geometry-and-trigonometry

[uncut grove]

can someone help me find x using the special triangles

im new at the subject

and dont quite understand

[upper karma]

sin(30degrees) = x /hypotenuse

right ?

or tan(30degrees) = x/20

use any of those

[uncut grove]

answr is 20 sqrt3 / 3 tho

[upper karma]

yes tan30 degrees is 1/sqrt3

you will get that as ans

[empty yew]

Is it $\sin(x)-\sin(y) = 2\sin(\frac{x-y}{2})\cos(\frac{x+y}{2})$

[somber coyote]

KingDanger

[upper karma]

yes

[empty yew]

Mm i don't know this but I will remember it for the next time.
Continue please

[upper karma]

yeah so that would imply

either one of terms

is 0

right

[empty yew]

Yes

[upper karma]

so solve for both

sin((x-pi/6)/2) = 0

whatdoes this imply generally ?

you know that sine is 0 at n(pi)

where n is any integer

right ?

[empty yew]

Mm

[upper karma]

so it basically means (x-pi/6)/2 = npi

[empty yew]

Yeahhh

[upper karma]

x = 2npi + pi/6

yes

this is one of the solutions

[empty yew]

Yes

[upper karma]

but

[empty yew]

We need both

To zero

[upper karma]

you also have cos((x+pi/6)/2) =0

[empty yew]

Yes

[upper karma]

both or even if either one of them is zero that will also make entire term 0

so we consider all solutions

[empty yew]

Yes yes understood

[upper karma]

when is cos(A) = 0 ?

when A is odd multiple of pi/2

[empty yew]

npi/2?

[upper karma]

no

odd multiple

(2n+1)pi/2

because cos pi is -1

cos 0 is 1

cos2pi is 1

but odd multiples of pi/2

cos is 0

i mean its 0 for (1)(pi/2) not (2)(pi/2)

[empty yew]

Understood

[upper karma]

as cospi is -1

and cos2pi is 1

[upper karma]

yeah so (x+pi/6)/2 = (2n+1)pi/2

(4n+2)pi/2 - pi/6 = x

this is the another solution

[empty yew]

(2n+1)pi?

[upper karma]

now think about how you will club both solution as one set

sorry

[empty yew]

Np

[upper karma]

now club both solutions

how will you do it

??

you can put n = 0 , 1 ,2 ,3...

for both solutions

and check

the pattern

you will get first one as the answer in this

[empty yew]

Mmm quite confusing but I will take few minutes to grasp it! Thank you so much!

[upper karma]

yeah there are always possibilities to miss solutions when solving trig equations

solve tons of those and you will be comfortable with those ideas

keep your mind open though

[upper karma]

towards using those identities

they can be any others you know

[empty yew]

I simplified the solutions you got\
$x = (12n+1)\dfrac{\pi}{6}\
x = (12n+5)\dfrac{\pi}{6}$

[upper karma]

did you not get npi + (-1)^n(pi/6)

yes

right

[empty yew]

I don't know how to club what is club

[upper karma]

if you think deep enough this is same as this

[somber coyote]

KingDanger

[empty yew]

Mmm gimme few minutes

[upper karma]

you can imagine graph of sinx

and a horizontal line parallel to x axis

which is y = 1/2

it cuts graph of sinx at two points

pi/6 and 5pi/6

and periodicity of sine function is 2pi that means it will repeat after 2pi

so this is basically 2npi + pi6

and 2npi + 5pi/6

so its also graphically verified

[empty yew]

Understood

[formal geyser]

Can anyone help me with this equation

I dont even know what to begin with

I used sin(2x) = 2sin(x)cos(x)

But it only complicated everything

[dark sparrow]

yeah it's counterproductive actually

expand sin(10x) as sin(4x+6x)

[formal geyser]

Ans then i should apply sine of sum formula, right?

[upper karma]

yea

[formal geyser]

Whats next

I am stuck

That's 2sin(4x)cos(6x) = 2cos(4x)sin(6x) + 1

[dark sparrow]

2(sin(4x)cos(6x) - cos(4x)sin(6x)) = 1

[formal geyser]

Okay, thanks

I got it

Now the whole equation doesnt seem to be hard

But first i didnt know what to start with

[upper karma]

thats what everyone struggles with

[formal geyser]

As far as i know, there arent any specific methods to solve trigo equations. Is it just experience and the ability to "see" formulas and identites?

[maiden brook]

the good problems require thinking

[upper karma]

intuition

manipulation

thats it

maths is abt basic calculations and amazing manipulations

[grave pond]

And the willingness to try things to see if they work, even if you aren't sure that they will work.

[empty yew]

How can we identify a function by just seeing it's graph(for x^2 graphs especially scaled graphs)?

[proud nest]

What is this asking for?

[sharp ruin]

It's asking for the trigonometric ratios with respect to their definition in a triangle.
SinØ=Perpendicular/Hypotenuse
CosØ=Base/Hypotenuse
....

[proud nest]

ok so not the actual values, just the fractions

[sharp ruin]

You can get the hypotenuse using the Pythagorean theorem
a^2+b^2=c^2

[proud nest]

but then also the unknown side is a wildly long decimal

[sharp ruin]

Yup with respect to that triangle

[sharp ruin]

I don't think you usually express them in decimals

It's just left out as a fraction

[proud nest]

well if I solve for the Pythagorean theorem

it comes out at sqrt317

which is a decimal

[sharp ruin]

First it doesn't evaluate to that
It should be sqrt(75)

[proud nest]

how? 11^2 is 121

plus 14^2 is 196

OHHH

[sharp ruin]

Here 14 corresponds to c

Not b

[proud nest]

YEAH that just clicked 🤣🤣

thanks 🤦‍♂️😂

[sharp ruin]

Not really an issue

It happens when you're new to these things

[proud nest]

yeah I have a whole packet I’m doing so I’m flipping back and forth between problems so I get lost in them

[sharp ruin]

Also for the square root; you can express it as such but just take care of the denominator not being a square root term. Make sure to rationalise it

[proud nest]

ok

[sharp ruin]

What courses are you in?

[proud nest]

actually there is a problem where I’m having the same issue of the Pythagorean theorem C coming out as a decimal

[proud nest]

Just trig for now

I have a midterm tomorrow

[sharp ruin]

Just don't take it to the decimal
The examiners don't really want that from you
They just wanna know if you understand expressing the ratios in the terms of sides of a right triangle which can easily be done by fractions

[sharp ruin]

All the Best for that 👍

[proud nest]

so just leave it as sqrt75 or whatever the answer comes out to be?

[sharp ruin]

You can simplify it a bit
Like take out sqrt(25) and write it as 5

Along with a sqrt(3)

[proud nest]

but sqrt75 is a decimal

[sharp ruin]

What is 75?
It's 25*3

[proud nest]

ohhhhh

I see

[sharp ruin]

25 is a perfect square

[proud nest]

simplify the square root I see what you mean

[sharp ruin]

Yes exactly

A suggestion from my side
Just try to keep things in fractions until you're forced to take them into decimals

[proud nest]

Ok

[sharp ruin]

Believe me it helps a lot 'cause many terms do cancel

[proud nest]

yeah

fractions are a lot easier than decimals

[sharp ruin]

Would you be able to do this one now?

[proud nest]

Yeah

[sharp ruin]

Just use the definitions on both the triangles and you're done

[sharp ruin]

Agreed

[proud nest]

Wait what do you mean

[sharp ruin]

Use Sinø=perpendicular/hypotenuse and all of that

[proud nest]

I need to figure out the side lengths tho

[sharp ruin]

Yeah but you do have the angles right?

[proud nest]

yeah true

[sharp ruin]

And the values of their trig functions are pretty standard

That's given just so that you can use it

[proud nest]

I get what you’re saying but how would that figure out the side lengths

like sin30 is 1/2 but how do I know it’s not 2/4

[sharp ruin]

Put it equal to what the sides correspond to in the triangle

Like Sin60° here corresponds to d/b

[proud nest]

yeah

[sharp ruin]

And use this in the triangles
You'll have some equations and some variables

Eventually some terms will cancel and you'll have your answer

[proud nest]

and sin60=sqrt(3)/2 but how does that give me the side lengths

[sharp ruin]

Put sqrt(3)/2 equals d/b

You have an equation

Use some other trigonometric ratio in the other triangle (The one with 45°)
You have another equation

[proud nest]

30/30=tan45

[sharp ruin]

Eventually, there will be a lot of common terms there and you can solve those equations by standard methods you must have learnt in algebra

[proud nest]

wait no I’m figuring out b

[sharp ruin]

Yeah that will give you a=30 though

[proud nest]

which I already have yeah

[sharp ruin]

You have one of the sides

Think of some other ratio

If it has a and something else

[proud nest]

a/b?

[sharp ruin]

You'll figure that 'something else'

[sharp ruin]

Yeah Cos45°

1/sqrt(2)

And you have a
So you'll have b too now

[proud nest]

wait but I’m still confused, how do I find the side lengths once I have these equations

[sharp ruin]

You have one unknown and you know everything else

Just solve the equation as you do in algebra

[proud nest]

ohhhh

[sharp ruin]

You get it?

[proud nest]

kinda, let me see if I can get the answer for b

[sharp ruin]

Sure
That's the best way to learn
Figuring out things for yourself

[proud nest]

can I switch this and make it sec45, that way the variable is on the top of the fraction and I can multiply instead of divide?

and it would only be sqrt2

[sharp ruin]

Yes you can if you're more comfortable that way

Just put the values correctly

Math won't lie if you do it correctly

[proud nest]

so b=30sqrt2

[sharp ruin]

Yes you're right!

Well done

[proud nest]

Nice one down 😂😂

[sharp ruin]

It's 2 down actually

[proud nest]

True, I already have a

[sharp ruin]

You figured a out too 😅

[proud nest]

And now do the same for d/b which we figured out earlier

[sharp ruin]

Yes

Okay then
I'm kinda in a hurry

[proud nest]

ok np, I can see if I can figure it out from here

[sharp ruin]

I hope I've helped you enough and you'll take the baton from here

[proud nest]

thanks for all the help!

[sharp ruin]

Yes you will for sure

Not a problem

It's my first time helping someone on this server

[proud nest]

you def explained everything pretty well

[sharp ruin]

It's just good that you got it

That's all that matters

[proud nest]

yeah trig is not my strong suit 💀

[sharp ruin]

It will be
Just be good with algebra and manipulations

[proud nest]

yeah I think eventually it’ll click

[sharp ruin]

Yeah sure

Okay then bye

[proud nest]

Alrighty, thanks so much!

[proud nest]

How do you figure out the VAs?

[autumn pelican]

How do I solve
cos(tan^-1 b)
I would understand how if I was given numbers but a variable confuses me

[cunning lion]

draw a right triangle

[autumn pelican]

I solved it, hypotenuse was just sqrt(b^2 + 1)

[queen fog]

im so stuck on this :/

(I need to verify the identity)

[upbeat helm]

Hint: what is cos(x+pi/2)

[queen fog]

cos x and cos 0

[upbeat helm]

?

[queen fog]

you can split it up into two parts no?

into cos x and cos pi/2

and cos pi/2 = 0

[upbeat helm]

Using like a sum rule?

[queen fog]

yeah

[upbeat helm]

Try writing that out

[queen fog]

got to this point, but from here i’m stuck

[upbeat helm]

You have made a mistake

The left hand side is 0 but the RHSisnt

[queen fog]

so it would be 0cos?

not sure what you mean

?

[frozen ocean]

idk why I sometimes think this is #precalculus

[frozen ocean]

are you familiar with reflect shift trigonometry?

[jagged wyvern]

Why is this wrong? My teacher told me if the number was a decimal, don't round it, and rather keep it in fraction form.

Here's what I did:
3x + 815 = 900
3x = 85
x = 85/3 or 28 1/3

System marked both answers wrong(??)

Am I seeing things clearly??

[trail tendon]

maybe it wants 85/3?

:l

[trail tendon]

mixed fraction form or improper fraction form?

or does it matter

[jagged wyvern]

I already tried both

It marked both wrong

[trail tendon]

uhh

oh

wait

theres 8 angles

lol

ahh and theres 8 sides

one looked flat so i thought 7 for a sec

cuz 180(7-2) would mean theres 900 degrees intotal

but 180(8-2) would be the actual total degrees

[jagged wyvern]

Oh

[trail tendon]

that would probably fix it

[jagged wyvern]

So 265/3?

Uhh it doesnt let me check no more..

[trail tendon]

😔

[jagged wyvern]

Someone help pls

[trail tendon]

with wut

hmm wait sum aint right

my brain aint right, nvm it looks good to me XD

[vocal bison]

How do u do dis

[trail tendon]

seems to be a volume problem

[vocal bison]

Is This right

[autumn pelican]

Could someone fact check my work?

Question

A 96-ft tree casts a shadow that is 150ft long. What is the angle of elevation of the sun?```

This is my work

Ignore the sqrt(17339) I somehow misread my calc to a large degree, everything else though as it is nonconsequential

[subtle crest]

one moment

@autumn pelican

[autumn pelican]

I discovered I misplaced my hypotenuse and adjacent

[subtle crest]

I'm getting ~ 32.6 degres\es

[subtle crest]

💀

Its alr

Mistakes happen sometimes

[autumn pelican]

At least I didn't submit it yet lol

[subtle crest]

U made the shadow the hypo lol

[autumn pelican]

yeah haha

Brain fart, it made sense in the moment

Was my process except for that fine though?

[subtle crest]

mmm no

you used sin, instead of tan

ig it was because u misplaced the hypo tho

It should be something like this

put that in a calculator u get 32.61

[autumn pelican]

Oh yeah, I am using tan next time cause I don't want to deal with the SQRT

[autumn pelican]

Don't I need to use inverse of tan though?

[subtle crest]

Well, yeah but I stopped here

That would be the next step tho

[autumn pelican]

Oh okay

Thanks for the confirmation, glad I asked here and didn't submit

[subtle crest]

You can't really inverse tan these type of fraction by yourself, you need a calculator

[subtle crest]

You're welcome :D

[autumn pelican]

Oh yyeah 100%, I'd go insane lol

[hoary totem]

wonder if i can try an manually calculate an inverse tangent without too much work
48/75 is 16/25, which is nice
we have 16^2 + 25^2 = 256 + 625 = 881
sqrt(881) is... 29^2 is 400 + 360 + 81 = 841
30^2 is 900, which is closer
so let's lets say 30, which is a nice number
so we can say cos(θ) = 25/30= 5/6 and sin(θ) = 16/30 = 8/15
this isnt consistent with each other but whatever
now we have to compute either arccos(5/6) or arcsin(8/15)
the cosine 5/6 is large so the angle is small, so maclaurin series should work well, cosine has a simpler maclaurin series so we use the cosine
1 - x^2/2 = 5/6
x^2 = 1/3
x = 1/sqrt(3) = sqrt(3)/3 ~= 1.73/3 = 0.5 + 0.23/3 ~= 0.58
converting to degrees, 0.58 * 180 / pi ~= 0.58 * 57 = 25 + 3.5 + 4 + 0.56 = 33.06 degrees

not too bad eh?

[lofty spear]

average Citruz post

[hoary totem]

not really

too few trig functions

[lofty spear]

I couldn't tell you the difference, to be honest

[hoary totem]

well tbf my posts are usually pure trig
not me trying to approximate trig

[lofty spear]

still, I feel obligated to say

average Citruz post

[hoary totem]

hey a lil trig never hurt anybody

whats wrong with a lil geometry

i dont think i deserve the reaction

why does everyone think trig is cursed

[lofty spear]

trig is not cursed, the trig you do is cursed

the geometry too

[hoary totem]

what does that even meannnn

[hoary totem]

😭 its just geometry 😭

[obsidian harness]

Alternatively using the Maclaurin series for tan, we need to find x + x^3 / 3 - 16/25 = 0
f'(x) would be 1 + x^2

If we guess x = 1, then we have x_new = 1 - (4/3 - 16/25) / (2)
Or around 1 - (4/3 - 16/24) / 2 = 1 - 1/3 = 2/3

Guess x = 2/3 and we have x is around 2/3 - (16/24 + 8/81 - 16 * (24/25) / 24) / (1 + 4/9)
x is around 2/3 - (8/81 + 1/(25 * 24))/(3/2) = ....
Welp it's still off

[hoary totem]

id rather not solve a cubic

also i dont have the maclaurin series for tangent memorized

[obsidian harness]

Newton Rhapson should work in theory

[hoary totem]

newton rhapson without a calculator?

i guess u could do like

approximate calculations like i did

[obsidian harness]

Yeah that's what I was trying to do

Unfortunately I get around 0.6 rad

[hoary totem]

yea

i mean

thats correct

but no better than what i got i feel

[deep yarrow]

can anyone help with with this cursed ass question

https://tenor.com/view/spongebob-squarepants-begging-pretty-please-beg-on-your-knees-pray-for-mercy-gif-26344462

[floral mulch]

for the love of God help me with this. I need to find the area of this triangle but the formula for this is "base • height / 2". BUT I ONLY HAVE THE LENGTH AND HEIGHT (⁠ノ⁠ಠ⁠益⁠ಠ⁠)⁠ノ⁠彡⁠┻⁠━⁠┻

[deep yarrow]

i gotchu

Half of the base would be 8.5

*2

[upper karma]

🤯

[floral mulch]

OMG THANK YOU SM YOU CAME IN CLUTCH

[deep yarrow]

just dont forget to multiply *2

[floral mulch]

Kk

Tysm

[deep yarrow]

112.2 i fink

https://tenor.com/view/tobey-maguire-sad-wah-wahh-cry-gif-15729067800335956216

[ruby wigeon]

I can only prove A is right

Urgent plz

[dark sparrow]

why urgent

[ruby wigeon]

cuz it's my homework

[dark sparrow]

do you have a definition of "triple equidistant point" on hand?

[vocal bison]

How do I solve this ?

Pls help something like this will be on my quiz tmrw

[silent plank]

apply factorial definitions

[hoary totem]

apply definitions then work thrru the algebra, also this isnt geometry

[vocal bison]

What ?

Oh

[lime dune]

5.2.2: what can you say about the base angles of the triangles?

[jagged wyvern]

Can someone tell me how do I can do this?

[scarlet robin]

im in algebra 1, what should i expext next year in honors geometry?

[trail tendon]

geometric shapes

angles ig

[scarlet robin]

🤯

[trail tendon]

😂

[scarlet robin]

NO WAYYYY

is there gonna be.... lines too?

[trail tendon]

nah

☠️ jk of course there will be lines

[scarlet robin]

🚫

[trail tendon]

yes there will also be diameters in circles how did you know

☠️

[scarlet robin]

will there be... exponents?????

[trail tendon]

prob not

[scarlet robin]

you lie

[trail tendon]

i mean there COULD be exponents but i can't think of any topic which has them/reason for having them

[scarlet robin]

uhmm, area of a literally any 2d shape?

[trail tendon]

oh yeah ur right

there definitely are exponents mb 💀

wait you learn area in geometry?

i guess

maybe like area of circle....

[scarlet robin]

$Circle: A=\pi r^2, Square: A=L^2$

[trail tendon]

yeh

i actually dont know

uhh

[somber coyote]

X

[scarlet robin]

uhmm not very many other shapes that i can think of off the top of my head

[trail tendon]

💀

but there isn't much exponents like manipulation

there might be exponents just for multiplying numbers

but that aint bad

at least

i think

i dont really remember 💀

i just dont think

[scarlet robin]

so, no more multiplying polynomials??????!!!!!?!?!!?!?!?

$\frac {(35^{-38x}+92837x^{345}+29^{2}+4{x}^2)^{-3x}}{35^{-5x}}=-2.5587987x$ solve for x in simplest form.

[trail tendon]

bro forgot x 💀

[scarlet robin]

it auto multiplies?

[trail tendon]

bro what did you do 💀💀

[cunning lion]

1. only math goes between $$, leave the text out
2. enclose exponents in braces {}

[trail tendon]

did you mean to put the x in the exponent 💀

[scarlet robin]

yes i did

i really said Xponent

[trail tendon]

are you gonna set it equal to something

[trail tendon]

🤣

[scarlet robin]

nope

simplify is what i mean

[trail tendon]

its not really simplifiable

i mean

you can do a little bit but

actually not really

i don't think its simplifiable 💀

[scarlet robin]

only thing that could make that more annoying is having a negative exponent

[trail tendon]

it aint annoying its not possible 💀

[somber coyote]

X

[trail tendon]

i mean that just flips the top and bottom

[scarlet robin]

☝️ 🤓 AKSHUALLY ITS CALLED A RECIPROCAL

[trail tendon]

er

you can say that too ig

[scarlet robin]

the fun part of being in a math class, you know all the terms youll never need to use: Linear Quadratic Cubic Quartic Quintic Sextic Septic 8th degree, reciprocal, frenulum, monomial, binomial, trinomial, polynomial

[trail tendon]

bro edited that in there there's no way 🗿

[hallow ferry]

hey guys I just wanted to ask what is the value of infinity / infinity

or we cannot calculate it?

[hoary totem]

this isnt geometry

and infinity doesnt have a value

[cunning lion]

infinity isn't a number, so the expression doesn't make sense on its face (except sometimes in the context of limits)

[old crescent]

Can i get some advice

Anyone here?

[trail tendon]

with what

[trail tendon]

no

[old crescent]

I want to learn trigonometry but i dont know where to start

[trail tendon]

you can probably type "khan academy trigonometry" and do that if you want

[old crescent]

I've learned the basics like sin cos tan and stuff where do i go from there

[trail tendon]

what do you want to know exactly

[old crescent]

Everything I can

[trail tendon]

i mean idk what you already know

but i would look at khan academy

[old crescent]

Ok thank you

[trail tendon]

see what you know and see what you dont know, and do the stuff that you dont

[old crescent]

I will ty

[hoary totem]

find problems to solve using trig

[inner vapor]

Why not solve for x for some trig functions

For example find x in R for which sin(x) = 1/2

[old crescent]

What is R

[inner vapor]

x is a real number

[dark sparrow]

R means the real number line

[old crescent]

Oh so it means that x is a real number

[inner vapor]

From my perspective yes

[old crescent]

How do i solve this problem

[inner vapor]

Do you know arcsin 1/2?

[old crescent]

Not there yet

[dark sparrow]

have you learned about the unit circle

as it relates to trig

[old crescent]

No I have just learned what sin cos and tan are and have done some problems with them

[dark sparrow]

ok in that case you have two next steps for learning trig:

• the unit circle, incl. definitions of sin, cos and tan for angles not between 0 and 90 degrees
• trigonometric identities such as angle sum/difference, double-angle, etc.

[old crescent]

Where can i learn this

I am watching khan academy and learning about radians rn

[hoary totem]

or

try to solve some geometric problems

[dark sparrow]

ok then just keep going with KA

[old crescent]

Ok ty for ur help

[dark sparrow]

"all this shit"?

you're best off going to http://khanacademy.org/

[strong hearth]

Can this be solved without representing tanx as sinx/cosx

[grave pond]

What do you mean by "solved"?

[dark sparrow]

there's nothing to solve

but if you meant whether this can be simplified w/o writing all the tans as sin/cos, then yes.

[desert dagger]

Eh, I you want a headache then go ahead

If

If you’re going with pen and paper for this I believe it will be easier

[dark sparrow]

it is not that much of a headache actually

[desert dagger]

Well depends on the context I guess

[dark sparrow]

it is a greater headache to write them all out with sin and cos.

[desert dagger]

You got me there

[scarlet robin]

Autocorrect

I mentioned vinculum

Lmfao

NOT the same thing

[ruby wigeon]

still nobody?

[obsidian harness]

What's the question

[obsidian harness]

Centroid is ((sum of x-coords)/3 + (sum of y-coords)/3)

So A is correct

Also 'triple equidistant' means that D lies on the perpendicular bisector of AB and the perpendicular bisector of AC for example

Right, you can take EF to be the base and DB to be the height

Yeah I was thinking of using coordinate geometry

So then B = (6, 6) and the slope is 2

So by point-slope form, EB is y - 6 = (1/2)(x - 6)

Similarly, FB = y - 6 = 2(x - 6)

And also EF is y = -x

Also it's symmetrical so you have EF = 2 * ED for example

[hoary totem]

its also doable with trig

[dark sparrow]

trig is probably overcaliber for this imo

coordinate method is very appropriate here

[hoary totem]

yea

[grave pond]

We don't even need coordinates. The red triangle here is super easy to find the area of -- it's just the square minus three easy right triangles.
The larger orange triangle we're being asked for is (4/3)² times larger than that (count half-diagonals along the altitude BD).

[dark sparrow]

oh true

[hoary totem]

oh right

i may be stupid

[grave pond]

Nah, I think it takes some luck to come across this easy way. The intended solution method is probably coordinates.

[obsidian harness]

Yeah I just had so much trouble finding the scale factor of similarity

That I said screw it, let's use coordinates cause ofc linear equations are the easy way out

That's so clever, as in geometry you just need a few extra lines

[faint pasture]

can yall recommend me some practice sheet for trig with like insanely hard problems that still require nothing but trig

[hoary totem]

@lofty spear ....do i do it

[lofty spear]

your choice, Citruz.

your choice.

[hoary totem]

[hoary totem]

tbh i think

my problems arent the kind of problems he's asking for

cuz they dont really require complex trig identities and stuff

its just a lot of trig functions

not very complex once u read thru it

[wind knoll]

can ya'll help with this pls

[hoary totem]

do you know the formula for the volume of a sphere?

[wind knoll]

yeah i used the volume formula and got this long number with decimals at the end, however when we submitted it, the answer wasnt supposed to be long or some

[hoary totem]

it says round to the nearest tenth

so round to 1 decimal place

[wind knoll]

this is what i got so the answer would be all of that big number and then .9 right?

[hoary totem]

yea

[wind knoll]

yeah thats what i put first but we got it wrong for some reason

[hoary totem]

that is strange

it should be 255435213686.9

[upper karma]

anyone know any where where i can get math work sheets to pratice on

[faint pasture]

yess thats exactly the kind of problems am looking for

[hoary totem]

a fun problem, not necessarily extremely hard, but fun, is proving cos(a + b) = cos(a)cos(b) - sin(a)sin(b)

prove this identity geometrically

[faint pasture]

i have done that though 💀

[hoary totem]

oh

[faint pasture]

yea my textboook had it

it did sin(a+b) and gave cos(a+b) as an exercise

so yea

[upper karma]

i dont hate geometry anymore

[hoary totem]

K is perpendicular to the ecliptic
P_0 is perpendicular to the lunar equator

M is at lunar longitude and latitude (L - Ω, 0)
Q is at ecliptic longitude and latitude (Ω + 180deg, 0)
E is at ecliptic longitude and latitude (λ, β)
E is at lunar longitude and latitude (l, b)

find an expression for l and b in terms of the other variables

solution: https://github.com/CitruzSquared/essays/blob/main/Spherical Astronomy for Worldbuilders/Chapter 4C - Geocentrism.md

[faint pasture]

wh- is that 😭_

[hoary totem]

@lofty spear i did it

i dont think they appreciated it

[hoary totem]

its possible with only trig

check the solution

[faint pasture]

i will try later 😭

[hoary totem]

also another fun problem:

predict the path of the total solar eclipse of april 8 2024

solution:

also possible with just trig + a lil basic calculus + a lil bit of vectors

u asked for "insanely hard" so

i give you insanely hard questions

[hoary totem]

explanation and method: https://github.com/CitruzSquared/essays/blob/main/Spherical Astronomy for Worldbuilders/Chapter 9A - Solar Eclipses.md

from chapter 9A to 9F

[faint pasture]

insanely hard as in still trig funcs formula based buttt these still loook fun

ill try

[lusty dirge]

Can someone please help me with this, I dont know where to begin

[faint pasture]

can yall guide me to the correct answer, i did this using basic calculus but i have no idea how to prove it without using calculus

Well you could draw a diagonal and you get an isoceles triangle but not much else

[lime dune]

draw the unit circle

and a ray making angle theta with the positive x axis

are there any segment/arc lengths you can find that correspond to those quantities you’re comparing?

[faint pasture]

is that the only way?

isnt there some more algebraic approach?

draw the other diagonal and see what happens ig it should help

um gimme a minute ill try but im not that good at geometry so i cant make promises

sry i couldnt figure it out

[lime dune]

the geometric one is the easiest imo

[faint pasture]

well i dont really like geometry for one, and then this book i am using hasnt introduced unit circle yet so i dont wanna use it since its not the intended way

well the other diagonal thing seemed promising at first but it didnt help i dont see anything else that can be done

[wispy jewel]

This might be the dumbest question in history but... Can someone explain this to me, I know how to calculate but can you explain

Where O is? 🤦‍♂️.

[dark sparrow]

,rccw

[dark sparrow]

O is the origin

no idea why they did not just write H(-3, -4) tho

[wispy jewel]

Still lost 🤦‍♂️

So O is 0( Zero)

[dark sparrow]

O is the origin, the point (0, 0)

[grave pond]

The diagram does mark the point with an O, but somehow the bottom part of the letter has disappeared.

[lime dune]

$\Omega$ clearly

[somber coyote]

elrichardo1337

[stark crater]

I’m I tripping or is sin+cos=1 because if you root both sides of sin^2+cos^2=1??

[obsidian harness]

You are tripping bad bro

$\sqrt{a^2 + b^2} \ne a + b$

[somber coyote]

south

[obsidian harness]

Cause $(a + b)^2 = a^2 + 2ab + b^2$

[somber coyote]

south

[obsidian harness]

https://mathwithbaddrawings.com/2015/07/08/everything-is-linear-or-the-ballad-of-the-symbol-pushers/

these are all wrong

[grave pond]

Uh oh, generally a bad idea to post those false identities where someone desperate-and-in-a-hurry can pick them um.

[stark crater]

Alr I’m kinda fucked I took my math AAHL test yesterday and this shit kept me up all night

I didn’t use it but it was always lurking in my mind

[obsidian harness]

Holy shit you did AAHL this year

I'm so sorry for you; with all the leaks and whatnot

IB this year is such a shitshow cause turns out... bio got leaked too which is like whaaaaaat

[stark crater]

I think I did well tho but h the e grade boundaries 😭

[obsidian harness]

coming from an IB alumni and yes I got a 7 in AAHL

[stark crater]

All they care about is money (the ib)

[obsidian harness]

unfortunately and sadly

[obsidian harness]

I love this post and Ben Orlin but yes

[stark crater]

If aps are the hub then the ib is onlyfans (they gatekeep everything including the fucking past papers)

[stark crater]

That’s great man how were the grade boundaries then?

[obsidian harness]

I had COVID grade boundaries so they were alright

I think it was like 68 for a 7 which is wild

Cause the paper was fairly hard too

[stark crater]

Yep I studied of these papers and they had their challenges

[obsidian harness]

Yeah I mean it was not fun doing IB during COVID

[stark crater]

But I’m pretty sure I did fairly well wish me the best tho

[obsidian harness]

Yeah honestly I hope your brain glitched or something

[stark crater]

Covid was a shit show for everything related to everything

[echo beacon]

yooo thanks

[obsidian harness]

no worries

[faint pasture]

Is there a non calculus way for this?

i did the calc one but i am wondering if there is some non calc way

[faint pasture]

not for binging anime 💀

the arguement isnt the same here

[stark crater]

And no I can’t relate

[faint pasture]

xD

[stark crater]

Oh wait o didn’t read the question

[faint pasture]

xD its fine

[slender flint]

What do you think?

Does this remind you of any identity ?

[obsidian harness]

Well cos theta and sin theta are between -1 and 1
So cos^2 (cos theta) is maximised when cos theta = 0, or when theta = pi/2 let's say
sin^2 (sin theta) is maximised when sin theta = pi/2, which is impossible

So we should focus on sin^2 (sin theta): this is maximised when sin theta = 1, or theta = pi/2
What a coincidence! This is the same theta we got before!
Hence the answer must be 1 + sin^2 1

[obsidian harness]

You're tripping

[faint pasture]

not really

[faint pasture]

but like wasnt this still more like hit and trial isnt there some more concrete proof

[slender flint]

right, just realised

[obsidian harness]

The concrete proof is by differentiation ofc

[faint pasture]

xD sure that is true but like this textbook assumes i dont know how to differentiate a function so i dont wanna differentiate anything

[obsidian harness]

The question is set up in such a way that you can argue like this
That the theta that maximises this whole expression is when cos^2 (cos theta) and sin^2 (sin theta) are individually maximised

It's a very special case

Literally anything else (I assume you know what happens with A sin theta + B cos theta) wouldn't work

[faint pasture]

i seee thanksss! problems like these is what makes me like trig even a lil bit anymore-

[faint pasture]

yepp

[stark crater]

I was gonna say find when the derricitive is 0 but it is 11am and I haven’t gotten any sleep for the past two days

[clear anvil]

non-calculus way

[obsidian harness]

Yeah I just can't find a way to use the Pythagorean identity directly, and also given the answer choices

[faint pasture]

i mean sure derivative is relatively easy but i dont wanna use calc

[obsidian harness]

Generally the idea is that sin theta <= 1 so sin(sin theta) is at max sin(1)

So the max of sin(sin(sin theta))) is sin(sin(1)) etc

cos(cos theta) etc is more tricky

But relies on a similar principle, yeah I must have seen somewhere with this kind of question

[faint pasture]

i seee damn trig can be beautiful sometimes

[obsidian harness]

It's interesting how x = 0 or x = pi/2 always give a turning point for these kinds of functions

[faint pasture]

btw would the minimum of original function be cos^2(1)

the fact that the sum of maximum and minimum is 2 is also fascinating for some reason

[obsidian harness]

It would actually

Yeah both are minimised at x = 0

[faint pasture]

yeppp

[hoary totem]

@faint pasture try and prove this:
let ABC be a spherical triangle on the surface of a sphere with center O
if we denote angle BAC = A, angle ABC = B, angle ACB = C
angle BOC = a, angle AOC = a, angle AOB = c

prove that cos(c) = cos(a)cos(b) + sin(a)sin(b)cos(C), a result known as the "spherical law of cosines"

[faint pasture]

😭 what even is a spherical triangle though?

[hoary totem]

first some terminology: a great circle is a circle along a sphere with center as the center of the sphere, you can think of it as the circle you get when u cut a sphere perfectly in half

the green and blue circles are both great circles on the blue sphere

a spherical triangle is the triangle formed by 3 of these great circles

[grave pond]

A figure on the surface on a sphere made from 3 segments of great circles.

(What Citruz said).

[hoary totem]

note that even i have not done this proof lol

i just took the law as gospel

but it doesnt seem too difficult

[faint pasture]

formed as in how? like the area enclosed between the circumference of the great circle and the surface area of the sphere?

[hoary totem]

yeah like this

[faint pasture]

i seeee okayy
am i just tripping hard enough or is this just a normal triangle bended a little
ie it follows the angle sum property and the triangle inequality?

[hoary totem]

it does not follow the angle sum property!

[obsidian harness]

The whole point of this geometry is that it doesn't obey (sum of angles = 180 deg)

[hoary totem]

the sum of the angles is always greater than 180

[obsidian harness]

And for a hyperbolic surface the sum will always be less than 180, in fact

Welcome to non-Euclidean geometry

[faint pasture]

damnnnnn it looks interesting can you give me a name for what studying about this is called? or even better if you could suggest some lectures/articles

[obsidian harness]

So like the parallel postulate is unique to a flat surface
That's why they couldn't prove Euclid's 5th postulate from the other 4

[faint pasture]

icccc

[hoary totem]

this is called spherical geometry

and is used extensively in astronomy

which is where i use it most

but it has some other uses

its also used extensively in navigation etc

[faint pasture]

i can see how that will be useful damn there is always something more to know!

[hoary totem]

for example take these three points on the earth's surface:

• the north pole
• the point on the equator with longitude 0
• the point on the equator with longitude 90

these 3 points form a spherical triangle, with every single one of its angles 90 degrees

so the sum of its angles is 270 degrees

[faint pasture]

hmmmmm, i suppose there is a maximum value for the angle sum too? if i am correct it should be just less than 540

[hoary totem]

if the triangle does not take up more than half the sphere then yes 540
this happens when all three points lie on the same great circle and therefore all its angles are 180

[faint pasture]

oh like yea kind of like a major/minor segment of a basic circle

[obsidian harness]

Yeah

[faint pasture]

iseee ill read up on it

[hoary totem]

apparently the maximum possible sum of angles is 900 degrees

if u allow triangles that are more than half the sphere

[faint pasture]

damn

[obsidian harness]

It's 360 * 3 - 180

https://math.stackexchange.com/questions/75569/maximum-sum-of-angles-in-triangle-in-sphere

[faint pasture]

wouldnt the triangle basically cover the entire sphere then?

[hoary totem]

yes

[obsidian harness]

By turning the inside out apparently, yes it would

[faint pasture]

since the angle sum of the other part of the triangle is 180, soo it has to be infinitely small

wait the trig rules for these oughta be different though

[hoary totem]

man u got that faster than i did lmao

[faint pasture]

💀 like i was trying to visualize it and i couldnt do it for the life of me so i just assumed it would be a single point and the angles made sense

[hoary totem]

no every trig formula still works

cos^2 + sin^2 is still 1 for example

cuz nothing fundamental actually changes

like ur not changing the definition of sine and cosine

[obsidian harness]

https://en.wikipedia.org/wiki/Law_of_sines#The_spherical_law_of_sines

[faint pasture]

the cos = b/h thingy still works too?

[obsidian harness]

Yeah spherical law of sines still works somehow

[hoary totem]

no

[hoary totem]

yeah there is a spherical law of sines

spherical law of cosines

theres also the "four parts formula"

[obsidian harness]

https://en.wikipedia.org/wiki/Spherical_law_of_cosines

[hoary totem]

and the "analog formulae"

[obsidian harness]

But cosine is different

[hoary totem]

this is what i presented to ender to try to prove

[faint pasture]

is there some special case when it does? except for the infinitely small triangle case

[hoary totem]

cos = b/h doesnt work cuz b and h are not lengths anymore

everything is angles

side lengths are defined as the angle from the center

and also because of the bend it just doesnt work

even if u computed the arc lengths

[faint pasture]

oh soo similarity 2.0 damnnn, the expressions oughta get pretty complicated while calculating though

[hoary totem]

the classic proof involves drawing tangents to the sphere

or drawing planar triangles inside the sphere

basically somehow converting it into a normal triangle

[faint pasture]

no not like for this specific problem, but for like general problems you encounter

like for basic euclidean geo you can visualize stuf to make it easier

[hoary totem]

its just a lot of trig

[faint pasture]

but human brain is shit at visualising 3d objects

[hoary totem]

astronomy is fun

i know of this stuff just to do astronomy

[faint pasture]

the fact this trig is relatively simple yet the crazy stuff you are doing with it is just fascinating

[hoary totem]

yeah exactly the trig is not actually that bad

@lofty spear

[hoary totem]

the prediction of solar eclipses is also done almost exclusively with trig

i learned how to do that over spring break, just in time for the april 8 eclipse

[faint pasture]

trueee like i could understand most of it and today was the second day id heard of this subject

soo it was relatively doable

[hoary totem]

calculating the lunar phase, calculating the time of sunrise, etc etc

all trig

[faint pasture]

damn it looks interesting

[faint pasture]

trig looks soo simple yet sooo interesting

[hoary totem]

true

anyway

https://proofwiki.org/wiki/Spherical_Law_of_Cosines

here is the proof if u need some help

[hoary totem]

btw the proof of this starts from the law of cosines

and is just a lot of algebra

https://proofwiki.org/wiki/Spherical_Law_of_Sines

ofc its also doable geometrically

[obsidian harness]

The trouble is that my mind won't let me accept the law of cosines

Oh well

[faint pasture]

umm i tried to read it but i guess i should try something simpler first-

[hoary totem]

https://proofwiki.org/wiki/Analogue_Formula_for_Spherical_Law_of_Cosines
https://proofwiki.org/wiki/Four-Parts_Formula

here are some other useful laws in spherical trig

idk if these have analogues in euclidean trig ngl

[hoary totem]

rip

[hoary totem]

maybe yeah

[steady orchid]

is this correct?
I had a leftover quiz and I haven't done this in months so i just want to know if I remember it correctly.

[obsidian harness]

Nope not correct

It's central angle theorem, then sum of arcs that make a circle is 360

And then the last one is definition of supplementary angles

There's no nice way to say this but having reading comprehension would get you a long way

Like you know number 3 already which is good at least

[steady orchid]

i misclicked the first one, didnt even notice. but i appreciate the help

[obsidian harness]

But you can figure out 2 and 4 pretty much from this
The first one I agree isn't trivial if you haven't seen an example of it before

[grave pond]

When you find yourself writing the first half of that ... perhaps don't say anything at all?

[obsidian harness]

https://external-content.duckduckgo.com/iu/?u=https%3A%2F%2Fwww.researchgate.net%2Fpublication%2F349743977%2Ffigure%2Ffig1%2FAS%3A997346678685698%401614797386690%2FThree-different-cases-for-the-Angle-at-Centre-theorem.ppm&f=1&nofb=1&ipt=08b113c2cf774b64a697f882e3fab3e6f4a1e2ec9c1f5530f0c7b63b156ae795&ipo=images

[obsidian harness]

I disagree, I think honesty is the best policy

And I totally agree that maths education in many places is shit, and that COVID has really disrupted a lot of people's education

[steady orchid]

also last one i didnt even read at all so it really is on me 💀

[obsidian harness]

I'm not trying to assign blame (although it can be interpreted like this I know)

Of course the problems are systemic

[steady orchid]

for me i had really poor math skills my whole life. im good at other subjects but math is nogo

[faint pasture]

not to mention a lot of tchrs just do it as a job instead of actually teaching

[obsidian harness]

Yeah that's nice that you have those other subjects right

[steady orchid]

yeah but the sat is a big problem

[obsidian harness]

Aren't most colleges test-optional now though

[faint pasture]

i didnt prepare for sat specifically but i gave a few practice tests a few years ago so iirc there is no way problems like this are gonna be asked-

[steady orchid]

not the ones i want to apply to, and im an international student

[obsidian harness]

Oh right that really sucks

[faint pasture]

the problems in the sat are so much easier than this lmao

[steady orchid]

yep. and i want to apply to a stem field so that is quite an issue

[steady orchid]

not that im aware of, but i did do a couple practice sats. its just a pain in the ass

[obsidian harness]

Well the good news is that you never touch this kind of geometry after doing that one course on it

[faint pasture]

my memory might be off sure

[obsidian harness]

Like focus on your algebra and functions skills

So basically precalc material

Those are much more important for a STEM field

And of course, statistics is also super useful

[faint pasture]

and trig!

[steady orchid]

well my stem field I want to major in is biology. so hopefully not too much of maths but honestly I think you're right

[obsidian harness]

mhm

[obsidian harness]

Oh sorry wrong kind of STEM

[steady orchid]

lol i forgot

[obsidian harness]

Yeah okay then you probably won't need calculus but your uni might make you take calc 1, it's stupid ik

[steady orchid]

Im aware. its just a science

[steady orchid]

I checked and yeah probably

[obsidian harness]

Yes it's just another hurdle unfortunately

[steady orchid]

but if I do go premed, at least I don't have to do it anymore..maybe only stats

[steady orchid]

just got to barely pass until i can declare my major. after that it shouldnt happen afaik

[obsidian harness]

exactly

Yeah good luck, if you're doing premed then you'll have to bear with organic chemistry and stuff

[steady orchid]

true but its way more bearable (at least for me) than math related subjects

anyways appreciate

[obtuse field]

To follow up on what Tropo said: If there's no nice way to say something, then don't say it, especially if it's vague and doesn't contribute anything to helping them understand. Consider instead giving specific constructive feedback or asking clarifying questions about what the asker does or does not understand.

[harsh iris]

Hi could someone please help me with this problem?

[dark sparrow]

well now subtract cos(x) from both sides and factor, hm?

[harsh iris]

so (cos+x)(cos-x)=0?

[hoary totem]

mm not quite

[clear anvil]

uh

[hoary totem]

like this:
cos x cos²x = cos x
cos x cos²x - cos x = cos x - cos x = 0
cos x (cos x - 1) = 0

[dark sparrow]

bad bad!!!

cos x does not denote the product of "cos" with x

[hoary totem]

🤔

this is probably enough of a crash course in geometry

[clear anvil]

A+B+C>= 180? Dang is this non-euclidean geometry?

[hoary totem]

yes

[dull pilot]

can i prove these triangles are congruent by asa?

[hoary totem]

no but u can with AAS congruence

[hard moon]

Yes you could prove them congruent by asa, but you could just do aas

[harsh iris]

@hoary totem like this?

[hoary totem]

i mean yes you could find side YZ and side BC, with the law of sines
since angle Y = angle A and side XZ = side AC, by the law of sines, side YZ = side BC
now u have ASA

[hard moon]

Yes because the sum of the angles add to 180 so you just deduce that the other pair angle x and angle a are congruent

[hoary totem]

or that too

lol thats simpler

[hoary totem]

yea
altho depending on what they want they might want u to write 0 instead of 2 pi

[dull pilot]

okay guys thanks

[harsh iris]

is 2 pi included?

[hoary totem]

oh yeah im stupid i misread the notes

its (0, 2pi]

😭

i thought it said [0, 2pi] and got confused

[harsh iris]

thats fine, so since it closed to 2pi, do I include 2pi for my answers?

[clear anvil]

yes

[hoary totem]

yes

[harsh iris]

thanks

[clear anvil]

also cos(x) = 0 -> x = pi/2 + pi * k

what you did is right as well but you can "consolidate" those 2 into just 1 thing

if u need it for later use

¯_(ツ)_/¯

[hoary totem]

spherical trig is awesome more people should do it

[trail tendon]

whats that

[hoary totem]

u do stuff with triangles on the surface of a sphere

instead of on a flat plane

[trail tendon]

oh

that seems a bit difficult lol

[harsh iris]

i'm rewriting now but got confused again which is the right factoring

[dark sparrow]

navigation and shit

[hoary totem]

navigation and astronomy and shit

my beloved

[hoary totem]

image is a bit small

lol

[trail tendon]

🗿

[hoary totem]

nah

[trail tendon]

oh it isn't?

[harsh iris]

hows that?

[trail tendon]

small and blurry

the same size actually

[hoary totem]

yeah u just gotta kinda learn these formulas

and then u can apply it to a whole bunch of things

[trail tendon]

oh theres like spherical trig identities? 😂

[hoary totem]

yes

[hoary totem]

big image

les go

[hoary totem]

um both are wrong

lets see why

[trail tendon]

he factored out cosine of one term but not the other

and then re-factored it out

[hoary totem]

you have this line here

then if u pull one cosine out

you should have 2 cosines left in the 1st term and no cosines in the 2nd term

so the correct factoring is this

think:
how would you factor a³ - a
it becomes a (a² - 1), right?

its the same but we're dealing with cos(x) not a

but still everything is the same

[hoary totem]

theyre more fun than the normal boring euclidean formulas

[harsh iris]

okay last try

[hoary totem]

ah

cos(pi) is not 1

cos(pi) is -1

check it on your circle diagram

only cos(0) = cos(2pi) is 1

[harsh iris]

i move it to the right to get the x

[hoary totem]

oh wait

im an idiot yeah its ±1

yeah ur right

[harsh iris]

lol

[simple prairie]

School makes me hate math

[trail tendon]

fax

[simple prairie]

Like genuinely makes me want to fail on purpose

None of these stupid videos are engaging at all

10 minutes of reading just to answer one question that's unrelated to what I just read

I'm about to fail cumulatives in a couple days

I'm thinking I'll just learn the formulas

[harsh iris]

can someone help me with this please?

[hoary totem]

this is pretty easy

say y = pi x
then now we have cos(y) = 0.5

what values of y give cos(y) = 0.5?

then once we have those, we can just divide by pi to get x

[hard moon]

Hint for 5.3.3: find a pair of triangles that are similar by SAS similarity

[low shuttle]

How do i solve this using the formula OM = n/M+n OP + m/m+n OQ

[obsidian harness]

The change in x is 1 and the change in y is -2

[obsidian harness]

This only works if your point B is between A and C I think

[low shuttle]

damn

[obsidian harness]

It's better to understand rather than relying on formulas

So in this case, the changes in x and y would be 3 times bigger, 1:3

So 3 and -6

So it's just x = -1 + 3 and y = 5 - 6

[low shuttle]

i dont understnad

[obsidian harness]

https://external-content.duckduckgo.com/iu/?u=https%3A%2F%2Fi.ytimg.com%2Fvi%2FM9jfAjw2KKc%2Fmaxresdefault.jpg&f=1&nofb=1&ipt=eb6ad19389a136c8fb53b53ab449445c004c05d20707120104b5957b8b6300dd&ipo=images

btw look at how messy this is

[obsidian harness]

Which part don't you understand?

[low shuttle]

whys it -1 amd 7

[obsidian harness]

I never said that

[low shuttle]

how do i use this using vector algebra

[obsidian harness]

What I said above

[low shuttle]

this how they did it

[obsidian harness]

Fair enough

[low shuttle]

why is it 3ac

[obsidian harness]

Cause OB is OA + AB right

But then AC:AB = 1:3

So AB = 3 AC, AB is 3 times longer than AC in the same direction

[low shuttle]

well if i made OC

replacement for OM

could i ues the formula

cuz i dont udnerstnad basic ratio

[obsidian harness]

If you don't understand basic ratios, then using a formula is only going to confuse you even more

A formula is not a way to understand something

It's a way to remind you of how to do something

Figuring out what to do in a question is the hard bit, cause it requires understanding the material

[low shuttle]

https://www.youtube.com/watch?v=Cx-_PxD4DJM&t=5s

just watched this now it makes sense

[obsidian harness]

Ah okay that's good

Yeah I should have suggested a video, no worries then

[faint pasture]

Can someone help with 57? Preferably not relying too heavily on sequence and series concepts

,rccw

[faint pasture]

I have a feeling on multiplication it should result in a whole lot of ones but i have no clue have to get to that

[dark sparrow]

geometric series

[faint pasture]

oh- OH

nvm sorry im dumb

[dark sparrow]

no cheating

<@&268886789983436800> we have an attempted exam cheater here, i think

[faint pasture]

you can ask any doubts here

[faint pasture]

(looks to me like she just wanna learn for the exam)

you can always open a help channel
#❓how-to-get-help

[dark sparrow]

30 minutes is not enough time to prepare for sure

[faint pasture]

what even is included in 10th grade though

hmmm try khan acad's lectures you can prolly read the transcripts and practice a few ques in 30 mins

[upper karma]

this is quite fun

the question has a trivial mistakr

it's supposed to be A, B, C are in AP

[dark sparrow]

a typo, more specifically.

[upper karma]

yess

[faint pasture]

wait i totally missed you message im sorry

lemme try it

[upper karma]

gl

[wispy jewel]

Yo can anyone explain 3.1.2?

Ik Sec is cos but how so you calculate this?

[frosty magnet]

helo

[wispy jewel]

Explain Mainly

[lime dune]

sec is not cos

[faint pasture]

i tried something but it became an absolute mess soo yea can you give a hint

[lime dune]

be precise with what you say

[wispy jewel]

Is it not it's reciprical?

[lime dune]

it is

but "sec is cos" is blatantly incorrect