/4 = pi/2 = 4 legs
Logic
But like, why does that oddly make sense
I never thought of pi and tau as having "legs" up until now
I propose a leg and a second V leg
What about "chi"?
It's like an "x"
χ
please star this message so i can get another hat :P
5
02:11
Hey, that's my line!
you get a hat for starred messages? o.O;
is it a nice hat?
Here's how I feel about degrees, by the way. Imagine that we'd never invented exponents, but we have a function called sqr that does something squaring-like. The idea is that sqr operates on distances, and its output is an area converted into a distance, so what it does depends on your units, and it's conventional to use units called squrbles with the property that sqr(180 squrbles) = 180 squrbles. [... continues]
And then any time you or I would just square something, in this hypothetical nightmare-world you're supposed to express whatever-it-is in squrbles, and since with everything in squrbles sqr(x) = x^2/180 you end up with a lot of gratuitous factors of 180 everywhere in what would otherwise be nice simple mathematical facts.
So why do we have angles and degrees? Doesn't it make like idk building things easier?
02:17
And your calculator has a button that does sqr, but by default it assumes you're working in squrbles, so when you enter 1 and hit sqr you get 0.00555555... or whatever it is.
And then one day someone says "wait, why don't we adopt units where sqr(x) just equals x times x?" but everyone says no, we're used to squrbles.
I think the reason we have degrees is that 360 is a nice round number that approximately equals the number of days in a year, and the orbit of the earth around the sun is roughly circular.
Anyway, I claim that although there isn't a formula for (say) sin or cos that's as simple as x*x, they are functions that behave much more nicely when you work in radians, so much so that we shouldn't think of angles as having units at all; they're just numbers, and the functions we use them with are defined "in radians".
I didn't know Gareth was a conspiracy theorist O_O
e.g. sin x = x/1 - x^3/(1x2x3) + x^5/(1x2x3x4x5) - x^7/(1x2x3x4x5x6x7) + ... -- the series goes on for ever, but for any value of x the terms eventually get small enough fast enough that it all adds up to a definite value.
Not as simple as x*x but still pretty nice.
Measurements are always best described from a base identity. Angles are "how far around the most basic circle do I have to move to push this line's intersection point to the other line's intersection point?" Degrees are that, as are radians. It's just that degrees are "how many 1/360ths of that circle do we pass?" and radians are literally "how far did we travel"
(if you're happy with factorial notation then write it as x/1! - x^3/3! + x^5/5! - ...)
anyway, it's 2:20 local time and I should really be in bed so I'll say good night now.
sleep well, dream nice?
02:22
incidentally, the thing about years and circles and whatnot isn't conspiracy theory, it's a perfectly serious suggestion. Blame it on the ancient Babylonians.
(the conspiracy goes waaay back ;p)
 
4 hours later…
05:58
Shhhhhh!
 
16 hours later…
22:21
@bobble Are you around?
The problem asks me to simplify (cos^2(x))*(sin^2(x)) using only linear functions of cosine
And I was able to simplify it down to (1-cos^2(2x))/4 before I got lost on what I'm supposed to do. I have no idea what my teacher did
@Deusovi Are you available?
drat
@GarethMcCaughan?
¯\_(ツ)_/¯
Don't worry, the extremely shaky mathematician is here to help! :D
who is definitely good at math!
(I always feel bad pinging people for tutoring)
Hold on, so I think I've got an idea
I'm gonna have to write this down on paper
22:36
Okay, sure
Okay, so we've got this equation and the goal is to use only cosine functions to simplify?
Not just cosine functions, linear cosine functions
One thing that'll be useful is that cos nx has n'th powers of cos x in it.
So if the thing needs to be linear it'll have to use cos^4 x.
You've already got a thing involving cos^2 2x. What do you know about cos 2x and cos 4x?
I was thinking of using cos^2(x)=1+cos(2x)/2 and substitute 2x into it
Ohh, yeah, I recall that
22:37
That sounds like an excellent idea. Why not try it?
Gareth is here to save us
Okay, let me try plugging some stuff in
Eek that's a hard fraction to type out
Let me try to simplify it down
But plugging in 2x for cos^2(x) gives me (1+cos(4x))/2
That looks to me like a thing with only linear functions of cosines in it.
Set common denominator for 1-((1+cos(4x))/2))
Hm, looks like this simplifies down to (1+cos(4x))/8 if I did my simplifying correctly, which is a linear function!
Oh wait the negative sign flips stuff
Hold on
(1-cos(4x))/8
Or as my teacher wrote, 1/8(1-cos(4x))
Yay!
Yay! :D
22:44
I got stuck on "what the heck is cos^2(2x)" until I remembered our little conversation about the "x can be anything"
I'm still not sure how my teacher simplified it down. @GarethMcCaughan And what did you mean by cos^4 (x)? That seems to be the route my teacher took, but I can't read her handwriting
@Sciborg Perhaps you can understand why I'm confused? :P
Well done!
cos^4 x means (cos x)^4.
It's a stupid bit of notation, especially since e.g. sin^(-1) x doesn't mean (sin x)^(-1).
Right
I have no clue what's happening in the third line of her work. What happened to all the squared stuff?
Third line is pulling out a common factor ((1 + cos 2x) / 2).
Amusingly, on the very first line she already had that common factor -- it's the "cos^2 x" at the start -- but she expanded it all out, and then decided to undo it a couple of lines later.
22:52
I skipped the first three lines by directly just turning sin^2(x) into (1-cos(2x))/2
And actually I'd be inclined to start it off a bit differently.
Remember that sin 2x = 2 sin x cos x. So sin^2 x cos^2 x = 1/4 sin^2 2x.
That was actually on the next slide
And it is much simpler :/
And now if you happen to remember how to write sin^2 2x in terms of cos 4x then you're pretty much done.
Thank you for throwing us a mathematical life preserver, Gareth :p
23:01
Thank you everybody :)
I always feel so bad pinging people for tutoring stuff ;-;
Hope it doesn't bother any of y'all
On the contrary, makes me happy to open my messages and see a North ping :P

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