geometry-and-trigonometry

[upper karma]

my bad

i was wondering why I got -b instead of -a

ty

[upper karma]

Need help please

I got 176 but it says it’s wrong

[fossil lotus]

Which side/angle do each of the values correspond to?

(Genuinely asking because I can't tell)

[upper karma]

@fossil lotus I assume angles

And the angle they are closest to

[fossil lotus]

Ah

Okay I see it now

[upper karma]

🙂

[fossil lotus]

Is PQRS a parallelogram?

[upper karma]

Yes

@fossil lotus sry I didn’t mention that

[fossil lotus]

How did you get 176?

(176 would be almost a straight line)

[upper karma]

Ur right

Wait

But that would make sense right

If x is equal

Wait no

What did you get?

[fossil lotus]

You're looking for the measure of angle RSP, right?

[upper karma]

Yea

[fossil lotus]

From the drawing it's clearly less than 176 degrees

[upper karma]

Yea

[fossil lotus]

I got 125

[upper karma]

Ur right

How?

What did you get as the value of X

[fossil lotus]

5

[upper karma]

And u plug that into 17x +3 right?

And multiply by 2?

[fossil lotus]

I plugged it into 17x+3 and then added 37

[upper karma]

Ohhh

Ur right

I get it know

I know what the other side is

Thx

[fossil lotus]

Np, glad I could help

[silk crown]

If I have a point (h,k), and I know that $h^2 = \frac{a^2(ma+b)}{ma-b}$ and $k^2 = \frac{b^2(ma+b)}{ma-b}$, how will I write the equation for the locus of the point?

[somber coyote]

CoolShot:

[dark sparrow]

??

[silk crown]

I know how the x value of the point varies(h) and I know how the y value of the point varies(k)

How can I write a single equation which would describe how it varies in x and y

[dark sparrow]

i

ok first off

why tf did you not write x and y

instead of h and k

second, which one of a, b and m is your parameter

[silk crown]

m is my parameter, and I did say that the point was (h,k) lol

[dark sparrow]

yeah but just

hhh

your notation, god

[silk crown]

Ok but how would I solve the problem lol

<@&286206848099549185> hey could I get some help with this

[silk crown]

Welp

[silk crown]

is it that hard of a problem why is no one responding

[upper karma]

Do the diagonals of a parallelogram always bisect angles

[dark sparrow]

no

[upper karma]

Ok thx

What are all the properties of a diagonal of a polygon?

[dark sparrow]

,,,

what do you mean "all the properties"

[silk crown]

I mean there are uhh

kind of a lot

[fast spindle]

2x/(2x+5)=32
How to solve this example?

[smoky mountain]

Multiply (2x+5) to both sides

And then solve for x

[fast spindle]

@smoky mountain what do you mean?

[smoky mountain]

@fast spindle

[dark sparrow]

@fast spindle this is not geometry

but yeah do what shufflespy said

[fast spindle]

This is about geometry

Without geometry draw

[dark sparrow]

the problem as written is an algebra problem

[fast spindle]

I am asking about ratio things in geometry

[dark sparrow]

whatever geometry problem you started with, you reduced it to algebra.

[fast spindle]

Why I have to show a drawing to try it is a geometry and someone answered to help me

[dark sparrow]

...

[karmic swallow]

https://www.fiverr.com/share/Egk3Rq

[solid thorn]

Help please

help @fast spindle

[robust socket]

,, 2sin^2x=1

[somber coyote]

Umma.Gumma:

[robust socket]

hi, I would appreciate some hints to solve this

first thing that comes to my mind is

[fast spindle]

I am really bad at math lol

[robust socket]

,, sin^2x = \frac{1}{2}

[somber coyote]

Umma.Gumma:

[robust socket]

I've found a couple or roots but apparently there are 4 of them

according to a solution book I'm using

what would be the best course here?

[silent plank]

what were the solutions you got?

[robust socket]

,, \frac{\pi}{4} + 2k\pi and \frac{7}{4}\pi+2k\pi

[somber coyote]

Umma.Gumma:

[silent plank]

sin is positive in 2 quadrants
sin is negative in 2 quadrants

[smoky mountain]

sin(x) = +sqrt(2)/2 and -sqrt(2)/2

[robust socket]

ah, the fact that it is squared compels me to check both cases I guess?

[smoky mountain]

so there'll be 4 angles with those sin values

[silent plank]

well you sorta did half and half

you found the gen solution for sin(x) = 1/sqrt(2) in quadrant 1
and the gen solution for sin(x) = - 1/sqrt(2) in quadrant 4

[robust socket]

it's getting clearer

so basically if I stumble upon something like this

the way to tackle is treating it as if it were a second degree equation by all means

,, sin(x)=+-\frac{\sqrt{2}}{2}

[somber coyote]

Umma.Gumma:

[smoky mountain]

and then find all x values where that is true

[robust socket]

great, thanks a lot

[silent plank]

didn't you have the +- when you did it the first time?

since you had a solution in quadrant 4?

[robust socket]

I had a solution in quadrant 4 because I actually made a mistake

,, 2\pi-\alpha \ instead \ of \ \pi-\alpha

is what I did

also, need to know how to put bloody spaces in latex

[silent plank]

for space use \

lone \ with nothing next to it

[somber coyote]

Umma.Gumma:

[robust socket]

brilliant

thanks

[silent plank]

$(2\pi-\alpha)$ instead of $(\pi-\alpha)$

[somber coyote]

ramonov:

[robust socket]

even better

[silent plank]

using $ signs instead of ,, lets you split text up more easily

did you mix it up with where cos was positive?

[robust socket]

yes I guess I got distracted somehow

I solved other eq with squared cos and tan now

it's clear

thank you

[civic sphinx]

Number 24

[upper karma]

would it be y = -4x + 5 ?

[silent plank]

how did you arrive at that?

[upper karma]

y = mx + b

@silent plank

[silent plank]

be more specific

[silent plank]

your y intercept isn't at 5

the point you are given is (-3, 5) NOT (0,5)

[glacial bane]

he's right

it's a point, not specifically on the x or y

i would recommend drawing a graph for this one if you can't do it

it's a much simpler way and you'd understand it faster

[upper karma]

was the y = -4x + 5 the correct equation?

[silent plank]

no

[glacial bane]

no

[upper karma]

can you tell me what is

[glacial bane]

uhhh

[silent plank]

if you're using the point slope form
substitue the gradient and the point into the equation to find b

[glacial bane]

i can tell you how to do it

[silent plank]

alternatively,
you should be using the point-gradient formula

[upper karma]

i have done it before if i know the answer i can remember the way i done it before

i have never heard of that

[silent plank]

y- y_1 = m(x-x_1)

[glacial bane]

i dont think im right and i havent done this in ages

[upper karma]

what is that

[glacial bane]

but i think it's y = -4x - 5

think im wrong idk

[silent plank]

that would be the point-gradient or point-slope formula, depending on where you are from

[upper karma]

yeah thats what i was asking, is y = -4x - 5 the correct equation of the straight line

[glacial bane]

---

carticatToday at 23:45
was the y = -4x + 5 the correct equation?***

[silent plank]

no

[glacial bane]

i mean, you didnt say it

[upper karma]

jaumn there is only 1 i could be reffering to if u look at the question

[glacial bane]

i said y=-4x-5

[upper karma]

y = mx + b is the equation u need to use for it

[glacial bane]

you said y=-4x+5

[upper karma]

then im obviously talking about the equation you need to find out

[glacial bane]

you said if y=-4x+5 is correct

I said no

i said I think it's y=-4x-5

[upper karma]

why minus?

[silent plank]

no those are both wrong

[glacial bane]

i know they are both wrong

but it fits the statement

wait no

[silent plank]

if its wrong it cant fit the statement

[glacial bane]

yeah im high

it doesnt fit the statement

[silent plank]

lets go back to your intended method

y = mx + b

substitute your gradient: m = -4
and your point: (x,y) = (-3,5) into the equation

[glacial bane]

isn't it y=-4x-7?

or am I still tripping

[silent plank]

||yes||

[glacial bane]

hmmmm

ok just realised I did the calculation wrong in my head

@upper karma give me another one

I'm actually high or I'm right

@silent plank
is it
y=mx+c
y=-4x+c
y=-4(-3)+5
y=-7

haven't studied this topic forever

[silent plank]

||y=-4x-7 was rihgt||

[glacial bane]

but was the working out right?

[silent plank]

not the way you just did it

[glacial bane]

yup it's a coincidence

[upper karma]

i dont see any more on my homework

but thank u for the help

[glacial bane]

no i didnt help

i just gave you the answer with the wrong working out

so I didn't help you in anyways in the long-term

[upper karma]

i always remembered those with having 2 points instead of 1

[glacial bane]

yeah same

[silent plank]

its the same idea with two points

[upper karma]

so i am confused with the y intercept

[glacial bane]

it is, you gotta find the midpoint of the two points

and do the same thing as a 1 point

[silent plank]

not midpoint

gradient

[glacial bane]

damn ok

you're right

[silent plank]

there are 3 forms
two-point
point-gradient/slope
gradient/slope intercept

[sinful marten]

Oh boy trying understand cosine in non-right triangles at 1am

[silent plank]

if you are given a point and the gradient, it is preferable that you use the point-slope form

[glacial bane]

remember this for a gradient and a point

[silent plank]

not the ideal example there. since the point IS actually the y intercept for that one

[glacial bane]

i guess so

[vivid wraith]

you could just do the x-coordinate times the gradient and subtract it to the y-coordinate to find the y-intercept. Because you are given the gradient, you can write it in y=mx+b form

[silent plank]

you mean subtract it from the y coord?

[glacial bane]

y - y1 = m(x- x1)
y-5 =-4(x-(-3))
y-5 = -4(x+3)
y-5 = -4x-12
y = -4x-7
@upper karma

hope that helps

if you want a simpler equation:
y = m(x-x1)+y1

[vivid wraith]

yea subtract*

[glacial bane]

ngl, this helped me too so it's a win win

it's 12:14 gotta go sleep

[vivid wraith]

bye!

[upper karma]

"let P = (a,b) be a point on the ray which forms one side of A... such that P lies on the circle of radius 1
So this means a = cos(A) and b = sin(A)

P' = (a',b') is the rotation of P by π/2. Which means a' = cos(A + π/2) and b' = sin(A + π/2).

Diagramatically, you can also see that b' = a, by looking at the distance of Q from the origin. i.e. sin(A + π/2) = cos(A), which is where the first formula comes from.
"

what angle are htey referring to

the cos(A+ pi/2)

is it the angle I circled?

[proper scaffold]

can someone lead me through the theorems of solving this in a two column proof?

[upper karma]

What's the sqrt format of 0.707106781 + 0.707106781?

what do you mean sqrt format?

square root

what do you mean square root format

sqrt of i

Can you send a photo of the question?

in your book or somethin

If it's too hard, then nevermind

thanks for actually helping, but.

I literally have no idea what you're lookin for

https://discordapp.com/channels/268882317391429632/326138757474680852/647275378674499597 @upper karma

[somber coyote]

Display results online and refine query

Requested by Archsys#2547

Display results online and refine query

Requested by Archsys#2547

[upper karma]

how are the operations defined by this matrix meant to be interpreted?

i am unfamiliar with rotation matricies but need to know how to rotate points

what are the operations to rotate point (x,y,z) around the x, y, and z axes?

[dark sparrow]

multiply the column vector [x; y; z] by these from the left

[upper karma]

are they then added together?

[dark sparrow]

what

[upper karma]

i dont know what to do with this

[dark sparrow]

to rotate a vector $\mathbf{x} = \begin{bmatrix} x \ y \ z\end{bmatrix}$ around the $x$-axis by an angle of $\theta$, you compute the matrix product $R_x(\theta)\mathbf{x}$

[somber coyote]

Ann:

[upper karma]

ok listen

idk how to use matricies

i just want to know the operations required to rotate a point

ik that for the standard cartesian xy plane, there is this set of operations you can do to rotate a point

what are these operations to rotate

[dark sparrow]

okay, i can write out the matrix products explicitly if you want

[upper karma]

yes please

[dark sparrow]

R_x(θ) (x, y, z) = (x, y cos(θ) - z sin(θ), z cos(θ) + y sin(θ))
R_y(θ) (x, y, z) = (x cos(θ) + z sin(θ), y, z cos(θ) - x sin(θ))
R_z(θ) (x, y, z) = (x cos(θ) - y sin(θ), y cos(θ) + x sin(θ), z)

[upper karma]

thank you

testing this in a 3d graph rn

[upper karma]

@proper scaffold hi , it’s your tutor from yesterday , consider posting this in one of the question channels

[upper karma]

Hey, so I'm studying Geometry of Space from a book, and in one example it shows how get the coefficients and the equation of a flat determined by 3 points. Thing is you have to make a system of equations , and there are 3 equations and 4 variables, and the way the book solves it doesn't make sense. Any idea?

[quartz edge]

Hmm if you could translate it maybe that would help

I can't see why they divided everything by d. Seems like a bad idea since we don't know what d is

But you could always solve for a variable in terms of the other three and substitute it into the equations so you only have three variables

@upper karma

[upper karma]

Could you elaborate on that last part?

I'll translate it

Find te equation of the plane determined by the points A, B and C

Every plane is determined by 3 non-aligned points. Since A, B and C belong to the plane, their coordinates have to verify the equation of said plane: ax + by + cz + d = 0

We can reduce the system of 4 variables to one equivalent to 3 variables, by dividing each equation for d =/= 0. In case in which d = 0 we can divide the equations with any other coefficient.

Considering a'= a/d', b'= b/d' and c'= c/d', the previous system is (the system with the 1s), and the solution is a'= 3. b'= 2 and c'= 1

The equation of the plane is 3x + 2y + z + 1 = 0

That's what it says

[vague berry]

Ans: 60/13 what’s the most efficient setup of the problem

Please ping me I love math

[upper karma]

@vague berry okay, so

It’s saying that DE is perpendicular to AC

Which implies there’s a 90 degree angle

[vague berry]

Yes

I’m here @upper karma

[vague berry]

Hello

[robust socket]

$ sin(x) \ = \ sin(2x) $

[somber coyote]

Umma.Gumma:

[robust socket]

hi, I'm not sure how to solve this, would appreciate any hints

[weary drift]

$\sin(2x)\equiv 2\sin(x)\cos(x)$

[somber coyote]

RokettoJanpu:

[robust socket]

yeah I'd like to solve this without recurring to the duplication formulas

the hint the book is giving me is to use associate and/or complementary angles

no success so far tho

[weary drift]

imagine not allowing yourself to use trig identities to solve trig eqns

glhf

[umbral snow]

Indeed, if you can solve this, I think you'll derive that trig formula

No, you won't nvm

[robust socket]

ok good to know :)

ty

[cloud merlin]

https://cdn.discordapp.com/attachments/326138757474680852/647560372643037216/221702448304226304.png

idk how sin of x can be sin of 2x

but like i can help you derive the formula for like sin of (a plusb)

unless it's like 0

[upper karma]

well that's a solution

[gritty sail]

just think about a unit circle

and when sine is equal

so 0 is a solutio yes and all 2pi multiples

also you know that sin(180-x)=sin(x)

so

180-x=2x ==> x=60

[cloud merlin]

Oh so here’s the proof

[gritty sail]

If you have a right triangle

And you draw a line from the right angle to the hypotenuse so that the line cuts the hypotenuse into 2 parts

If the line is the same length as one of the parts

Is it the same as the other one as well?

[smoky mountain]

it doesn't mean it's equal to both parts

it's only equal to both parts of the hypotenuse if the triangle is a 45-45-90 triangle

[gritty sail]

That's not true try a 3-4-5 triangle

[smoky mountain]

well

by "draw a line from the right angle to the hypotenuse so that the line cuts the hypotenuse into 2 parts"

it can only be equal to both if it splits the line splits the hypotenuse in half

[gritty sail]

Nvm

When it spits the hypotenuse in half the line that splits it is also the same length

Easy to prove

[smoky mountain]

which is only true if, for a right triangle, it's a 45-45-90 triangle

[gritty sail]

Nope

The line doesn't have to create a right angle

Buddy I literally j proved it it's rly ez to prove

I think u misunderstood my q it's all good m8

[smoky mountain]

you sure?

[silk crown]

@gritty sail do you know what a centroid is

[smoky mountain]

@silk crown it's essentially the center of mass, if you were to think of an object in 3d

[silk crown]

no I'm just asking him

[smoky mountain]

o

thought you were pinging for help 😄

[silk crown]

If he "proved" that, let him try to figure out the centroid of the triangle he proved it on

[upper karma]

https://cdn.discordapp.com/attachments/326138757474680852/647260732555264048/downloadfile.png

Hi, I'm assisting a student in Geometry but it's been years since I've had to work with two-column proofs. Can anyone assist us in this problem? It should be a trivial one. My guess is that it should involve demonstrating that AB and DC are proportional, that angle ACB is equal to angle E, and that angle A is equal to angle D, or something along those lines?

@proper scaffold

[cloud merlin]

Look at the lines on the left side and the hypotenuse

They are parallel

If those two lines are proportional to the triangle then the third always has to be

[upper karma]

does gimbal lock occur with rotation matricies?

[proper scaffold]

hello @upper karma

[upper karma]

Can someone please help me with that?

[upper karma]

@ helpers

Um

[random glacier]

<@&286206848099549185>

[wooden compass]

I would like resources to help me study and learn geometry in highschool

[sudden locust]

This is, in my opinion, the most intuitive way of proving the pythagorean theorem! https://youtu.be/BmUjXSP5-3o What your favourite proof? 🤓

[sour totem]

@wooden compass Kahn all the way

[vagrant steeple]

can anybody tell me how to get theta?
i tried doing arctan(b/a) here

[vivid wraith]

for c> the range for ceta is between 0 and 2pi, your anwer is negative

[vagrant steeple]

do i just take the negative off or add 360

bc coterminal angles right?

[vivid wraith]

add 260

but its asking for radians

360*

[vagrant steeple]

so then convert it to pi radians after adding 360

[silent plank]

you angle will depend on the quadrant

[vivid wraith]

^

[silent plank]

which quadrant is -5 + i in?

[vagrant steeple]

i've no idea

[silent plank]

arctan ( 1/5) will give you the reference angle
and then do the proper angle conversion

plot it on the complex plane if you don't know

[vivid wraith]

if j was i

idk why it says j instead of i

[silent plank]

same for b), if you didn't figure that one out yet

[vagrant steeple]

so for the graph a=x and b=y

and the hypotenuse, r is sqrt26

[silent plank]

in the complex plane they are the
Real and Im axes instead of x and y

[upper karma]

what is the correct order in which to rotate a 3d point?

is it yaw, pitch, roll?

i think so

each one rotate around a different axis

yaw is z, pitch is y, roll is x

correct

https://www.desmos.com/calculator/aqelcrwqfz

i discovered a 2d rotation formula a while ago

finally got it to work for 3 dimensions

interesting.

goddamnit

it suffers from gimbal lock

[upper karma]

How can gimbal lock be solved

[oblique tiger]

nvm im dumb

[mighty narwhal]

Use quarternions 🤡

[robust socket]

,, 2cos(2x)-\sqrt{2}=0

[somber coyote]

Umma.Gumma:

[robust socket]

I've got a doubt on this one

the two solutions I am getting are:

,, \frac{\pi}{8}+k\pi \ and \ \frac{7}{8}\pi+k\pi

[somber coyote]

Umma.Gumma:

[robust socket]

in particular, I obtained the second one recurring to

,, 2\pi - \alpha + 2k\pi

[somber coyote]

Umma.Gumma:

[robust socket]

where alpha is the angle of interest

the solution of this one should be + and - 25°+k180°

only the first one seems to be correct then, what am I doing wrong here?

[mighty narwhal]

What is 2π-π/8?

[robust socket]

yeah I know, but if I try to convert 7/8 to degrees I get 157.5°

[mighty narwhal]

No

Just answer my question

What is 2π-π/8

[robust socket]

15/8pi

[mighty narwhal]

15π/8+2kπ is the second answer

If you substract -2π you will get the -25°

[robust socket]

I'm not following how do you get to 2pi-pi/8

[mighty narwhal]

Its what you wrote

$2\pi-\alpha$

[somber coyote]

DarK:

[robust socket]

alpha is pi/4 actually

before diving by 2 at least

[mighty narwhal]

2alpha is pi/4

Alpha alone is pi/8

Alternatively
Cos(x) is even

So if Alpha is a solution, -alpha will also be a solution

So you got the first solution a.k.a π/8
So -π/8 is the other one

Now add +2kπ and you got all solutions

[robust socket]

yeah but I really wanted to get through that 2pi-alpha thing

[mighty narwhal]

Well it works that way too

2π-π/8+2kπ =-π/8+2nπ

I can just "remove" that 2π because it has a period of 2π

$2\pi-\alpha+2k\pi=2\pi-\alpha+2(n-1)\pi=\cancel{2\pi}-\alpha+2n\pi\cancel{-2\pi}$

[somber coyote]

DarK:

[mighty narwhal]

I just shifted the k to be 1 less

[robust socket]

do you mind if I go step by step, that part only?

for the sake of clarity, we start from:

,, 2cos(2x)-\sqrt{2}=0

[somber coyote]

Umma.Gumma:

[mighty narwhal]

You can put $ around it to latex inline

[robust socket]

I get to

,, 2x =2\pi-\frac{\pi}{4}+2k\pi

[somber coyote]

Umma.Gumma:

[robust socket]

then

what

[mighty narwhal]

Wait nvm
Its good

Sry

[robust socket]

no worries, then I'd do:

,, 2x = \frac{7}{4}+2k\pi

[somber coyote]

Umma.Gumma:

[robust socket]

which is

,, x=\frac{7}{8}\pi+k\pi

[mighty narwhal]

You forgot the π after 7

[robust socket]

right

[somber coyote]

Umma.Gumma:

[robust socket]

now I would expect this to be -22,5°+k180°

but it's not

therefore something is clearly wrong I guess

[mighty narwhal]

,w 7π/8 to degree

[somber coyote]

Display results online and refine query

Requested by DarK#2404

[robust socket]

exactly

how is this possible

[mighty narwhal]

,w 157.5-180

[somber coyote]

Display results online and refine query

Requested by DarK#2404

[mighty narwhal]

There

[robust socket]

can I say fok me?

[mighty narwhal]

Yea

But also, I dont recommend that 2pi-alpha stuff
It gets the period shifted which will result in this

Just do -alpha

π/8+kπ is one solution
So -π/8+kπ is the other

[robust socket]

yes, well thanks very enlightening

still, I'm not completely grasping the 157.5-180 thing

what exactly happens there

in terms of period

[mighty narwhal]

Its the same as your method, but yours shifts the period
So you have to shift back if you need is as a negative number

Well
The period is 180°

[robust socket]

yep

[mighty narwhal]

So you can just substract 180°

[robust socket]

as simple as that

[mighty narwhal]

Yes

Proof:

$\pi-x+k\pi=\pi-x+(n-1)\pi=\cancel{\pi}-x+n\pi-\cancel{\pi}$

[somber coyote]

DarK:

[mighty narwhal]

K is any integer, and n-1 is still any integer
So I can just shift the period like that^

[robust socket]

didn't know this proof interesting

well thanks a lot, much appreciated :)

[mighty narwhal]

Your welcome

[silk crown]

Can i use integration to get the area of the ellipse x²/a²+y²/b²=1?

[gritty siren]

Sure, there's no problem with that

[fringe dirge]

Well yes okay

[unkempt jay]

are you asking how to do so?

[gritty siren]

I think $\int_0^{|a|}\sqrt{b^2-\frac{b^2}{a^2}x^2}\d x$ is very doable

[fringe dirge]

You're right I'm trolling

Elliptic integrals are for the perimeters of ellipses

[somber coyote]

Tuong:

[silk crown]

basically I'll do y=something and int y dx right

[gritty siren]

on the upper half of your ellipse, you have
y=sqrt(b²-b²x²/a²)

[silk crown]

Alright thanks I'll try it out

[gritty siren]

that integral above is 1 quarter of the area

[silk crown]

If I'm not wrong, if I change the lower limit to -a I should get one half of the area, right?

[gritty siren]

Yes

[silk crown]

I can basically use this approach for any curve right

[gritty siren]

it's a bit more complicated

you can't integrate everything

[silk crown]

Hmm why is that so?

[fringe dirge]

I mean, sometimes you just get things you can't integrate

Try finding an antiderivative for e^{-x^2}

[dark sparrow]

[silk crown]

Any second-degree conic can be integrated though right

[gritty siren]

You're limited by several things, for example if you exclusively work with Riemann integrals, you can only hope to integrate Riemann-integrable functions

then you can allow more stuff by introducing improper Riemann integrals...

and then there's still no guarantee of finding a nice formula

[silk crown]

I'm a lowly highschooler I won't be using those anytime soon lol

[gritty siren]

haha x')

[silk crown]

I used calculus to prove the area of a circle wow

I'm sure there are probably uhh

Simpler methods but

calculus is lit

[rare isle]

I don't know why I am confused over this one but how do I write the general solution

For cos2x= -cosx

What do I do with the negative sign??

[silent plank]

consider doing an angle shift

alternatively you can also write cos(2x) in terms of cos(x) and solve a quadratic

[rare isle]

I mean I can, but I wanna do it by general solutions

[silent plank]

i mean the general solution just involves adding 2kpi

what happens when you shift cos by pi?

[rare isle]

-cosx

oh gosh, thank you so much.

I wonder, if i was lost lol

[silent plank]

you should get 2 sets of solutions for this

[robust socket]

,, sin\frac{x}{3}=0

[somber coyote]

Umma.Gumma:

[robust socket]

I'm having quite a hard time on this trivial one

in particular I don't get how the result is k540°

I get k1080°

[gritty siren]

if you look at the circle, you can see that sin(t)=0 of and only of t is in πZ

so sin(x/3)=0 if and only if x/3 is in πZ

i.e. if and only if x is in 3πZ

and 3π is 540°

[somber coyote]

Umma.Gumma:

Umma.Gumma:

Umma.Gumma:

Umma.Gumma:

[robust socket]

ty

[silk crown]

well, is there a way you can write arccos(1/4) in terms of arctan?

[fossil lotus]

arccos(1/4)-pi can be simplified

[silk crown]

You're not gonna get the answer without trying it, mate

[fossil lotus]

Think about how cos(x-pi)=-cos(x)

[silk crown]

Try harder then

[fossil lotus]

Uh not quite

It’s not cos(x-pi)=cos(x)

Wait

Maybe I was thinking about this the wrong way

Ah

I was thinking you should simplify arccos(1/4)-pi to arccos(-1/4), but that would give you the same result

[silk crown]

@fast swift try drawing it

I mean

A right angled triangle

[fossil lotus]

Yeah, think about the unit circle

[silk crown]

||the Pythagorean theorem will help||

[fossil lotus]

Do you understand how sin, cos, and tan correspond to the unit circle?

And what arccos means?

[silk crown]

If an angle is arccos(1/4), what is the cos of that angle?

indeed

and what is cos?

well

in a right angled triangle

if you take an angle, what do you define the cosine of that angle as?

You know it's a ratio of some sides, right?

Which two sides is it a ratio of

I mean what is the very first definition of cosine you learnt?

indeed

Now draw a triangle with adj/hyp = 1/4

And figure out its other side somehow

mmhm

np

it's pretty common to switch the function using which you write the angle, it's useful

np np

[cold tiger]

Heyooo

Ok so basically, I have to verify trig identities but idk what to do. Like I have the trig cheat sheet pdf, but idk what identities to use and when to use them.

[silk crown]

what do you mean

[cold tiger]

So you know how in verifying trig identities, you gotta use multiple to make one side the same as the other side?

Idk how to do that. Like I’m never sure if I use them correctly or not.

[silk crown]

What multiple

I mean you basically just practice lol, you will see general trends in how you solve the problems and how to approach one

Having a formula sheet but not knowing how to use it won't help

[upper karma]

How do you call this in english?

its supposed to be cube sliced using 3 points on edges

[dark sparrow]

cross-section

[upper karma]

what should i type in google in order to find instructions on how to construct the shape?

is this at least the correct channel for this question?

[dark sparrow]

yes it is

[robust socket]

got a question on this equation (lower half of the picture)

the second result I get should be, according to the solutions, -11.25°+k90°

which happens to be the 22.5°-33.75°

I'm not quite understanding what is going on here, I am pretty sure I did everything good

[celest crag]

Hi all this is my first question here. I'm trying to work through Moise/Downs Geometry and the absolute first question has me sort of stumped. It goes like this "Imagine a suspension bridge 400 ft long. on a hot day the bridge surface expnds lengthwise by two feet. The bridge surface remains rigid, except that it rises at the exact middle so that cars have to go up and over a hump. Each half is now 2001 feet long. How far above its normal position does the bridge rise?" It says the problem can be solved with the pythagorean theorem -- but I'm confused by that because we only are given the length of the bottom short side and not the other short side or the hypotenuse? Thanks very much for helping!

[cerulean parcel]

how am i supposed to solve 3A and 3B?

[supple badger]

What do you know about the two triangles in 3b?

Looking at just the angles

[zinc crow]

your basically given the answers for 3a

[vivid wraith]

try HG instead as the numerator for the second fraction

[formal raptor]

Trigonometry help?

[fringe crater]

What have you tried so far?

Think about the amplitude, vertical shift, and period of the graphed function. Each of these are given to you within that graph

Amplitude is the difference between your maximum and minimum, divided by 2. Period is how long it takes you to go from one minimum to the next minimum, and equals 2pik, where you have sin (kx). Hope that helps

[hollow harness]

Amplitude is the difference between your maximum and minimum
i wouldnt exactly phrase it that way, as by that definition the amplitudes of sin and cos are both 2

[fringe crater]

Ah whoops, I meant to add in a divide by 2. So you want to take the difference and then divide by 2. Good catch!

[hollow harness]

👍

although i personally prefer defining it as the vertical stretch of the original function kek

[fringe crater]

That's a good definition too

[hollow harness]

i would say Angle Angle Side

um sure

[icy acorn]

Guys help i need help with this HW

"Proof that the triangle created by the tangent to a hyperbole and the asymptote has a constant area" theres no data given on this one just that, i know it has something to do with getting an area with a determinant

[cloud merlin]

I’m pretty sure you do 15 over 20 is equal to 2x - 2 over x +9

Then solve for x

[split otter]

"Without the use of a calculator, find the exact value of the expression: csc 37° sec 53° - tan 53° cot 37°." Not sure where to start. May I get a tiny hint?

Ah wait, I think I got it. I'm dumb. I can use the cofunction identities to get sec² 53° - tan² 53°, which according to the Pythagorean identity, is 1, right?

[dark sparrow]

yes

[wild hamlet]

Hm

[proper scaffold]

@mossy vine can you help me with this problem? https://cdn.discordapp.com/attachments/326138757474680852/647260732555264048/downloadfile.png

[mossy vine]

no I'm busy sorry

[proper scaffold]

oh it's ok

@vale inlet could you assist me in solving that problem?

[silent plank]

what have you tried?

[proper scaffold]

yes but I have to make sure it correct

[silent plank]

what did you do?

[proper scaffold]

i did

1. Angle E = Angle ACB
2. Because AB||DC, Angle ABC = Angle DCE
   Both triangles are therefore similar

or I need something to reword it because my teacher needs a two column proof with theorems and postulates

[silent plank]

be more specific about the rules you are applying

[proper scaffold]

if I can't be more then i may not know how to do it

[silent plank]

e.g for 1
the statement is, angE = angACB
the reason is "given"

for 2,
the statement is
Angle ABC = Angle DCE
however the reason is insufficient. just "parallel lines" by itself isn't enough justification to state that 2 angles are equal

you need to state the exact property related to parallel lines

[proper scaffold]

im not sure

let me check another example

[silent plank]

do you know the terms
alternate, vertical, corresponding, co-interior, complementary, supplementary

[proper scaffold]

yes

[silent plank]

and which one would angABC and angDCE be?

[proper scaffold]

corresponding

[silent plank]

yes. and that is exactly what you would need to state

[proper scaffold]

oh ok i see

[silent plank]

they are equal because they are corresponding angles on parallel lines

and for 3.
you also need to be more specific about the reasoning

i.e. and how you came to that conclusion

[proper scaffold]

ok i see

thanks

[robust socket]

I am expected to get -11,25° as second result here, instead I get 78.75°

can someone explain me how I should change approach?

78.75° is not completely wrong: 78.75°-90° actually is -11.25°

<@&286206848099549185>

[gritty sail]

Not understanding what u did in ur second one

[robust socket]

,, f(x)=[2\pi-g(x)]+2k\pi

[somber coyote]

Umma.Gumma:

[robust socket]

that's what I did there

[gritty sail]

k is any integer so it should be the same thing

[robust socket]

thing is, I sistematically get this issue in equations including cos

so I wonder if I'm doing anything 'out of the book'

can anyone confirm this is correct/wrong please?

[civic sphinx]

Please explain number 3

[civic sphinx]

<@&286206848099549185>

[vivid wraith]

im not sure if this is right but you could call one side of the triangle x and find the other side in terms of x, you could then use triangle inequality to find the minimum value of the side, giving the maximum altitude.

[dire obsidian]

hi

apologies if im interrupting something

Im doing trig rn in alg.2

and these are what u would call quadrantal angles correct?

and say if I was told to find all six trignometric functions for the angle: 330º

i would use the fourth quadrant?

[vivid wraith]

yes

at point (1,0) it is 0 degrees and 360 degrees

[dire obsidian]

ah ok

[vivid wraith]

whatever angle they ask for, use the quadrant where the angle is in

dang my grammar sucks

[dire obsidian]

would i draw a triangle?

right triangle

30-60-90

since 360-330=30

[vivid wraith]

yes that would be the fastest way

[dire obsidian]

ah i see

[vivid wraith]

or just get a calc :))

[dire obsidian]

haha i have to do this without a calculator

):

[vivid wraith]

unlucky

well good luck!

hope i helped

[dire obsidian]

ty!

idk why

but i just get confused as to which is the θ angle

cause ik i did some problems earlier

and 60 was the θ angle at some point

[vivid wraith]

just go counter clockwise and you should be fine

[dire obsidian]

like if it was in quadrant 3

[vivid wraith]

if it is in quadrant 3, add 180 degrees

[dire obsidian]

your y value would have to be xroot3

[vivid wraith]

yep thats right

[dire obsidian]

it was this problem. i figured it out

but i was just a bit confused

so 60 would be the reference angle correct?

and so does that mean the reference angle is considered the θ angle?

which helps me find which value is y,x,r

oof

i mightve just drawn the triangle wrong

or is it like that

[wild hamlet]

You should draw the circle that indicates the start and end lines that you take the angle through

For better clarification

Your 60 and 30 are switched anyway

[dire obsidian]

for which

which image

is right

its the first one?

[wild hamlet]

Both

They are both wrong

I can't accurately know where you mean 30 and 60 to be

You need to draw the angles and not just the numbers

[dire obsidian]

@wild hamlet something like that?

lol sorry if its wrong

im not the best when it comes to bringing out the things i envision in my head onto paper or a drawing

[vivid wraith]

@dire obsidian what angle are you trying to draw

from your picture it seems like you are drawing 240, but your angles make it 330

[dire obsidian]

OOOOF

im trying to draw

240

[silent plank]

that looks alright
try make that 240 look less like 270

[unkempt stag]

Hey guys I have this one problem and Its really confusing, how do I find the permitted and area of a trapezoid with Numbers that are square roots

[scenic geyser]

can you be more specific?

the permitted?

and what has square roots

[split otter]

I'm guessing they meant perimeter?

[unkempt stag]

Perimeter sorry

Stupid autocorrect

[split otter]

Can you give an example?

[unkempt stag]

So it's asking me to find the permiter and area of a trapezoid (it has missing sides but I've gotten them) and I'm trying to find the perimeter but It's difficult bc the height is a radical square root

[split otter]

You can still do it. Can you show the working thus far?

[unkempt stag]

Want me to send a pic?

[split otter]

Sure.

[unkempt stag]

That's what I mean lol

[split otter]

So your perimeter would be P = 8 + 11 + 13 + 4√6, right?

[unkempt stag]

Yes but how would I add in the 4root 6

[split otter]

You just leave it as P = 32 + 4√6?

[unkempt stag]

Oh ok thank you

[split otter]

Did your teacher specify to obtain an approximate decimal? If not, that's fine.

[unkempt stag]

No, he told us to use that type of radical thingy lol it's confusing too I'm not even sure if that's the right height tho too

[split otter]

Yeah. The height is correct.

Then the area A = 1/2 × 4√6 × (13 + 8).

[unkempt stag]

Alright dude thank you so much I've been struggling with this for like a half hour

[split otter]

No worries.

[unkempt stag]

Hey guys i have another question

How would I solve a problem like this one?

[silent plank]

do you notice anything special about the triangles?

[unkempt stag]

They have 90 angles?

They share the same height?

[silent plank]

do some angle chasing

[unkempt stag]

Angle chasing?

[silent plank]

which angle would angCAD equal to?

[unkempt stag]

DCB? They look somewhat similar

[silent plank]

is that a guess?

[unkempt stag]

No lol am I wrong?

[silent plank]

they are indeed equal

your additional comment indicated a bit of uncertainty

[unkempt stag]

Oh ok

[silent plank]

now your triangles both have a right angle and another equal angle
the remaining angle would also be equal
what does that tell you about them?

[unkempt stag]

Oh ok

I think I get it

[plucky jacinth]

i need some help with this. Solve for x and y given triangle J K L approximately equal to triangle M N P, JK=12, LJ=5, PM=2x-3, m angle L equals 67, m angle K equals y plus 4 and m angle N equals 2 y minus 15.

<@&286206848099549185>

[silent plank]

!15m

[dark sparrow]

"approximately equal"

[silent plank]

anyway, have you drawn a diagram yet?

[plucky jacinth]

ya

[silent plank]

where are you stuck?
were you able to find K?

[plucky jacinth]

hold on

how to find y

[silent plank]

wdym? what's your K?

[cobalt path]

Hi I am a little confused on SAS for triangles

[silent plank]

what trig rule do you think is appropriate here?

[plucky jacinth]

x=4

i just need to find what y=

[silent plank]

uh what? how are you getting x = 4?

[silent plank]

the triangle JKL is unique,
and from that the value of y is fixed

[plucky jacinth]

i got what it is now

[silent plank]

y is part of angle K.
and you will need to determine that first

[plucky jacinth]

i figured it out

y=19 x=4

[cobalt path]

Hi can someone explain the triangle SAS rule

I am Extremely confused about it

[silent plank]

the inclusion of the JK=12 and 67° makes the question a little inaccurate

what's your issue with the SAS rule?

[dark sparrow]

no you didn't

sin(β) cannot be sqrt(3) no matter how much you wish it was

sqrt(3) is greater than 1

sin(β) is always between -1 and 1 for all β

sqrt(3) is greater than 1

sin(β) is always between -1 and 1 for all β

well you know what sin(β)/cos(β) is

and you also should know that sin^2(β) + cos^2(β) = 1

and since you know what quadrant β is in, you also know the signs of cos(β) and sin(β)

do you not see that sin^2(β) + cos^2(β) = 1 is true always?

regardless of what else you might know about β?

you can write sin^2(β) as tan^2(β)cos^2(β)

then solve for cos(β)

sin(β) = (sin(β)/cos(β))*cos(β)

sin(β) = tan(β)cos(β)

does this need explaining

tan^2(β)cos^2(β) + cos^2(β) = 1

4cos^2(β) = 1

cos^2(β) = 1/4

all that work for such a small part of the problem
i'm trying to get you both to stop overthinking and to not be so clueless when it comes to basic trigonometric facts

was that a sarcastic remark

i was being 100% serious

[dire rampart]

[dark sparrow]

...

@signal hornet please don't delete your half of our dialogue.

[queen python]

[dark sparrow]

they probably want you to leave it at that

maybe write it as -arctan(1/2) instead of arctan(-1/2)

but that's it

[heady eagle]

Hey, can someone help me with math? :3

[dark sparrow]

don't ask to ask...

just ask

i.e. post your question

and someone should come and help you

[heady eagle]

ooo okay thanks 😄

[dark sparrow]

so what did you need help with

[heady eagle]

This 😁

[dark sparrow]

uh

okay i'll let someone else take over. i'm too tired for this.

[heady eagle]

haha it´s all good:3

im tired too :0

[gritty sail]

is there really no information other than a picture?

[shell pendant]

How do I calculate the sides of triangle if I have all the degrees of it and the area?

[median crown]

Well

You’ll probably have to pull off

Some sort of

idk proportion

of sorts

But the method I thought of uses some tougher algebra

let me see if I can pull it off

[upper karma]

$\cos{a}=b^2+c^2-2bc$
$\text{Area}=\frac{1}{2}bc\sin{a}$

[somber coyote]

C-O:

C-O:

[upper karma]

And say area is 4 and θ is 6

(b-c)^2=sin(6)

8/bc=cos(6)

Then one can work out b,c

[median crown]

Yeah that works

Mine is probably the same amount of work

mine just set a coefficient of proportion x, and then 3 sides that form a triangle simile to the one you haven

And then you just solve for x, the coefficient of proportion using Heron’s formula

And then solve for the sides

[signal hornet]

what is the next step?

I re-wrote the left side, that is all I got

[dark sparrow]

are you asked to prove $\sin(\theta + k\pi) = (-1)^k \sin(\theta)$ for $k \in \bZ$?

[somber coyote]

Ann:

[signal hornet]

yes

I tried dividing both sides by sin(theta) but that was no good

[dark sparrow]

uh

i see you wrote $\sin(\theta)\cos(k\pi) + \cos(\theta)\sin(k\pi) = (-1)^k \sin(\theta)$

[somber coyote]

Ann:

[dark sparrow]

i don't understand what you're confused about

[signal hornet]

either side needs to look like the other

[dark sparrow]

$\sin(k\pi) = 0$ and $\cos(k\pi) = (-1)^k$...

[somber coyote]

Ann:

[signal hornet]

ahh

I see

$\cos(k\pi) = (-1)^k$

[somber coyote]

Jason_Bjorn:

[signal hornet]

That is what I got now, I see how this is always true

but I wonder if there is more work to do

so they look the same

[dark sparrow]

do you accept that $\cos(k\pi) = (-1)^k$ for all $k \in \bZ$

[somber coyote]

Ann:

[signal hornet]

yes

[dark sparrow]

do you accept that $\sin(k\pi) = 0$ for all $k \in \bZ$

[somber coyote]

Ann:

[signal hornet]

yes

[dark sparrow]

then no there is no more work to be done

[signal hornet]

thanks @dark sparrow

I don't understand how the author got these domains

cos is only postive on the right side

and sin is only positive on the top half

did the author flip the domains?

[silent plank]

those are the range of arccos and arcsin

[signal hornet]

Do you mean range of arccos and arccsin? @silent plank

[silent plank]

whoops yeh

[signal hornet]

ok

I will use that

so cos = x/r

and sin is y/r

why do both equations below say that 1 is the hypotenuse (r) ?

[silent plank]

they're applying
sin^2(x) + cos^2(x) = 1

[signal hornet]

I don't see how

but ok

I will just use the triangle method

[silent plank]

its pretty much the same thing

you recognise that sin^2(x) + cos^2(x) = 1 is an identity right?

[signal hornet]

yes

[silent plank]

and rearranging the equation
sin^2(x) = 1 - cos^2(x)

[signal hornet]

yes

[silent plank]

and then take the square root

[signal hornet]

I follow

why can't this textbook show all the steps

[silent plank]

in the question, everything is positive, we ignore the negative solution

we'll application of that identity should be simple enough that it shouldn't require an explanation

[signal hornet]

I'll try to use it on my problems

[signal hornet]

...

gonna stick with the triangle method

why the different brackets?

I know what they mean with domain and range

but this isn't like that

why not just use bigger round parentheses?

[upper karma]

Can gimbal lock be solved like this?

Say you want to rotate around Z by 45 degrees.

You take the gimbal, and rotate it 45 degrees, the plane included.

Once that rotation is completed, you remove the plane in its current orientation, reset the gimbal, and then re-insert the plane now rotated by 45 degrees relative to the gimbal to the center.

You then perform the other rotations like this, avoiding the stacking of rotations which leads to gimbal lock.

Is this a solution to it?

Can someone help me out, i get 1.37m squared but textbook says 1.08m squared

@proven belfry

i just saw that ur a helper and i already used my helper

[silk crown]

Don't tag people

How did you get your answer

show your working

[upper karma]

NVM

coolshot

i found the answer

i am sorry mono, i used the helper tag already and the helper said he is going to bed

Someone pls answer my question

Does performing the rotations individually affect the characteristics of the rotations?

[dark sparrow]

,,,,

what

[silk crown]

what

[upper karma]

Bruh

I explained it simply

[silk crown]

bruh not really

[upper karma]

Perform each rotation individually, as opposed to subsequently

Thus avoiding gimbal lock

Rotations wouldn’t be stacked

There’s no significant defined order because only one rotation is ever performed at a time

Would this affect the outcome of the rotations

I don’t think that it would but testing it out would take too long on desmos

[tardy junco]

Where have I gone wrong? What's the reason?

1 - sinx is definitely not always zero.

[dark sparrow]

$\sqrt{t^2}$ is in general not always equal to $t$

[somber coyote]

Ann:

[tardy junco]

I had to integrate root(1-sinx). They hadn't given the domain. If I did it by completing the square then I would get root t squared.

So should I not complete the square for such problems?

The answer didn't have mod or anything

@dark sparrow

[dark sparrow]

uh

what answer did they give you

[tardy junco]

I forgot

I don't have the book with me right now

I might be able to figure it out. Gimme one min.

2*root(1 - sinx) I believe.
They multiples and divided by root(1-sinx)

The numerator root cos squared x they took as cosx (so is doing this wrong?)

@dark sparrow

[dark sparrow]

uh

[tardy junco]

Why is math so confusing 😭

Should I.. Assume that x>=0 for whatever reason?

Pls @mention me if you reply to me

.
But how would you convert sin(x/2) - cos(x/2) to root(1-sinx).
It could be either root(1 - sinx) or - root(1-sinx).

What to do?

Is it + in one interval and - in another interval like modulus?

[sudden locust]

You have probably heard about the "Pythagorean Theorem, but did you know it applies for semicircles aswell?"
https://youtu.be/Ir3MUW_VkgQ

[tardy junco]

Somebody pls explain. 😭

[quiet mason]

square root always means the principal square root

or |√x|

[silk crown]

√(x²)=|x|

[at least for real numbers afaik]

[tardy junco]

@quiet mason umgh yeah that's the square root I was using..?

If had used -root() it would have come up to x = x.

Thanks coolshot

@silk crown so what would sin(x/2) - cos(x/2) be equal to?

[silk crown]

...it would be equal to sin(x/2)-cos(x/2)

what do you need exactly

[hollow harness]

Half angle identities

Apparently

[minor coral]

Hello I need some help with half-angle identities

[silk crown]

cos(a-b)=cos(a)cos(b)+sin(a)sin(b) how would I prove this?

[uncut nova]

Hi.
I have such equation.
sin(x+pi/6)+cos(x)=3/2
Can i split left side to
sin(x+pi/6)+cos(x)=1 + 1/2
and then find those x'es that way?
(sin(x+pi/6) = 1 and cos(x) = 1/2) or (sin(x+pi/6) = 1/2 and cos(x) = 1)
???

[dark sparrow]

no you can't

[uncut nova]

@dark sparrow Can you explain why? This method gives me correct answer though

[dark sparrow]

why split it as 1 + 1/2 and not, say, 3/4 + 3/4? or 5/6 + 2/3? or 8/7 + 5/14?

you made a completely arbitrary choice

and just happened to get lucky

[silk crown]

I mean you just got lucky

The solutions just happened to be that way

[vast dune]

@silk crown Ive recently seen this proof