geometry-and-trigonometry

[somber coyote]

aleptian

[vagrant ore]

17.386

[errant salmon]

ok that’s what u got right

for area of triangle?

[vagrant ore]

Yes

[errant salmon]

and u previously calculated the area of sector right

[vagrant ore]

Yes

[errant salmon]

ok

what do u think is the next step

[vagrant ore]

So subtract the two

[errant salmon]

yep

[vagrant ore]

Ok

29.739

[errant salmon]

check if that’s correct

[vagrant ore]

How?

[errant salmon]

by plugging the answer in

and clicking enter

on the question

[vagrant ore]

Oh it doesn't tell me until the end

[errant salmon]

ohhhh

it should be correct

[vagrant ore]

Ok

[errant salmon]

i suck at geometry so i’m not 100% sure

maybe if one of the helpers is lurking they can verify

[vagrant ore]

Rounded to the nearest tenth would be 29.7

Right?

[errant salmon]

yes

[pallid frigate]

@unique flower @opal carbon
i've thought about the problem quite a bit today. Firstly, one issue with the two circles approach is that by selecting a point inside the red intersection, you are forcing the lengths to satisfy the triangle inequality theorem. However, in the original problem, they didn't say that it necessarily had to make a triangle. When you pick two random numbers from 0 and 1, and then having 1 as the last side, in many cases it won't be able to form a triangle at all, for example, 0.1, 0.2, and 1.

I've also thought about what del did with the diagram showing how some parts of the circle are more dense to other places, and upon experimenting with digrams it's clear to see why that is true.

Assuming we were to first randomly pick a number (the purple and orange circles), then decide the next one (somewhere on the circumference of purple/orange circle), it is clear that for the orange scenario would have more options for the second number; which means more possibilities for the second one.

This explains why with the two circles method, the graph of it is not evenly spaced.

[opal carbon]

my assumption was that the solution would involve somehow the ratio of the area of the purple circle to the area of the two large circles combined (so something like 2pi-area of red, since that would represent all possible picks of the two numbers, even non triangle ones), not ratio of purple to red, but yeah i still dont see how to make it work 😄

[pallid frigate]

i guess rectangles reign supreme over circles lol

[unique flower]

its the even spacing

but now you understand :)

[steel ibex]

Hello everyone,
Do you have any resource for the proof of the Triple angle formula

[dark sparrow]

just derive it yourself from the angle sum formula and the double angle formula @steel ibex

[nocturne remnant]

Or, use De Moivre’s Theorem and skip through half the trouble

[ember oasis]

could anyone check my solution

[dark sparrow]

,calc 400*pi/60

[somber coyote]

Result:

20.943951023932

[dark sparrow]

not 6.67

[ember oasis]

oh shoot i missed adding the pi to my calculator, thank you

[gloomy shuttle]

can someone help with these

or could someone do these and show the steps please

[silent plank]

what have you tried?
@gloomy shuttle

[lilac whale]

For number 8, you must remember that if you have a supplementary angle, the two other angles are equal to the bigger one.
For number 3, just add the equations for the angles together and make them equal 180. Use algebra after.

[real wyvern]

queston: how in the world does this work for area of all quadrilaterals ?

s is half the perimeter

why though?

[upper karma]

looks like herons formula twice using a law of cosines in between

[icy berry]

can some1 help me understand the r formula

[valid harbor]

@icy berry you mean the r equation?

well if so it is a parametric equation

it is equivalent to

x=13+2t , y=7+2t , z=1-t

[icy berry]

So a given set of coordinates on the line is equal to t ?

[valid harbor]

that is not what i said

give a value to t and you will get coordinates of a ointt that belongs to that line

*of a point

[fast pulsar]

Technically they use lambda instead of t, but they are just variables, and they represent the same thing.

[icy berry]

can some1 explain why AP * (2,2-1) = 0

i know hes using the dot product

Why does he use 2,2,-1 instead of 3,-2,6

[fast pulsar]

Couple things. If the dot product of two vectors is 0, then they are perpendicular (or one of them is the zero vector).
Next, we need (-3,-2,6) to define the vector that connects (-3,-2,6) with (-p,0,2p)

[real wyvern]

Question:has anyone ever needed to use brahmguptas formula for area?

[real wyvern]

This is what I mean

[torpid mountain]

helpppppp

anyone?

[nocturne remnant]

Try to draw a diagram to visualise things

[torpid mountain]

i tried

and messed up

[nocturne remnant]

What did you do

[torpid mountain]

i drew a hole

then 7 meters to the left

and 25 meters up

[nocturne remnant]

It’s not 25 meters up
It’s that the distance between the holes is 25 meters

[torpid mountain]

i still dont get i

it

:\

[nocturne remnant]

The holes are at opposite corners of the yard right?

Therefore 25 is actually the length of the diagonal

[next flame]

hey guys. i'm trying to find a quick and dirty solution to a problem i'm having.

assume you're watching a 3D object that is in front of you. its projection onto your visual plane is a square. what can we say about the size (area) of this square as a function of distance?

i'm tempted to say area drops proportionally to the square of distance

[dark sparrow]

perspective projection?

yeah, should be inverse square

[next flame]

can you give a rough argument as to why?

[dark sparrow]

the surface area of a sphere grows quadratically in its radius, and the area you see that's at distance r from you is a constant fraction (corresponding to the solid angle of your FOV) of the sphere of radius r centered at your eyes

if you make the same argument but one dimension lower, the (linear) angle that an object of fixed size subtends in your FOV is about inversely proportional to its distance from you, for much the same reasons except that the circumference of a circle is linear in its radius instead

[next flame]

cool so the angle of fov is the determining factor here

[dark sparrow]

yeah you could say that

[next flame]

so, is it exactly inversely proportional to the square, or approximately?

[terse flume]

One day i want to be smart like u

[dark sparrow]

approximate, but refining that seems to depend on the shape of the object.

[next flame]

actually that was inaccurate

it's the angle the object makes with your eyes, and how it varies with distance (that sounds more correct)

[dark sparrow]

yes

the apparent angular size, as i believe it's called

[next flame]

thanks Ann 😄

hmmmmm. so i'm trying to do this in the low-dimension case.

seems to me that θ = arctan h/r

and arctan(1/x) looks very much like 1/x

[dark sparrow]

well that it does of course

the power series expansion of arctan(t) at t=0 begins with t

and the next term is -t^3/3

[next flame]

which decays very fast and we can therefore say arctan(1/x) ~ 1/x

cool stuff

[dark sparrow]

yeah

i'd imagine that if there's an error term in the two-dimensional case it would need to decay at least as fast as r^-4

[upper karma]

Have been struggling with this question lately and would really appreciate it someone could help me understand this situation. Trying to get the right equation, but I'm not getting the right answer.

[nocturne remnant]

That paper looks like it’s been through both world wars

But you can drop an altitude from A to BC

To get two right triangles with all angles known

The rest is just length chasing

@upper karma

[upper karma]

Is it something like this? I've tried it before about it wasn't the right answer. The statement I'm most confused about is the angles of depression to each end of the bridge to be 54° & 71°.

[nocturne remnant]

why did you change the diagram?

the first one was correct

also, i feel like your original approach was correct

[upper karma]

I mean, I'd still end up with the same answers, so i don't think any of my diagram are correct at all. I feel like the angles of depression are throwing me off, which is what I'm confused about.

[nocturne remnant]

I think AC is 270

[grave pond]

It looks like you never completed the calculations in your \emph{first} answer from the war-torn pencil drawing that ended up with $\frac{270·\sin 55^\circ}{\sin 54^\circ}$. The two calculations with red marker on whiteboard are indeed the same as each other, but the pencil one ought to give the right result if only you finish calculating it.

[somber coyote]

Troposphere

[upper karma]

Oh! I just got it! Turns out the first approach was correct, I just needed to split it in half and use my trig ratios to find the lengths.

[nocturne remnant]

Wouldn’t the answer be 273. something? You have to add up the bases of both triangles.

[upper karma]

The answer said it was 97.6m so that was just different the two adjacent lengths, I was thinking about adding them as well. However I did some more research on this

And saw a very very a similar question like that had subtracted the two angles

Wait lemme show it to you

This was it, the placement of the angles of depression, were really confusing, so that's what was throw me off.

[nocturne remnant]

Ahhhhh

Tbf the question did not make is clear which configuration it was

[upper karma]

I didn't like this approach, and wasn't learning anything form it

I talk to my teacher about it, and see if the way I did it was reasonable or not

[nocturne remnant]

You mean this?

[upper karma]

Yes

But thank you, for the feedback @nocturne remnant @grave pond , really appreciate you guys 💖

[nocturne remnant]

Uh

[upper karma]

Yes

[grave pond]

I find it difficult to claim the balloon is "flying above the bridge" in that diagram ...

[nocturne remnant]

I don’t think your approach was “correct” if you started off with a different configuration and then subtracted the sides instead of adding them to obtain the correct answer

[upper karma]

Me neither, but I don't necessarily I understand it

[nocturne remnant]

In the other configuration the right triangles are still there

Just arranged differently

[upper karma]

Wait let me try and incorporate this into my original approach, because that other method was just something I had researched.

[upper karma]

I keep, stopping up to this part, how does one of the triangle become obtuse? Shouldn't both sides stay as right triangles? Also, to find the full length of the bridge, why did you do 270sin71° ÷ tan54°?

[nocturne remnant]

The right triangles I was referring to are the orange and pink ones that I drew

[upper karma]

Ooh so they're place on top of each other

So then, since on of the my first the light if the bridge would be the different of my two adjacent lengths, due to one being longer than the other

Is that how we'd end up with that obtuse triangle, which is the has the length of the bridge?

Let me try redrawing this

[knotty yoke]

Would it be possible for someone to help me in a separate channel with Trig Identities?

[upper karma]

Now this, this makes more sense. My values were correct from before, however my diagrams needed a newer perspective, in order for me to figure it out.

[twin plover]

how do u find the base of a trapezoid and how to u find the height of a trapezoid

[sudden anchor]

do you need help with 1 or 2

[twin plover]

mostly 1

both is preferred

cuz I’m confused

[sudden anchor]

well, look how do you calculate the are of a trapezoid

I am gonna dm-you because it's my handwriting(really bad on computer)

[twin plover]

a+b / 2 * h

[sudden anchor]

yes

[twin plover]

oh k

[twin plover]

anyone know how to do this?

<@&286206848099549185> sorry

[upper karma]

do you know what dm:cm=2:1 means?

[twin plover]

kinda but idk what the numbers are

[upper karma]

it means that the length of DM is twice the length of MC

[twin plover]

yea

[upper karma]

the total length is 10 (by adding the ones on the bottom)

[twin plover]

I need the find the area but idk how

[twin plover]

yup

[upper karma]

find the area of the whole rectangle and minus the area of the shapes inside,

[twin plover]

yes I know but idk how to find the small quarter of a circle and the triangle

[upper karma]

well the radius of that small circle is 4

[twin plover]

mhm

[upper karma]

that is a circle right?

[twin plover]

yea

but it’s like cut off

[upper karma]

its 1/4 of a circle

so its 1/4 of the area

[twin plover]

mhm but like it’s cut off?

[upper karma]

if NB is 4 then BK must also be 4 because the circle is the same radious all around\

[twin plover]

yes but idk what the measurements for the triangle

[upper karma]

well
BK + KC = AD
BK = 4 (by circle with radius NB (4))
AD = 6 (by circle with radius AN (6))
so
4 + KC = 6

KC = 2

understand how to get that?

[twin plover]

ohh ok but what’s MC

[upper karma]

DM:CM = 2:1
and DM + CM = AB (which is 10)

so theres a total of 3 parts (2:1 is 3 parts spits in a 2-1 ratio)

[twin plover]

so what is it sorry lol

[upper karma]

like this

if you devise the line us into 3 sections, DM gets 2 and MC gets 1

so MC is 1/3 of DC

[twin plover]

yes but what’s the number

[upper karma]

AB = DC so DC = 10

MC = 10/3

[twin plover]

is there a number for that or just 10/3

[upper karma]

well, in decmial it would be
3.3333333333333...
so i recoemnd jsut using leaving it as 3/10

[twin plover]

oh ok

[twin plover]

could u help me again?

[twin plover]

<@&286206848099549185> so sorry! Hopefully I can tag u guys :’)

[upper breach]

<@&286206848099549185>

Ignore the BDA

[grave pond]

Name the major arc
I think I'll call it Kenny.

[limber marlin]

this is a code red '

my teacher hasnt been here for a bunch of clases and hasnt taught us anything

i have a test tm and i am taking a practice test rn but i need desperate help ngl

if anybody is willing to help

[grave pond]

hmm, my guess would be that the diameter of the circle is supposed to be sqrt(8²+2²). But that is just a guess -- it doesn't even look like it is exactly centered on a grid point ....

If we assume that, you can compute the area of the circle and of the rectangle, and divide.

[final stag]

Can someone help me with this?

[grave pond]

Draw radiuses from E, F, G. That gives you three quadrilaterals where you know three of the angles in each...

[final stag]

So then what do I do

😭

[dark sparrow]

radiuses

radii

[dark sparrow]

angle-chasing

[final stag]

Huhhh

[dark sparrow]

perhaps you could give a name to that center point - O is a common choice (but it is by no means obligatory if you want to go with something else)

[final stag]

I did but idk what to do after

[dark sparrow]

so let's look at PEOG shall we

what angles do you know in quadrilateral PEOG?

[final stag]

G

P

E

[dark sparrow]

give their measures too, not just their names.

[final stag]

G and e are 90

[dark sparrow]

uh huh

[final stag]

Oh and every quadrilateral was like 360 or something right

[dark sparrow]

the angles in every quadrilateral add up to 360 degrees, yes.

[final stag]

Tyy

[zealous vector]

Can anyone help me with this

@spark stag Im sorry for the ping but I am so desperate

you know what

@everyone

[gusty wasp]

fuck off

[analog mango]

it's like a Challenge

[wanton edge]

Yeah. It is like a challenge. ||I solved it but I don't want to give the answer||

[mossy crater]

how do I even start? (need to calculate area)

[maiden hearth]

It seems like you have to use Riemann integrals

[upper karma]

Hello friends I need help with this annoying trig excersise

Why is:

Cos²(45°+a) = (1/2) - sin(a)*cos(a) ?

[dark sparrow]

$\cos^2(x) = \frac{1+\cos(2x)}{2}$

[somber coyote]

Ann

[upper karma]

Here's most of the formulas u need

I tried and tried but this is just.. too much

[dark sparrow]

have you tried applying the formula i quoted to the left-hand side

[upper karma]

Yes

(1+cos(90+a))/2

But i don't really know how to go from that point on

[dark sparrow]

(1+cos(90°+2a))/2

and cos(90°+2a) can be written as something different

[upper karma]

Such as?

Cos(2a) because cos(90) = 0?

[dark sparrow]

no

cos(x+90°) = -sin(x)

[upper karma]

What about the 2a?

[dark sparrow]

well i expected you to infer that cos(90°+2a) = -sin(2a) as a consequence.

[upper karma]

I'm sorry bro it's just that I've been stuck for an hour on this god damn maths excersise

That nobody can do

[dark sparrow]

please don't call me bro.

[upper karma]

Ok

Sorry

[dark sparrow]

also saying "nobody can do this exercise" can range from being an exaggeration to being false outright

do you understand why cos(90°+2a) = -sin(2a)

[upper karma]

Yes

Atleast that I can understand

[upper karma]

But I haven't found a way to apply any of the formulas to the right side of the equation

[dark sparrow]

yeah so now you have (1 - sin(2a))/2

that won't be necessary

we are almost at the point where we have the right-hand side

there's one more identity that you should recall

[upper karma]

What do you mean with identity? Sorry I'm not a native English speaker

[dark sparrow]

well when you posted that big list up there you called them formulas instead

[upper karma]

You can use a formula on sin(2a)

[dark sparrow]

so if it makes you happier i can call them formulas and not identities

[upper karma]

Ooh

[dark sparrow]

and yes that is exactly what i wanted you to do

[upper karma]

Ok let me do that on paper right now

Alright

[dark sparrow]

and there you have it

[upper karma]

Huh

Are these equal to eachother

[dark sparrow]

yes...

$\frac{1-2\sin(\alpha)\cos(\alpha)}{2} = \frac{1}{2} - \frac{2\sin(\alpha)\cos(\alpha)}{2}$...

[somber coyote]

Ann

[upper karma]

Holy shit

[upper karma]

Wait

💀

No nvm

Ok I got ot

@dark sparrow thanks so much

[teal kayak]

guys so i made the spiral of theodorus a few days ago and today decided to do something similar but instead of the height of each triangle remaining at one, i made the height and hypotenuse always equal

what is this shape called

[uneven axle]

A cool cyclone

[teal kayak]

ok so i've noticed that the hypotenuses here (the line segmentss that go to point A, excluding line AB) follow a pattern like sqrt(x) where x doubles every next hypotenuse

sqrt(2), sqrt(4), sqrt(8), sqrt(16), etc

[analog mango]

with a similar process you can make sqrt of 5 and then make the golden ratio or a pentagon

[teal kayak]

oh that's cool

i know how to get to sqrt 5

[analog mango]

You could gave me a hint? i want to solve it but i dont know from where start

[teal kayak]

i also know this formula that i think gets u golden ratio spiral in desmos r=\phi^{\frac{2\theta}{\pi}}

bruh

it didnt do the thing

r=\phi^{\frac{2\theta}{\pi}}

!!?!?!?!

\phi^{\frac{2\theta}{\pi}}

[analog mango]

hahahha

[teal kayak]

AAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAA

[analog mango]

but i understand

[teal kayak]

commits murder

[analog mango]

https://latex.codecogs.com/eqneditor/editor.php

use that

and take screenshot

[teal kayak]

r=phi^((2theta)/pi))

tbh i dont even know what theta is i just saw the formula online

[analog mango]

i trisected an angle of 90 🤪

[teal kayak]

$r=\phi^{\frac{2\theta}{\pi}}

ok thats not how u summon the bot

rip

[analog mango]

the bot is sleeping zz

[wanton edge]

1. Find 30-60-90 triangle
2. Use law of sines

Sorry I'm late. I was sleeping

[analog mango]

i found the 30-60-90. but the law of sines didnt come to mind!, thank youu

[analog mango]

teorema de thales?

[wanton edge]

you're welcome

[analog mango]

||is sqrt of 2?||

[grave pond]

That's what I get too.

... but without using the 30-60-90 triangle.

[plush quest]

Can someone help me with 12

Im a little stuck

[grave pond]

Square root?

[plush quest]

Yea big problem idk how to square it

[grave pond]

You mean you don't know how to find the square root of 45?

[plush quest]

Yes exactly

Not trolling

[nocturne remnant]

the square root of 45 is not rational. the best you can do is to simplify it

[grave pond]

Your calculator will have a button for it. Or leave it as $\sqrt{45}$ or $3\sqrt{5}$ for an exact answer.

[somber coyote]

Troposphere

[plush quest]

Alright thanks big dawg

[wanton edge]

||Yes||

[magic lichen]

which channel does vectors come under?

[nocturne remnant]

probably #linear-algebra

but #geometry-and-trigonometry is probably fine as well if it doesnt involve 'real' linear algebra concepts

[twin sail]

Someone help

[wanton edge]

Use Pythagorean theorem when two sides of a right triangle is known

[tight lagoon]

It anyone could dm me some reminders about triangle congruencies to help me study for a test that’d be nice

[wooden epoch]

do you know if this is .64?
https://cdn.discordapp.com/attachments/326138737606262786/976000926265516042/unknown.png?size=4096
https://cdn.discordapp.com/attachments/326138737606262786/976000996184576061/unknown.png?size=4096

[uneven axle]

what

why is it such a mess

[lusty stratus]

let's take a look at the upside-down triangle at the right

we know it's hypotenuse(8) and one of t's other side (4)

so

let y be this side

8^2 = y^2 + 4^2

64-16 = y^2

42 = y^2

y = sqrt(4)

==============

for the triangle in the middle, we know the hypotenuse and the sqrt(48) side

6sqrt(3) can be written as sqrt(36*3) = sqrt(108)

hypotenuse = sqrt(108), and one of it's other side is sqrt(48)

108 = 48 + z^2

z^2 = 60

z = sqrt(60)

=============

now we solve for the triangle in the left

we know the hypotenuse(sqrt(60)) and one of it's other sides(sqrt(50))

60 = 50 + x^2

10 = x^2

x = sqrt(10)

================
@twin sail

that is the answer for the fourth question

in the same way, solve the 5th question

[uneven axle]

that looks very easy tbh

[lusty stratus]

because it is

[teal kayak]

this is kinda cool for some reason, i took the first five points on the golden spiral and drew lines between them and it kinda sorta lines up with a square root spiral

[icy berry]

the solution is to complete the square

but i dont see why

[upper karma]

Oh hey

A or b?

I dont get it btw but damn looks intriguing

[upper karma]

What is your course called with that exercise may i ask that?

[icy berry]

me ?

[icy berry]

b

i know the solution but i wanted to know geometrically what would happen if i chose a t value that was slightly less/more than the maximum point

OHHHHHHHHHHHH

IGET IT NOW

if you solve for the roots there is still a lot of distance between them

so it doesnt do anything

but as you get closer to the maximum point you would get to a point wheres theres almost no distnace inbetween the 2 points

which is why you solve for the maximum

[turbid thunder]

je fais mes devoirs

[barren agate]

Quick question, I'm taking an online course to relearn math to get ready for college in the fall. I'm having trouble understanding this problem.

Basically, I just want to know if there's a way to solve this problem without the reference images. When I begin the actual problems, the reference images aren't shown and I'm just left with cos(5π/6 ). Am I maybe suppose to be saving these reference images and referring to those to do the problems or am I suppose to know how to do this without them? I haven't a clue since it just randomly started throwing Trig stuff at me and I'm honestly not ready for this section lol

[nocturne remnant]

(I'm using degrees here for my own convenience)
there are three non-obvious values you should definitely remember:
sin(30)=cos(60)=1/2
sin(45)=cos(45)=sqrt(2)/2
sin(60)=cos(30)=sqrt(3)/2
Then you can derive the rest of the common angles just by properties
e.g. sin(150)=sin(30)=1/2

[lyric spire]

solve for the area of parallelogram

pls help

[gusty wasp]

what are your thoughts?

[lyric spire]

well I just got till the diagonal being 15

n thn Im just blank

[static jolt]

some notes on parallelogram: A parallelogram is a quadrilateral in which both pairs of opposite sides are parallel A diagonal of a parallelogram seperates it into two congruent triangles The opposite angles of a parallelogram are congruent The opposite sides of a parallelogram are congruent The diagonals of a parallelogram bisect each other Two consecutive angles of a parallelogram are supplementary In a parallelogram with unequal pairs of consecutive angles, the longer diagonal lies opposite the obtuse angle

i guess you could use trigonometry to find the height

but im not sure that is the best way to approach this problem

maybe someone else can help

[barren agate]

thank you, took a minute to click, but i get it now

[upper karma]

@dark sparrow hey, sorry I pinged you, but can you tell me how you found -sin(a)? It is not in the formula list, and I cant seem to find the logic behind it? How do you find cos(90°+a) = -sin(2a)?

[dark sparrow]

it's not cos(90°+a) = -sin(2a), it's cos(90°+2a) = -sin(2a)

and the cosine addition formula is in your list

cos(90°+2a) = cos(90°)cos(2a) - sin(90°)sin(2a)

[upper karma]

Oooh

I see

Thank you again

[cursive fossil]

Sum of area of both triangles, use pythagoras theorem to find the diagonal

[lusty stratus]

the area of a parallelogram is base*height

we already know the base is

17

now find height

[lusty stratus]

notice: both hypotenuses have the same length and in a parallelogram, and opposite sides are equal as well

[lusty stratus]

you are correct

the diagonal is 15

we get this because the opposites sides of a parallelogram are equal

we know the hypotenuse is 17

and one of it's other sides is 8

so naturally, the other side should be 15

the area of a right angle triangle can be calculate as

(adjacent*opposite)/2

so

8*15/2

= 60

that is the area of the triangle in the left

by this, the other triangle must also have the area of 60

because it also has the same dimensions

60+60 = 120 total

[lusty stratus]

theta is a variable, which usually represents an angle

it's like "x" but for an angle

[magic lichen]

and also the two triangles are congruent

[lusty stratus]

yes, i forgot to mention that

thank you

[magic lichen]

np

its a property of parallelogram anyways

[wanton mantle]

A trapezium is enclosed by the straight lines y=0, y=6, y=8−2x and y=x+k where k is a constant. Find the possible values of k given the trapezium has an area of 66 units^2.

[lapis moon]

plz show us your work, so that we know where you're stuck

[wanton mantle]

? oh no, this is just a random question a friend gave me

[lyric spire]

@lusty stratus oh tysm
Makes sense now

[lusty stratus]

you're welcome!

[somber coyote]

Alex88

[upper karma]

How do I do this? When I try applying all the formulas to the right side of the equation, I get 0. The question is "prove this equation is true"

[stoic jay]

is the second term sin ?

[upper karma]

I don't know how to deal with this excersise when the left side of the equation has everything to the power of 4

[upper karma]

Yes

Here are most of the formulas needed for the trigonometry we learn

Try as I might, I just can't seem to get them on equal values

[stoic jay]

is the answer 0

[nocturne remnant]

Try to apply difference of squares on the left

[upper karma]

I don't think so, no. But that is the value I got in the right side of the equation

[nocturne remnant]

It should be true for all values

[upper karma]

?
Sorry, not a native English speaker

[nocturne remnant]

a^2 - b^2 = (a+b)(a-b)

[stoic jay]

pi, 2pi, 3pi and so on

[upper karma]

The answer doesn't have anything to do with pi fortunately

[upper karma]

Hm

I'll try,

[nocturne remnant]

The equation is true for any angle bruh

[upper karma]

but we haven't seen that getting used in trig in school

[stoic jay]

sry, i thought we had to solve for alpha

[nocturne remnant]

Let this be your first time then

[upper karma]

Lol

Ok

[upper karma]

It's fine

@nocturne remnant I did it

I got 0 for the left side of the equation in the end

[nocturne remnant]

???

How??

[upper karma]

Well I'm an idiot with a negative iq

[nocturne remnant]

Cos^2 alpha plus sin^2 alpha is just 1

[upper karma]

Oh yeah you're right

[nocturne remnant]

And cos^2 alpha - sin^2 alpha is on the formula sheet

[upper karma]

Damn I forgot about that

[upper karma]

But what I wrote down is different than the formula on the sheet, isn't it?

It is getting divided by 2, or am I allowed to ignore that

Damn I'm so bad in trigonometry

[nocturne remnant]

It isn’t, check the formula for cos 2a

[nocturne remnant]

I meant check the formula sheet

[upper karma]

Haha sorry

[upper karma]

Well I applied that formula, but what about the division by 2?

[nocturne remnant]

?

[upper karma]

Fuck

Sorry

[nocturne remnant]

You shouldn’t have to divide by 2

[upper karma]

No

[nocturne remnant]

Lol

[upper karma]

I'm just dumb

[upper karma]

I'm sorry bro but there's like this big test tmrw about trig and no one understands anything about it

[nocturne remnant]

Yeah trig is weird

[upper karma]

Especially when you're learning about it in any other context except basic Pythagoras theorem

[nocturne remnant]

So did you get the solution

[upper karma]

Yes

Thanks to u

[nocturne remnant]

Oh cool

Good luck on that test

[upper karma]

Tyty

[rose tendon]

How can I represent the midpoint between the orthocentre of a triangle and one of the vertices as a Trilinear coordinate?

I need to figure out what the first input is, because (x:sec(B):sec(C)) gives a point on the altitude with side a as the base

[strange harbor]

Hello guys

i need help

does anyone know how to solve it ?

[tranquil lark]

where do i go from here

wtf do u do with cos 4pi/15

[static jolt]

do you have a caclulator?

[tranquil lark]

yea

[static jolt]

does it have a cosine button?

[tranquil lark]

yes

but im not finna have a calculator in class

[static jolt]

ahh

[tranquil lark]

when i take the test

ima use photo math

[static jolt]

nice

ummm

well

4pi/15 is 48degrees

i dont think you would have to memorise cos(48)

so maybe something is wrong earlier

usually it would be in 15 degree increments

so 0,15,30,45,60,75,90

[tranquil lark]

exactly what jm thinking

im so lost cuz

I think they did this problem in class the last time

i just dont remember exactly what they did

going crazy

[static jolt]

i dunno if can do pi/3 to 4pi/3

[tranquil lark]

some of my classmates saying u just leave it like that

well shit

imma have to ask the teacher

[lilac whale]

#calculus

[static jolt]

@tranquil lark i guess its right going from pi/3 to 4pi/3 but im not 100% sure

Reference angle for 240°: 60° (π / 3)

but idunno what it says after you find tan

maybe you can type it

[harsh gull]

Hi all, I'm working on a proof and all I need to show in order to prove the theorem is that sin(b+c) < sin b + sin c for b + c < 180 degrees.

I have tried this on desmos using a number slider and it seems to be true which is promising for my overall proof, but i don't know how to prove this. Also neither b nor c can be equal to 0, otherwise it obviously wouldnt be correct

[nocturne remnant]

||Consider a triangle with angles b,c, and 180-b-c||

@harsh gull

Alternatively, see that
sinb cosc + cosb sinc < sinb + sinc

[harsh gull]

yup this is actually what i have already done because its part of my proof

[harsh gull]

i dont see how this is true though

[nocturne remnant]

both cosc and cosb are not more than 1

[harsh gull]

because it isnt true if you go beyond b + c =180

[nocturne remnant]

Just apply law of sines on the triangle

[harsh gull]

i see

that makes sense

thank you

[nocturne remnant]

This is a nicer proof tho

[harsh gull]

well the reason why im not doing that

is because im trying to prove that 2 sides of a triangle is longer than the third

for which i am already using the law of sines

[nocturne remnant]

Oh is that

Ok lol

[harsh gull]

xd

[nocturne remnant]

You should say that cosb and cosc cannot simultaneously be equal to 1

So that the inequality is strictly less

[harsh gull]

yeah makes sense

ty

[round oracle]

A lighthouse, 50 metres high, is on a cliff. From a ship, the angle of elevation from the bottom of the lighthouse is 44° and the angle of elevation of the top is light house is 49. Find the height of the cliff.

[upper karma]

Maybe just visualize it

[round oracle]

Can someone tell me what cotangent means

[dark sparrow]

cotangent as in the trig function?

and wdym by "means"? are you looking for a definition of cot(x)? it's cos(x)/sin(x).

[round oracle]

I meant what it means when cotan is the reciprocal of tan

[dark sparrow]

you should be more precise when asking your questions

do you know what the word "reciprocal" means?

[lofty depot]

how do I start this , whats the formula ?

[dark sparrow]

no such thing as "the" formula

however you might find it of use to note that angle BAO is a right angle

[grave pond]

If you know the concept of "power of a point", then this is very quick, and you can ignore O and y.

Otherwise the only way in I can immediately see is that the cosine of angle AOC must be 1-2·8/11 = -5/11.

In fact I would name the point at the other end of the diagonal D, and express it as cosine of angle DOC is 5/11.

Then the law of cosines gives you |OC| = |OD| = y.

[lapis moon]

I'm responding to @upper karma's comment at #help-19 message, which is now closed.
Euclid showed a proposition (Elements Prop. III.16) that gives an equivalent definition of the tangent of circle without using #calculus.
This relevant MSE answer https://math.stackexchange.com/a/55645/290189 links to an online notes
https://mathweb.ucsd.edu/~ashenk/Section2_8.pdf, whose section History of the derivative starts with Euclid's definition of a tangent of a circle.

[upper karma]

thats just calculus

and derivatives

even if what is a tangent line was defined how will u prove that the tangent is perpendicular to the vector from the center to the boundary of the circle

which is needed for solving the problem ig

[lapis moon]

Euclid shld hav a proposition for that in Book III of his Elements

[upper karma]

is it there in that pdf?

oh ig u pinned the wrong answer its given on the top answer lemme see

[lapis moon]

no but you can ctrl+f to search for Euclid inside the PDF

the linked MSE quesiton has only one answer. the "Prop III.16" quoted is in the question body

[upper karma]

Euclid (ca. 300 BC) stated that a line is tangent to a circle if it intersects the curve at one and only
one point.(2) He also used this definition for ellipses, but it could not be applied without modification to
the other conic sections—parabolas and hyperbolas—since there are two such lines at each point on a
parabola and three at each point on a hyperbola. For the parabola y = x
2
in Figure 4, the two lines are
the tangent line and the line parallel to the y-axis. For the hyperbola y = 1/x in Figure 5, the three lines
are the tangent line and the lines parallel to the x- and y-axes.

this part right?

even if it is how do u prove thats its equivalent to the definition of a tangent using calculus or a touching line

without calculus

[lapis moon]

in the linked Discord question it's about the tangent of a circle. His Prop. III..16 showed that the tangent (in the sense of Euclid) is perpendicular to the radius joining the center and the point of tangency

The straight line drawn at right angles to the diameter of a circle from its end will fall outside the circle, and into the space between the straight line and the circumference another straight line cannot be interposed, further the angle of the semicircle is greater, and the remaining angle less, than any acute rectilinear angle.

[upper karma]

uh but

how do u prove that

[upper karma]

it will never touches the circle

i mean

meets once

[lapis moon]

straight line (AE) drawn at right angles to the diameter of a circle (AB)

[upper karma]

same question was whats written here

[upper karma]

ya

but my question was why will it touch only once

which was his first claim

like how do u prove it

rigorously

without using calculus or the fact that its at right angles [ coz u derived it from the first claim]

[lapis moon]

no the right angle is from the construction

and the proposition's claim is that AE only touches the circle once at point A

the following part gives the equivalence

[upper karma]

i mean the fact that the tangent is perpendicular

[upper karma]

yes but how on earth do you prove it

without calculus

[lapis moon]

Euclid's proof doesn't contain calculus

[upper karma]

ya but how will you prove the claim then

[lapis moon]

space between the straight line and the circumference another straight line cannot be interposed

[upper karma]

why?

[lapis moon]

AF is a line between AE and AB

you can observe that it intersect with the circle

[upper karma]

i do

but how will you prove thats a problem

like why would it intersect at only one point

[lapis moon]

if the straight line makes a right angle with the radius, it touches the circle once.
otherwise, it touches the circle twice.
that's the equivalence between "making right angle with the radius" and "touching the circle once"

[upper karma]

ya but euclids proof isnt rigourous then

[lapis moon]

the former can be constructed without calculus

[upper karma]

coz its defining it

not justfying each step

tangent line in simple words is the touching line

[lapis moon]

why not rigorous? there's a logical equivalence between the two

[upper karma]

yes

but why will it make 90 degree angle

[lapis moon]

you mean AE? that's by construction

[upper karma]

yes AE is making 90 degree angle i can see

but why will all tangent lines have 90 degree angles

with the radius

[lapis moon]

that's a logic problem. we wanna prove $P \iff Q$. first we prove $P \implies Q$. Then we prove $\lnot P \implies \lnot Q$.

[somber coyote]

vin100

[upper karma]

i am not so familar with the symbols

[lapis moon]

nvm lemme explain:
P and Q are propositions (statements that can be proved either true or false)
P ⟹ Q means "P implies Q"
P ⟺ Q (P iff/if and only if Q) means "P implies Q" AND "Q implies P"

[upper karma]

interdependent?

[lapis moon]

here P is the condition that "AE ⟂ AB"; Q is the condition that "AE touches the circle once"

[lapis moon]

sorry i dun see a math definition for this word.

[upper karma]

makes sense but thats still not rigorous , coz then u are just stuck in a circle

coz to prove P u need to prove Q and to prove Q u need to prove P

isnt that circular reasoning

[lapis moon]

if we have "P ⟺ Q", we can say that P is logically equivalent to Q.

[upper karma]

ok but how do we have it?

also i have to go now coz i have work

[lapis moon]

the validity of P (AE ⟂ AB in the above diagram #geometry-and-trigonometry message) comes from construction

[upper karma]

thats just a drawing a visual not rigoruous thats my entire point

[lapis moon]

you can construct a line perpendicular to the diameter

with classical compass and straightedge construction

[upper karma]

I have just came across those identities for the first time, I managed to figure out the reasoning behind sin(arctanx), cos(arctanx), and tan(arccosx) from triangles after assuming theta = arctanx/arccosx but i couldn't really figure out a way to find out the last one?

[wise pawn]

might be easier to look at a right triangle and look at cos(x) = sin(pi/2 - x) and look at how the two non right angles give rise to the same ratio for one being sine, the other being cosine

sort of wordy, it's obvious in a picture, just label one angle x and the other angle must be pi/2 - x and then look at the ratios you get from cosine and sine of those angles respectively

[upper karma]

Oh interesting

Good to know, will test it out tomorrow

[undone cove]

can someone pls help, idk how to do this

[wanton edge]

For 6 and 7, check using Pythagorean theorem.

For 8 and 9, use Pythagorean theorem for finding the value of r

[hardy plinth]

Can someone explain how to get the answer a ≈ 448 from sin(27°34')=a/968 I can't find the explanation anywhere

[grave pond]

Multiply both sides by 968, then punch the LHS into a calculator?

[hardy plinth]

Sorry, still lost. I don't have a calculator with the key LHS

Are you referring to ")^In"?

[grave pond]

__l__eft-__h__and __s__ide

of the equation after you multiply it by 968.

[hardy plinth]

HAHA oops,1sorry

I will try that now

Thanks!!

Punching in sin(27.34)*968 is getting me 444.57318045 How could it be approximately 448 from this

[grave pond]

27°34' is not the same as 27.34°.

minutes of arc are 60ths of a degree, not 100ths.

[hardy plinth]

Right. 1° = 60', correct?

so 34' would be 0.6°

Do I add 0.6° to 27°

sin(27.6°) * 968

448!

THANK YOU!!!!!

[upper karma]

Did I do this correctly?

I am not sure if it is the most efficient method though

sorry about the colours btw lol

oh nvm

thats completely wrong

[wise pawn]

you have the right idea but yeah you worked it wrong

[upper karma]

i should've just assumed 1/3 as x in the beginning

and did the theorem with that and then applied the numbers

[wise pawn]

your placement of 1 and 3 are good

but 4^2+1^2 is not equal to 3^2

looks like you calculated 4^2 to be 8, but that's 4*2

[upper karma]

ah no

it was the other way around

oh i do see my mistake

but yeah u r right

i just assumed the square root of 8 is 4 lOl

this always happens and it is sad

I did write a more general way to approach this anyways

[wise pawn]

go slightly slower, just be a bit more careful with each step and then before too long you'll get faster at being meticulous

[upper karma]

yeah thats a good idea

[wise pawn]

at least, that's what worked for me personally, I remember the change of attitude after working through the same physics homework problems like 3 times over because of stupid mistakes lol

[upper karma]

oh god, physics is going to be way worse once i get into that haha

I am hardcore learning math right now since most of the physics stuff I am learning requires some good basic calculus foundations

so it is better to get that out of the way first

[wise pawn]

personally physics made me appreciate trig and calculus a lot more, because they were useful tools and you get to see what it's used for, before that it was kind of just abstract logical trivia

[upper karma]

i remember my physics professor telling me integrating is not about finding the area under the curve, but about infinitely summing infinitely minuscule things

i tried to conceptualise what he said

i still don't really understand it

which is why my main struggle points in calculus are mainly with integrals

[wise pawn]

yeah, that's good, the "area under a curve" is sort of an easy way out as a special case of a double integral which confuses people cause you have to sort of reconceptualize it in multivariable calculus

[upper karma]

yeah

i just have no idea how integrating a position vector would lead you into getting a velocity vector, and integrating that would get you an acceleration one

it is understandable from the derivative view point

i can see it from that point of view

but integrating it? how and why

i guess it is because the integral is the antiderivative so it makes sense that it would undo what the derivative would do

[wise pawn]

well, in the simplest case distance = rate*time

[upper karma]

oh so u would be considering the rate instead

interesting

[wise pawn]

so we could write this as $x=vT$ and so at each point of time, our rate is the same so we could also write it as $$x = \int_0^T v dt$$ because $vdt$ is a little displacement $dx$

[somber coyote]

Merosity

[wise pawn]

but now because we're looking at the velocity at each point of time, we could put any function there, not a constant, and we are still adding up the individual little displacements

[upper karma]

Interesting

Will look more into it once I get to that part of things

[wise pawn]

as a separate example that I find it's a bit easier to think of a 1D rod as the interval [a,b] which has a density f(x) with dimensions of mass/length at each point, then f(x)dx is the mass at a point, so to get the total mass, $$m=\int_a^b f(x)dx$$

[somber coyote]

Merosity

[upper karma]

so basically integrating is like cutting a stick to many many parts and adding those parts together until u get the 'full' thing again

well, not the greatest analogy

but i guess it makes sense in this case?

[wise pawn]

yeah exactly

[upper karma]

also can i ask you about a problem i am trying to conceptualise? @wise pawn

[wise pawn]

sure

[upper karma]

it basically asks to find the domain and range of this function and its derivative

for the domain, i kinda just assumed 0 <= e^x <= 1 because it cannot be higher than 1 or lower than -1

thus -infinity < x < 0 after taking the natural log

[wise pawn]

sure sounds reasonable, although 0 < e^x <=1 since you can't get 0=e^x

[upper karma]

yes

[wise pawn]

sounds like you have the right idea so far

[upper karma]

for range, i assumed [0, pi/2]

considering the ranges of both

but im not sure

e^x cant be lower than 0 so we can remove the negative

arc sin is only between the negative and positive of pi/2, so i am guessing it would be the intersection with it and e^x which is [0, pi/2]?

[wise pawn]

it will be the intersection of the range of e^x with the domain of arcsin(x) yes, which is slightly different than what you wrote, one minor difference

[upper karma]

wait you consider the intersection of the range of e^x and the domain of arcsin(x)?

hm

that does make more sense

since you are 'inputting' the e^x into the arcsin

so it would really be [-pi/2, pi/2] after all

oh yeah that's definitely way more sensible, why did i shortsight myself into seeing it only with the ranges of both?

[wise pawn]

nooo that's worse now

[upper karma]

ahaha seems like i am falling down a hole of misconceptions

[wise pawn]

not every value in [0, pi/2] is attainable by plugging in a value of x to arcsin(e^x)

[upper karma]

really?

[amber lagoon]

I would add, the equation looks like this:

Since e^0 = 1 and arcsin isn't defined for values above that, to my knowledge.

[wise pawn]

hint @upper karma

[upper karma]

well, since it cannot be 0, then it cant be including the 0 in the overall equation either

so

(0, pi/2]?

[wise pawn]

yup

[upper karma]

very interesting, i was on the right track but just missed the 0

[upper karma]

this graph does make it more sense as well

what can be sum of alpha beta and gamma equal to if this is a rectangle?

[wanton edge]

https://tenor.com/view/could-you-just-give-me-a-little-more-detail-about-that-neil-degrasse-tyson-startalk-can-you-give-me-more-information-about-that-i-want-to-know-more-about-that-gif-20162364

[dusty lark]

If w is the width of the rectangle, as $w\to 0$, $\alpha+\beta+\gamma\to \frac{3\pi}{2}$.

[somber coyote]

PhysMan

[vocal badge]

a^2 + b^2 = c^2 😎

[upper karma]

Is there a shorthand way to refer to the midpoint of a line? Like a symbol or something?

[silent plank]

capital M

with the name of the line segment in subscript

$M_{AB}$ would refer to the midpoint of segment AB

[somber coyote]

ℝamonov

[upper karma]

gotcha, thank you

[soft marten]

help

[sleek mirage]

How do i solve these two

5 and 6 go together bte

[ember stratus]

A triple of points on a plane is called lucky if one of them lies exactly in the middle of the segment connecting the remaining two. Mark multiple nodes on the checkered plane so that each of them is included in the same number of lucky triples, more than $1$. help me visualise

[somber coyote]

erictheeonicpizhao

[fair vapor]

guys what is the name of the subject that we learn ratio theorem in it ?

[nocturne remnant]

A 3x3 square of 9 points work except the centre point which lies on both diagonals as well; is there a way to “distort” the square so that the centre doesn’t lie on the diagonals anymore?

[full lagoon]

can anybody help me with my geo hw please its due in an hour and i have no clue wtf im doin

[mild sleet]

#❓how-to-get-help

[ember stratus]

brute force 😄

[jaunty cosmos]

Hi, I want to ask about trigonometry. I've solved tan(π/12), but why is the answer different when I solved by simplifying it to tan(π/3-π/4) (the ans is √3-1/1+√3) and when I solved by simplifying it to tan(π/4-π/6) ( the ans is 3-√3/3+√3). Can somebody explain it to me?

[nocturne remnant]

It’s the same answer

To see why, divide both the numerator and denominator of the second answer by root3

[dark sparrow]

parentheses!!!!!!!!!!!!!!!

[jaunty cosmos]

But then I got 3√3-3/3√3+3

[dark sparrow]

you get $3\sqrt{3}-\frac{3}{3}\sqrt{3} + 3$?

[somber coyote]

Ann

[nocturne remnant]

Bruh I wouldn’t tell you to do it if you just simplified it back the same way

Try to simplify each side of the numerator and denominator separately

[jaunty cosmos]

ahh I get it, thanks

[tawdry garnet]

guys can you tell me middle diagonal lenght formule in hexagon i cant find it anywhere

[dark sparrow]

is that a regular hexagon?

if so, then the diagonal is just twice the side

[fair vapor]

hello guys

guys what is the volume of tilted rectangle around its diameter

[tawdry garnet]

yes thats regular

[dark sparrow]

.

[tawdry garnet]

ok thank you so much

[rare siren]

is there a formula for the radius/diameter of a circumscribed circle around a regular polygon knowing the radius of the inscribed circle is 1

[wanton edge]

$\sec!\qty(\frac{\pi}{n})$
% Where n is the number of sides

[somber coyote]

Si Arya

[rare siren]

Thank you

[wanton edge]

np

[upper karma]

Does this function have any global extremas?

I sketched it and came up with this which would tell me (pi/4, 1) would be the global maximum but when i put it in a calculator it tells me no extremas exist on the domain

so im confused

[next flame]

does the calculator restrict the search to (-π/2 , π/4]

?

also, it says f(x) but is actually a function of θ

that's probably bad notation tho

[upper karma]

i kinda used symbolab so idk lmao

[next flame]

alright the calculator is incorrect then

critical points are used to determine extrema in the interior of the domain

you found that the function attains a maximum on the boundary of the domain

[upper karma]

oh critical points are not meant for endpoints?

oh that's interesting

good to know

i guess it makes sense

[next flame]

critical point means a point where derivative is zero

[upper karma]

can it not also mean where the derivative is undefined?

since you can get cusps i would assume that would give u a local maxima/minima

[next flame]

yea that is correct

When dealing with functions of a real variable, a critical point is a point in the domain of the function where the function is either not differentiable or the derivative is equal to zero

[upper karma]

i have a random question but

how can u exactly determine if the critical point is a minima or a maxima after figuring it out?

i guess i can just solve the function and find out

[next flame]

by the way i asked wolfram alpha and it gave the right answer

[next flame]

you can do that using the "second derivative test"

second derivative is positive it's a minimum, if negative it's a maximum

if zero "inconclusive"

[upper karma]

oh i have heard of that!

concavity i think it was called?

[next flame]

bingo

[upper karma]

still haven't gotten to that yet

i think i will after i am done studying extreme value theorem

[grave pond]

In practice, your first approach should not be to go directly to second derivatives, but to try if you can understand the behavior of the function intuitively.

Such as the form of the expression telling you what the rough shape of the graph will be.

Second derivatives are useful to prove that your intuition is accurate, or as a last resort if the situation is too complex for you to imagine reliably.

But don't let the availability of a formula-churning procedure mislead you into thinking you can do in the long run without developing an intuition for the shape of functions.

[upper karma]

yeah i think it isn't impossible if u can determine the domain and range

after that i guess just see what happens at certain points during the function to determine it i suppose?

but i dont understand how the second tangent line to the tangent line of the function would be able to give us the 'sign' of that specific point though

hard to visualise in my brain

[grave pond]

That's a somewhat indirect fact.

[upper karma]

ah i see

[grave pond]

Basically, the function near f(x0) when f'(x0)=0 will look roughly like a parabola with (x0,f(x0)) as its apex, namely
f(x) is approximated by f(x0) + (f''(x0)/2)·(x-x0)².

and if f''(x0) is nonzero, the approximation error cannot overshadow the parabola shape until you get some distance away from x0.

(This is Taylor's theorem).

So the sign of f''(x0) tells you whether the parabola points up or down.

[grave pond]

Note that "domain and range" are also not something that you should think of as the outcome of formula-churning procedures. In the ideal case they're something you see directly from your mental model of the function's behavior, and symbolic calculation is just something you do do confirm that mental model.

[next flame]

wikipedia's usually pretty good for that stuff

https://en.wikipedia.org/wiki/Second_derivative#/media/File:Animated_illustration_of_inflection_point.gif

just imagine what a minimum looks like roughly, it's like a cup:
\ /
\ _ /

so, before x0 the function is decreasing, and the derivative is negative. at minimum the derivative is zero, and after x0 the derivative is positive. this means that the derivative is increasing, and you get the minimum at the point where the sign change takes place

[upper karma]

Yeah

[upper karma]

Oh damn, that's a lot of things to consider. But I guess it makes sense for understanding purposes

[upper karma]

Ohh that's actually pretty crazy

good way to look at it

[grave pond]

The sketch you started with here is exactly the kind of stuff I'm talking about.

[upper karma]

I tried doing your method with a function such as this one

I couldn't do it that well in my brain but i did do it physically which does give me this

but i am not sure how to go about doing all of the operations intuitively though

[grave pond]

Right, that one takes quite a bit of experience to have an intuition for.

[upper karma]

It really sucks to do math in a formulaic way since it just always feels like i am just memorising concepts rather than understanding them

I try my best to research the origins of everything but it gets very complex really quickly

I guess that's the point of math after all

[grave pond]

Right. Some students get the impression that they're supposed to avoid imagining things and only do everything symbolically. I can now see you're not one of them. Apologies for jumping the gun and hitting you with a lecture that's not for you.

[upper karma]

I try my best haha, I am just a college freshman dealing with relatively pretty simple concepts for the first time

Calculus is really fun if you try and understand what the hell is going on with it all

[next flame]

yep. also 100% recommend 3blue1brown videos on youtube

they''re amazing

[upper karma]

I did watch his video on the essence of calculus

very informative

the organic chemistry is also amazing in teaching concepts, albeit it does seem heavily formulaic in some of his videos

[upper karma]

Also, I am curious regarding this, but is there a way to find out whether the critical point is the global extrema without computing the endpoints as well?

I know you can do the first derivative test or the second derivative tests to find out if it would be a local minimum/maximum

I don't see a way to find out other than comparing it to the value of the other endpoints/critical points

[next flame]

nope, you have to check the boundary as well

[upper karma]

ah i see

Nvm

[queen cradle]

can you use Pythagoras theorem on a square or circle if so can you leave a link to a video on how to do it?

[next flame]

combine the relation with pythagorean identity

and also use sin2x = 2sinxcosx

[magic lichen]

guys

in vectors

how do u derive

$\tan \alpha = \frac {B\cos\theta}{A+B\sin\theta}$

[somber coyote]

Alpha | Sync

[magic lichen]

where alpha is ur angle of deviation of vector C from vector A

[full dome]

help?

[pliant roost]

is A the center of the circle?

if yes then because the MA and AO are radiuses they're equal, and since AN is height (because AN is perp. to MO) it's also median, meaning NO = 3.5 ft
and then you just have congruence of triangles ANO and APO by hypotenuse and leg congruence, and you can finish that off yourself

[full dome]

so the final answer isnt 3.5?

[pliant roost]

it is

:/

[static jolt]

i got -.28

theta is -53.1301 degree

then did cos(-53.1301*2)

[thin mantle]

does anyone know why alternate interior was invented and what it was used for

[full dome]

help

[upper karma]

Hints: connect A with C, use the theorem of Pythagoras and the "intersecting cords" Theorem

Hi I am new here @upper karma*

[upper karma]

Hint: you can use the formula of cos(2theta) that is written only in terms of sin(theta).

[sweet whale]

https://cdn.discordapp.com/attachments/359052581022203914/978177050034716692/unknown.png

help

[full dome]

help

[dark sparrow]

@full dome do you still need help with this?

[full dome]

yes

[dark sparrow]

have you ever worked with equations of circles before?

[full dome]

yes but not backwards

[dark sparrow]

can you tell me in general what the equation of a circle looks like?

[quaint heart]

if you just want the answer, its ||(5,0) and r=7||

i think

[dark sparrow]

don't just give out answers.

[quaint heart]

thats why i blurred it

[dark sparrow]

better to just not post it at all

[quaint heart]

ok

[dark sparrow]

is there a reason why you chose to ghost me on the last question?

[full dome]

(x-h)^2+(y-k)^2=r^2

it appears as y^2 because y-0^2 is redundant

[nocturne remnant]

[full dome]

help?

[amber cobalt]

a) you can use midpoint formula b) you can use distance formula

Midpoint = ((-10+4)/2),((-2+6)/2)

-3,2

Which is the center of the circle

Radius is square root (49+16)

Square@root 65

Yea

I guess that’s what it is ?

a) (-3,2). b) square root of 65

c) (x+3)^2 + (y-2)^2 = 65

I am 95% confident that I am wrong

[full dome]

where does 49 come from

[fallen sleet]

Is there a simple way to prove that sin is a computable function?

[amber cobalt]

(-10-(-3))^2

( X1-X2)^2

[last magnet]

can someone help me find the

[grave pond]

First, draw a diagram.

[last magnet]

ok

[last magnet]

you still here?

[grave pond]

Do you have a diagram?

[placid parrot]

how do i solve this?

[pliant roost]

should we use the coordinate stuff?

if yes then you could find angle and so arc measures via some arccos (without + 2πk or ±) and etc.

[lucid aurora]

Distance formula to find the lengths of the sides and then Law of cosines to find the angles then Law of sines to get the rest of the angles

[harsh panther]

Once you get the individual lengths could you take the ratio of one length and the total length then multiply with 2pi

Would that work?

[atomic summit]

I have a problem that goes like this:

Imagine an obtuse triangle, and put perpendicular lines in the center of the triangle's sides. Wherever they intersect is the circumcenter of the triangle. Why is this so?

I've searched all over Google, and couldn't find an answer that covers this... Any ideas? And if so, please walk me through them.

[opaque gull]

have you tried drawing it?

[silk wasp]

is anyone willing to help me with geometry stuff

[atomic summit]

well, i have, but i'm not sure why it works

[atomic summit]

sure, what's your question?

[tulip minnow]

can someone help me with #2

I know i need to use B=1/2(10)(h)

but i don’t know what the height is

8 or 6?

[wanton edge]

Height is 6 m

[tulip minnow]

so would v=36m^3

[wanton edge]

Sorry, I thought the height of the pyramid 😅

There are 2 perpendicular line on the triangle whose lengths are 8m and 6m respectively

@tulip minnow So it isn't a great idea to use 1/2(10)(h) in this situation

[tulip minnow]

oh ye ye