geometry-and-trigonometry

[vague berry]

Ty

[upper karma]

measure of M and Y

[keen aspen]

We dont need to know it

We can subtract y

So M=23

[upper karma]

Oh

And we know that 2 is 23 + y and 3 is y

[keen aspen]

Ya

[upper karma]

So 180=23+y +23 +y

180=46+2y

Y=67

[keen aspen]

Mhm

[upper karma]

Ohh thank you so much!

[keen aspen]

Np

[upper karma]

So all I have to do is change what I know to the same variable?

[keen aspen]

Yep

[upper karma]

Oh Tysm

[keen aspen]

Working with 1 variable makes it very easy, establish a relation between one to another

Np, I'm heading off to sleep

Goodnight

[upper karma]

Night

[signal hornet]

hi

Apparently 2pi/3 is also a solution

But I don't see where or how I missed it

[keen aspen]

You didnt factor correctly

x(2x+1)

[signal hornet]

wow

I keep messing up the simple stuff

thanks @keen aspen you are a hero

[keen aspen]

Lol

Np

[signal hornet]

couldn't go to sleep without one more math problem 😉 ?

[keen aspen]

Lol you posted it same time I said that and it was an easy problem

But yes

[signal hornet]

rest well

[keen aspen]

Thanks catch ya later

[signal hornet]

cya

[signal hornet]

All I got is x = theta - 1

I know one trig propery kinda related to this

1 + tan^2(x) = sec^2(x)

but I don't see how that can help me here

[umbral snow]

What is x?

Wait wha

[signal hornet]

just a substitution

[dark sparrow]

uh

[signal hornet]

in the property one it's just and angle

[dark sparrow]

why would you do that

[umbral snow]

That's tan(θ) - 1

[dark sparrow]

this is not $\tan(\theta - 1)$

[somber coyote]

Ann:

[signal hornet]

oh

[dark sparrow]

see, this is why you should always put parentheses!!!

[signal hornet]

Yeah, good point

I did put them though

just in the wrong place

Yeah, that didn't make it anymore clear

still have no clue where to start

now I can't even substitute anything

[dark sparrow]

you don't need to

look at your equation

it says that the product of two things is equal to zero

[signal hornet]

one of these or both have to be 0

[dark sparrow]

well

yes

are you able to solve the equations tan(θ)-1 = 0 and sec(θ)-1=0 in isolation

[signal hornet]

let me try

I got pi/4 + n*pi

and 0

I guess the 2nd should be 0 + 2n*pi

[dark sparrow]

should it?

[signal hornet]

yeah

actually no

my homework limits the domain

so just 0 is okay

[dark sparrow]

well then.

[signal hornet]

thanks @dark sparrow

[obtuse tapir]

30-xroot3/2 * 2/x

@vague berry

By inverse I mean reciprocal

multiplicative inverse

https://www.mathsisfun.com/definitions/multiplicative-inverse.html

and then

x and x would cancel

and 2 and 2 would also cancel

leaving you with 30-root3

which is not equal to 30/16

[manic crown]

I doubt a statement in my textbook.

*If two co-terminal angle $\theta$ and $\alpha$ and $0^{\circ}\leq \alpha \leq{360}^{\circ}$

[somber coyote]

Raftaar:

[manic crown]

then $\sin(\alpha)=\sin(\theta)$

[somber coyote]

Raftaar:

[manic crown]

is this treu?

true?

[silent plank]

what's your definition of co-terminal angles?

[manic crown]

The angles with different amount of rotation but same position Initial and terminal ray

[silent plank]

what's making you doubt that statement?

[manic crown]

I mean

The way trig functions are defined

[silent plank]

well the position is the same
the relative angle is the same,
quadrant is the same,

[manic crown]

If the intial and termnial rays coincide then the tri f are equal

[silent plank]

if $\theta$ and $\alpha$ are co-terminal then:
$\theta = \alpha + 2k \pi$ where $k$ is an integer

[somber coyote]

ramonov:

[manic crown]

$sin{360n+p}^{\circ}=sin(p),n\in {\bZ}$

[somber coyote]

Raftaar:

[dark sparrow]

$\sin(p + 360^\circ) = \sin(p)$

[somber coyote]

Ann:

[dark sparrow]

@manic crown

[manic crown]

yes

[somber coyote]

Raftaar:
Compile Error! Click the reaction for details. (You may edit your message)

[dark sparrow]

$\sin(p + 360n^\circ) = \sin(p), n \in \bZ$

[somber coyote]

Ann:

[upper karma]

I understand an Isosceles triangle is a triangle with two equal sides

But would those triangles have four sides

[dark sparrow]

what triangles

[upper karma]

In the image above

[dark sparrow]

what triangle do you think has 4 sides

despite being a triangle

[upper karma]

ACd

So it’s asking for isosceles triangles right

But in the first answer, it takes the two sides on the bottom, which are congruent

If those are the two congruent sides, then the two other sides would make that four

[dark sparrow]

why do you think ACD has four sides

can you name the four line segments that you think are its sides

[upper karma]

DC DA C to the mid point A to the mid point

When it says triangle

I ssume that the segments become CA, DC, DA

But then it wouldn’t be an isosceles triangle

[dark sparrow]

the sides of ACD are CA, DC and DA, yes.

and yes it would be isosceles. AD = AC.

letting the midpoint be O... CO and AO are not sides of ACD.

[upper karma]

Ok

[obtuse tapir]

couldnt you just prove all these triangles equal

[upper karma]

How do you know AD=AC?

[obtuse tapir]

*congruent

[upper karma]

No?

We don’t know any angles

Oh wait

No we need to know if they are isosceles

[obtuse tapir]

AD=DC

[upper karma]

I don’t understand how AD=DC

[obtuse tapir]

Congruency

[dark sparrow]

AD=AC by pythagoras

[obtuse tapir]

There are 2 sides equal

[upper karma]

But it’s not a right triangle

[obtuse tapir]

and an angle too

It's right angle

triangle

[dark sparrow]

@upper karma ADO and CDO are right triangles.

i'm applying the Pythagorean theorem to those.

[obtuse tapir]

You could use the congruency criterias

[upper karma]

How do you know ADO is a right triangle

Is it because perpendicular lines create four right angles?

[obtuse tapir]

Linear pair my dude

[upper karma]

Ohhhh

[dark sparrow]

@upper karma well, yes

[obtuse tapir]

the sum angle AOD and AOB is 180

[upper karma]

So then if that’s the case

[dark sparrow]

if any of the four angles created by a pair of intersecting lines is right

then so are the other three

[upper karma]

That means DC and DA are the congruent sides tho right?

[obtuse tapir]

yes

[dark sparrow]

congratulations you just repeated what i said earlier

[obtuse tapir]

Side-Angle-Side

[upper karma]

U said AD=AC by Pythagorean’s

[obtuse tapir]

You will achieve the same result

If you take one of the side as x and other as y

root x^2+y^2 will be the hypotenuse

and since all the x and ys are equal in all triangles

or

you could use

SAS criteria for congruency

[upper karma]

I’m just gonna stick to the linear pair way

[obtuse tapir]

what?

[upper karma]

Congruency

[obtuse tapir]

ahh

[upper karma]

Proving all of them are congruent

[obtuse tapir]

Its the same thing

using pythagoras theorem is more intuitive

just that

BCD, ADB, ADC, ABC

[dark sparrow]

#calculus.

[upper karma]

What is the base angles congruence theorem

[fringe crater]

@upper karma that's the theorem that says if two sides of a triangle are congruent, the angles opposite those two sides are also congruent

[upper karma]

Alright thx

So it’s for isosceles triangles right

[fringe crater]

Yes

[sudden locust]

@fringe crater Look whos here! Hey buddy! 😊

[fringe crater]

Hey, how's it going?

[sudden locust]

Its going good, made a video about dimensions today!

What your next video going to be about?

[pure rain]

i need help finding this in order

this is sss

[pure rain]

<@&286206848099549185>

<@&286206848099549185>

[pure rain]

oh sorry

mybad

do you think you can help me

[pure rain]

<@&286206848099549185>

[summer spire]

are you happy with the order in your screenshot? What are you having trouble with?

[fringe crater]

@sudden locust that sounds cool, I'll check it out later. My next video is going to be about a math olympiad geometry problem

[signal hornet]

Hello

I don't understand step 5

the identity calls for a squared cos, the problem only has a cos to the 1st power

so how does that identity even apply?

[umbral snow]

Do you see how the identity was used, first of all?

[signal hornet]

no

I don't see how it can be used

[umbral snow]

There's a sin² in the equation. Replace it with 1 - cos²

[signal hornet]

that's genius

[umbral snow]

Lel glad it clicked. Yeah they didn't explain that very well

[signal hornet]

THank you

[umbral snow]

They just want everything in terms of cos, and you can always convert between sin² and cos² easily

[signal hornet]

saying to replace it helped

Now that you said it, it's obvious

[signal hornet]

why is my calc so bad at this?

what is this n# stuff it keeps doing

[dawn drum]

why dont you use symbolab or something

[signal hornet]

I do for HW

but I can't use my phone on tests or quizes

[upper karma]

uhhh @signal hornet are u in a test or quiz right now? Or just trying to learn how to use calculator for a test or quiz

[signal hornet]

Learn

[wet ice]

Not sure how to do them 100%

45-51

[buoyant spruce]

Hey quick question

[queen python]

@buoyant spruce just look

Qr =4

Use distance formula is my guess

[maiden rain]

??

this is my answer..

I don't see it..

[dark sparrow]

did you rationalize the denom in 1/sqrt(x^2+1)

to get that

@maiden rain

[maiden rain]

@dark sparrow Yeah I solved it in a weird way, I just solve it by drawing a graph and a triangle and the answer is (b) I think..

[dark sparrow]

a triangle is exactly how you're supposed to do it anyway

[maiden rain]

It's not possible to do it algebraically ?

[dark sparrow]

what do you mean algebraically

[manic crown]

If $x \to {\frac{\pi}{2}}^{+}$ then $\tan{x} \to \infty$ or $-\infty$

[somber coyote]

Raftaar:

[manic crown]

??

[keen aspen]

What are you confused about

Notation?

[manic crown]

Lol I got it

sorry

[keen aspen]

Lol np

[upper karma]

Ok I think this place is appropriate

This is physics but

How solve

[gritty siren]

depends on how you modelize friction and stuff

it can get quite complicated

[upper karma]

Bruh epic troller

[upper karma]

Any ideas on how to do Question 6? I have two unknowns I'm completely lost

This is where i got to pretty much

I've tried everything

[keen aspen]

Okay

Write l in terms of h and r first

@upper karma

[upper karma]

L?

[keen aspen]

The slant height

[upper karma]

Yep, it's at the bottom right of my board

csa/pi * r

[keen aspen]

Yeah but write it in terms of h and r

With pythagoras

[upper karma]

Ohh yeah I did try that but I got stuck

Okay

done

[keen aspen]

Yeah

[keen aspen]

So set both equal to each other

[upper karma]

Set my old l formula and my new one equal to eachother?

[keen aspen]

Csa/pir=sqrt(h^2+r^2)

[upper karma]

Okay

I could get rid of the square root or get rid of the fraction

[keen aspen]

Now you know h

So solve for r

[upper karma]

Uhhhmm

Do I know h?

[keen aspen]

3.6

[upper karma]

Oh wait oops lol

Duh

Yes so

[keen aspen]

Where did the pi r go

[upper karma]

oops i forgot tto

i forgot tto erite it

there

ok

so should i deal wjth the sqrt first

[keen aspen]

Now square both sides

Ya

[keen aspen]

Yep now move over the r^2 on the LHS

And evaluate 90.5/pi squared

[upper karma]

Huh? Algebra works like that?

No wait

[keen aspen]

(90.5/(pir))^2=90.5^2/(pi^2r^2)

[upper karma]

$(\frac{90.5}{\pi r})^2 = \frac{90.5^2}{\pi^2 r^2}$

[somber coyote]

Nemesis:

[upper karma]

This?

[keen aspen]

Ya

And you can calculate 90.5^2/pi^2

[upper karma]

Wait hold on

I'm lost

We moved an r^2 to left side

And it turned in to that?

[keen aspen]

You didnt remove it yet

Itll turn into a quadratic soon

[upper karma]

Ew

Okay hold on

Heres what i think

Right?

[keen aspen]

No no

Move that r^2 over to the right

You want it all the the rivht

Right

[upper karma]

so undo what i just did

[keen aspen]

Yes

Move it all the other way

[upper karma]

so we're here

so should i expand the fraction

thats what i did the last time

[keen aspen]

Are you able to vc

[upper karma]

Ehh okay but i cant talk

[keen aspen]

Ok

[upper karma]

Yea

Everything on the left move to the right

But wont that eliminate the equals sign

Ohhhh

Okay

So expand it?

And you can just make it 1 fraction

yes

i expanded the bottom

No but it expands out because we dont know r

Shit

Ok

Yeah its a 10 i rounded

[keen aspen]

,wolf 8190.25/pi^2

[somber coyote]

Display results online and refine query

Requested by PJS#5494

[upper karma]

829.8r^2?

oh

so i replace the square fraction

with the new fractioj

ok

it was a negative

so

my fraction will be negative right

But i thought it turned to r^4

No it was a plus, the r^2 is a seperate term

[keen aspen]

r^2(12.96+r^2)

[upper karma]

r^2 = x?

so i just write r^2 = x

done

yea

Yeah ill just use the formila

ok

im gonna wipe my board

this is messy

ok so

does that look right

ok ill get two answers though

so the positive value

is the answer

yep ok so

yes

right so now i have r

i can solve the original problem

ok lets see how this goes now

another board wipe

@keen aspen

Thank you so much

That was awesome

[keen aspen]

Lol np

Kinda messy but it worked

[upper karma]

Jesus

That WAS messy

[upper karma]

Anyone have any idea how to solve 7?

Sorry I can't get it.. uhh straight

[silent plank]

,rotate 90

[silent plank]

mmm, by standard form they mean vertex form right?

factorise (-1) out,
complete the square and then redistribute

[upper karma]

😅 I have to be honest. I have no idea what that means.

[silent plank]

were you able to do #4?

[upper karma]

To be honest I'd not know if I should divide by 6 or just do x(6-x)

[silent plank]

well x(6-x) is equivalent to 6x - x^2
but it doesn't seem to be what the question wants

they seem to want you to write it in the form
f(x) = a(x-h) + k

[upper karma]

does that form have a name?

[silent plank]

vertex form

[upper karma]

ahhhhh

[silent plank]

the form presented is actually standard, some places call vertex form standard which can make it mega confusing

[upper karma]

thank you for the help

i think i can do it now

[leaden belfry]

I'm planning a comms network in Kerbal Space Program, and I want to know the distance between the satellites in the constellation because they have a limited antenna range. What's the easiest way to get that distance? https://puu.sh/EBEui/e2e6e3055b.png

Equatorial radius of the planet is 700,000m

The pic is slightly misleading because it shows the resonant orbit, but the satellites are in a circular orbit of 250,000m

[leaden belfry]

If anyone's interested the answer is 2π(radius of planet + orbital altitude) / (number of sats)

[ivory vortex]

how do you do b?

i found the related accute angle which is pi/4

[dark sparrow]

exactly the same as the other five... via a unit circle

[ivory vortex]

im not given one

[dark sparrow]

what do you mean not given one

draw one yourself lol

[ivory vortex]

how

[dark sparrow]

do you not know what a circle looks like

...i mean ok fine you found the reference angle

is cos(π/4) something you are able to evaluate

[ivory vortex]

https://cdn.discordapp.com/attachments/490315911044136961/641873708415320076/Surdsintrigonometry.png

i was told to use these triangles

pi/4 is 45 degrees

and it's cos so it's 1/sqrt(2)

but that's not the answer

[silent plank]

what quadrant is 5pi/4 in?

[ivory vortex]

3

[silent plank]

is cos positive or negative in that quadrant?

[ivory vortex]

negative

[silent plank]

are you able to determine the answer from that?

[ivory vortex]

umm idk

[silent plank]

the reference angle was pi/4
the reference ratio is cos(pi/4) = 1/sqrt(2)
the sign of that will change depending on where your angle is

[ivory vortex]

but the answer is -sqrt(2)/2

[silent plank]

rationalise the denominator

[ivory vortex]

not -1/sqrt(2) or 1/sqrt(2)

what does that mean

[dark sparrow]

sqrt(2)/2 is the same thing as 1/sqrt(2).

[ivory vortex]

how

[dark sparrow]

$\frac{1}{\sqrt{2}} = \frac{\sqrt{2}}{\sqrt{2} \cdot \sqrt{2}} = \frac{\sqrt{2}}{2}$

[somber coyote]

Ann:

[dark sparrow]

@ivory vortex

[ivory vortex]

how did you get the numbers in the middle expression

i forgot everything about radicals

[dark sparrow]

i multiplied num and denom by sqrt(2)

a/b = (ac)/(bc)

[ivory vortex]

oh i see

how come when you multiply two of the same radicals they lose the square root

[dark sparrow]

...

do you know what a square root is

$(\sqrt{x})^2 = x$ is literally the definition

[somber coyote]

Ann:

[ivory vortex]

ohh

why is it preferred to be written like that

sqrt(2)/2

and not 1/sqrt(2)

can you not have radicals in the denominator

[dark sparrow]

you can

it's mostly an arbitrary requirement to make things easier to grade

[ivory vortex]

oh i see

[tepid vine]

i dont understand my math question
There is a 1440-cubic foot rectangular prism. The prism has a square cross-section and height 10 ft. What are its length and width?

[dark sparrow]

what don't you understand about your question

[tepid vine]

the use of "rectangular prism"

we're doing a square and cube root unit

im assuming its a cube but im not sure

also its worded strangely

not entirely sure what its asking

[dark sparrow]

a rectangular prism is a box

not necessarily with all the same dimensions

[tepid vine]

right

[dark sparrow]

do you know how to find the volume of a box

[tepid vine]

yes

[dark sparrow]

so how do you find the volume of a box

[tepid vine]

length•width•height

[dark sparrow]

\* so discord doesn't eat your asterisks

but yes

so the height is 10 ft and the volume is 1440 ft^3, so this means length * width = ?

[tepid vine]

🗿

[dark sparrow]

.....

???

can you answer my question

[tepid vine]

no

[dark sparrow]

$lwh = 1440 \ h = 10 \ lw \cdot 10 = 1440 \ lw = ; ?$

[somber coyote]

Ann:

[tepid vine]

12

x12

x10

=1440

[dark sparrow]

...

[tepid vine]

lw = 144

[dark sparrow]

you've jumped way ahead.

but yes, lw = 144, and since the base is a square we have l = w

[tepid vine]

sweet bro thanks for the help

ok but what was this mention of a cross section

i think that confused me some too

why not just give me height i dont understand

[dark sparrow]

don't call me bro

[tepid vine]

why

[dark sparrow]

i am uncomfortable with that term for multiple reasons

also you were given the height

[tepid vine]

in the same sentence as the unnecessary use of cross-section

[dark sparrow]

so what

[tepid vine]

why cross-section

im not dissecting an animal

im finding lengths of a rectangular prism

[dark sparrow]

maybe to make you THINK a bit

[tepid vine]

this textbook mad wack

[dark sparrow]

which i know you're one of those people who don't like that much

but it's quite a necessary skill

don't you find

[tepid vine]

good one dude

good one

[dark sparrow]

i'm not a dude.

[tepid vine]

ah i see

that is where the problem with bro stems

[dark sparrow]

i thought it was obvious but i guess i don't speak neurotypical after all

[tepid vine]

interesting

i'll let it slide just this once

just figured out the square cross section part too, nice

[manic crown]

$4\sin^{2}{\theta}-2(\sqrt{3}+1)\sin{\theta}+\sqrt{3}=0\ implies \sin{\theta}=\frac{\sqrt{3}}{2},\frac{1}{2}\ \implies \theta =(2\pi+\frac{\pi}{6}),(2\pi +\frac{pi}{3})$

[somber coyote]

Raftaar:

[manic crown]

$\theta $ in radians

[somber coyote]

Raftaar:

[manic crown]

can someone tell if the above is correvt

[dark sparrow]

,w 4t^2 - 2(sqrt(3)+1)t + sqrt(3) = 0

[somber coyote]

Display results online and refine query

Requested by Ann#0413

[manic crown]

,w 4sin^2x - 2(sqrt(3)+1)sinx + sqrt(3) = 0

[somber coyote]

Display results online and refine query

Requested by Raftaar#1096

[dark sparrow]

$\theta = \frac{m\pi}{3} + 2k\pi, , k \in \bZ, m \in {1, 2, 4 ,5}$

[somber coyote]

Ann:

[manic crown]

So i was wrong

[dark sparrow]

woah wtf

how tf is that emoji so gigantic

[manic crown]

No i think

idk

lol ye

[ivory vortex]

how do you do f

theta should be between 3pi/2 and 2pi

i found the value for all the trig ratios and got -0.17, -0.17, and 0.17

i think i get the answer by subtracting 0.17 from 2pi but i'm not sure why

how come i didnt straight up get the answer when solving the trig ratio

[dark sparrow]

arcsin and arctan give you a value between -pi/2 and pi/2

arccos gives you a value between 0 and pi

[ivory vortex]

do the ratios not give values greater than pi

[dark sparrow]

??

[ivory vortex]

why do i not get a value between between 3pi/2 and 2pi

considering thats where the point is

[devout harbor]

you dont get that from the calculator, you have to do it yourself

find the angle of the triangle thats formed with the x axis, and then subtract that from 360

[ivory vortex]

do i not get it from the calculator because the ratios cant give values greater than pi (or 180 degrees)

[devout harbor]

dont depend on the calculator, think yourself

did you get the angle of the triangle forming with the x-axis?

[ivory vortex]

well no

what exactly do the ratios give?

is it the angle to a point, but the angle is always less than 180 degrees

like that

[devout harbor]

now find theta

[ivory vortex]

ohh i see

[devout harbor]

so find theta

[ivory vortex]

so it really cant be greater than 180

[devout harbor]

It will be. Find theta, subtract that value from 360 degrees.

because

[ivory vortex]

oh yeah, but i mean theta

the theta within that triangle

[devout harbor]

We need to go anti clock wise, otherwise the clock wise angle gives us negative.

Yes, it will be less than 180.

[ivory vortex]

okok thats what i was talking about

ok i seee

thanks that really helps

[devout harbor]

np

[sudden locust]

Just made a video of a proof of the Pythygorean Theorem. Would really appreciete any kind of feedback from you guys. Hope you will have a great Thursday! 😊👍https://youtu.be/BmUjXSP5-3o

[sly marlin]

and here I'm still wondering what sort of syllabus and target audience you are targeting

[wind crypt]

Cool video!

[sudden locust]

Thanks! 🙂 @wind crypt

[idle bloom]

@sly marlin he's targeting people who are taking ACT and SAT tests

[sly marlin]

hmm maybe I'll see based on that

[maiden rain]

Why the range of Arcsin =

It should be from -infty to +infty

Yeah..

It is a one to one function..

Please read my quesiton.

[dark sparrow]

arcsin isn't multivalued

[maiden rain]

The domain of sin = -infty, +infty

[dark sparrow]

yes but arcsin isn't the true inverse of sin

[maiden rain]

so the range of arcsin should be the domain of sin.

[dark sparrow]

no

sin doesn't have a true inverse

sin is not invertible if considered on the entire real number line

[maiden rain]

Sin is like x^3 ?

[fringe dirge]

Here, if I ask you for arcsin(0) what do you think this is

[maiden rain]

0

[fringe dirge]

Why

Why isn't it 2pi?

Or pi?

You have that sin(pi) = 0

And sin(2pi) = 0

And sin (-pi) = 0

Why'd you pick 0 when there are so many other options

[maiden rain]

well it's 0 in a Real numbers graph..

[dark sparrow]

what's a "real numbers graph"

[maiden rain]

In the Radian graph it's different..

[dark sparrow]

what's a "radian graph"

why shouldn't arcsin(0) be 19π

sin(19π) = 0 after all

so why shouldn't 19π = arcsin(0)?

[fringe dirge]

If you want degrees, sin(360 degrees) = 0

So why isn't arcsin(0) equal to 360 degrees?

[dark sparrow]

why shouldn't arcsin(0) = 3420°?

@maiden rain

[maiden rain]

Well, 0 is infinity.

[dark sparrow]

big brain moment

[maiden rain]

xD

[dark sparrow]

but actually you're making no sense

[maiden rain]

So can I say that the domain of sin is -pi/2 , and , pi/2 ?

[dark sparrow]

no you can't

the domain of sin isn't {-π/2, π/2} and it is not [-π/2, π/2] either.

but arcsin is not the true inverse of sin.

arcsin is the inverse of the RESTRICTION of sin to [-π/2, π/2].

[maiden rain]

Oh

wow amazing!!

Now it make sense !!

You're good thanks for explaining that @dark sparrow

[maiden rain]

I need an explanation here, Apparently the answer is -infty...
the thing that confuses me is that there is no (x) values beyond pi/2 so how am I supposed to approach it from the right?

[mint sandal]

tan is defined on the whole unit circle except (0, +- 1)

[maiden rain]

That's not true, it has 2 vertical asymptotes at -pi/2 and pi/2

[mint sandal]

yeah you just paraphrased what I just said

(0,+-1) are points on the unit circle

which can be parametrized by an angle as +-pi/2

[dark sparrow]

@maiden rain are you claiming tan(x) is undefined for x >= π/2?

because that would be very wrong

[maiden rain]

Look I made a drawing, Tan(x) has 2 vertical asymptotes at -pi/2 and pi/2 ... SO look at the blue arrow I'm trying to approach it from an undefined area that doesn't make any sense to me. how am I supposed to approach it from the right?

So in my opinion, I think this should be "Doesn't exist" ^

[dark sparrow]

the domain of tan is not (-pi/2, pi/2)

it is not the case that tan(x) is undefined whenever |x| >= π/2.

[maiden rain]

Ok so the limit of 1/x when x approaches 0 from the left = -infty , an example of that would be x=-0.00001

Can you give me an example of the limit of tan(x) when x approaches pi/2 from the right ?

"what's so close to pi/2 from the right "

[dark sparrow]

π/2 + something small and positive, evidently

[maiden rain]

Great, I was thinking of tan as a restricted function..

[upper karma]

How do I prove this

[shell imp]

Can someone please explain this?

[devout harbor]

explain what?

@shell imp

[dark sparrow]

,rotate -90

[silent plank]

structures a bit weird.

its a combination of:
difference of 2 squares
double angle identity for \cos
sin^2(x) + cos^2(x) = 1

do you understand all those concepts?

[shell imp]

To an extent, yes. I just don’t understand how those connect and like how you piece it together

[somber coyote]

ramonov:

[silent plank]

wdym by piece them together?

if you want to write cos^4(x) - sin^4(x) in an alternate/simplified form,
the difference of squares would be a good place to start.
and then think about what you can use after that

[median crown]

$(cos^2(x)-sin^2(x))^2= cos^4(x)-sin^4(x)$

[somber coyote]

maleb1964:

[median crown]

[devout harbor]

lol

[median crown]

Jk don’t do that

Wrong

Don’t do that

[devout harbor]

@shell imp You piece it together by looking at what angle you want. Since we had cos(2x), so we know we have to simplify it and make something that equals cos(2x) or in other words, find its double angle identity.

[shell imp]

Thank you

To both people

[median crown]

No problem

[median crown]

I’m not sure about this drawing

It looks sketch

No pun intended

The drawing doesn’t agree with itself

Is anyone else seeing this?

Wtf

I see a 3 4 5 triangle for the entire thing, meaning the whole drawing is a right triangle

But when I cut a the whole thing into a similar triangle with the leg of length 12 becoming a leg of length 6, I am supposed to get a smaller triangle with leg length 6, 8, 10

the funny thing is 10 is supposed to make up the entire hypotenuse of that triangle

?

Wtf

[silent plank]

hmm

[median crown]

Lmao who did this

@upper karma uhh if you made this drawing you should check your stuff

If it’s from a book or an assignment and you didn’t make it then idk bro

Ooooof

ok

But is it from a book?

Oooof

Maybe you drew it wrong

You should post the problem

[silent plank]

diagram just isn't to scale

make that middle shape look less like a square and more like a rectangle

[median crown]

That still won’t help tho

the numbers won’t match

[silent plank]

the vertical line won't be to the vertex but somewhere on the side

[median crown]

I guess

I suggest checking the 10 measurement

[median crown]

[silent plank]

and then solve using similar triangles

[scarlet halo]

is it possible to create a equation for a parabola with a x intercept and a point ?

[crude pelican]

Doing a geometry Problem where i need to find x here

I'm setting up with proportionality by doing AB over BE = BC over CD

I keep getting a decimal and i'm expecting a whole number for the result

but why would i do x+60? and what makes this different from other questions? is it because its like a combined triangle sort of, instead of it being mirrored

yeah

oh that makes alot of sense

right

Thank you

[scarlet halo]

is this right ? https://media.discordapp.net/attachments/536995777981972491/642177251986636810/642177162501292042.png?width=630&height=473

[thorn linden]

Does anyone know how to do this

[formal raptor]

This is extremely confusing to me and I can’t get any farther than the part where I need to find the hypotenuse BUT I CANT FIND THE HYPOTENUSE pls help

[ivory vortex]

umm im not completely sure but cos = x/r right

r = sqrt(x^2 + y^2)

r = sqrt(61)

[formal raptor]

It’s taught to me like cos = adjacent/ hypotenuse

[ivory vortex]

yep it's the same thing

then uhh

doing

arccos (-7 / sqrt(61) = principle angle

then you have to find the related acute angle from that

not quite sure how to do that so you get a radical tho

[formal raptor]

Oof

Trig is annoying

[vagrant steeple]

don't you just put (-7,2sqrt(3)) into a^2 + b^2 = c^2

-7^2 + (2sqrt3)^2 = c^2

[ivory vortex]

what does that do for you

oh wait

[vagrant steeple]

finds the hypotenuse

[ivory vortex]

ya i already said that

hypotenuse is equivalent to r

[formal raptor]

I don’t understand how 2sqrt3 can be squared

That confuses me so much

[ivory vortex]

well you can just put it in your calculator and not worry about it

but if you really want to know

(2sqrt(3))^2

the square removes the square root of the 3

and squares the 2

so you get 12 as your final answer

as you're left with 4*3

[formal raptor]

Oooooh

Ok I see

[mellow lava]

hi

can someone please help me with this problem

im revising for exams and i cant find a solution

how did they get those answers?

[mighty narwhal]

Hmm
The first thing you should see is that
C+d=a+b=1/2

You can rearrange A and B to get half of the square
So A+B=1/2, so C+D is also =1/2

A and B are also equal, so both are 1/4

You can see that C is the half of a quarter-square, so 1/2*1/4=1/8

And because C+D=1/2, D must be 3/8

[tepid vine]

i concur

[leaden belfry]

Please excuse my bad paint skills... this is a 2D diagram of a 3D problem. I have a satellite around a body with radius R, and a dish on the satellite with a projection cone of θ. The satellite has an altitude of D.

I already know R and θ, and I want to know what value of D is required for the cone of projection to cover the whole sphere. Any help appreciated, thanks.

https://puu.sh/ECuc7/e267cd24e3.png

[sly marlin]

it's a 2D problem, isn't it?

Try extending the lines from the satellite until it goes past the sphere

@leaden belfry

How should the rays interact with the sphere?

[leaden belfry]

@sly marlin So I've solved the 2D problem, what I want to know is if the solution to the 2D problem translates into 3D

Intuitively I think it does, but I know cones can be unintuitive, so I just want to be sure

[sly marlin]

well this is a cross-section, isn't it?

Here we are assuming the cone points directly at the sphere

[leaden belfry]

i.e. does the cone covering 50% of that circle (with the other half being occluded) translate directly to one hemisphere of coverage in 3D

And the cone does point directly at the centre of the sphere

Yes, it's a cross-section

[sly marlin]

what do you mean "covering 50% of that circle?"

[leaden belfry]

The area inside the cone will contain some portion of the circumference of the circle, but since it cannot penetrate the circle (or the sphere) in reality, the other side occludes the projection

So the maximum you can actually cover is 50%

Only the side facing the satellite has line of sight with it

[sly marlin]

So your answer for this question is?

idk if it's correct, from what I interpret it looks like a slightly better than fair chance of being correct

[leaden belfry]

Actually... I think I made an assumption and an error, but I will show you...

I figured I could use pythag to solve the 2D problem

https://puu.sh/ECvPk/f0cd6d92eb.png

Obviously this is a terrible diagram but I think the problem is still there

[sly marlin]

I think I know what's going on

Note the tangency point

[leaden belfry]

Yeah

[sly marlin]

draw a line segment from there to the centre

[leaden belfry]

Sure

Okay, now I have two triangles and I can use trig to solve it?

[sly marlin]

What information do you have?

[leaden belfry]

I know theta and I know R

Sorry, theta is the angle next to the red dot

Thanks for your help by the way, my math is super rusty and I know I'm making basic mistakes

[sly marlin]

Any other angles?

[leaden belfry]

The angle at the origin of the sphere/orbit is 90

[sly marlin]

maybe that triangle with that angle shouldn't be used

you have an expression for one side, but that's all

maybe use another triangle?

[leaden belfry]

Not quite sure what you mean, here's what I have https://puu.sh/ECvSx/2750911c7b.png

[sly marlin]

look at the tangency point

[leaden belfry]

Sure

[sly marlin]

what angle is there?

do you have enough information for any triangle in your diagram?

[leaden belfry]

I think I can work out the angle that's directly above 90 degrees using trig

But I'm not sure how to do that, it's been a long time 😄

[sly marlin]

where is your right triangle?

(there are 2 right angles, one which is more useful)

[leaden belfry]

There's right angles at the tangent point and one at the origin

Maybe I should give some context, I'm an adult (not a student) and this is a personal project, so I don't have a trig book in front of me or anything

This is memory from like 13 years ago

Aren't there three right triangles here? There's the big one containing the other two, which has a right angle at the origin, then that's divided into two triangles which have their right angles at the tangent point

[sly marlin]

yeah there are 3 right triangles

but which one should you use?

[leaden belfry]

The one containing the angle theta

Okay, you've given me something to work with, thanks... I'll do some googling on trig and see if I can figure it out

Thanks for your patience

[sly marlin]

there are 2 triangles containing theta

[sudden locust]

https://youtu.be/Ah7lP4Rhwoo In this video, we are going to take a look at dimensions. Hope you can get some value out of it, and if not, any feedback is really apprecieted! Hope you have an awesome weekend guys! 😊👍

[sly marlin]

let's take a look

[chrome sinew]

Why is Benedict Cumberbatch in the thumbnail?

[sly marlin]

idk, probably just their style

putting a hypercube model without mentioning it in the video too lol

[upper karma]

@sudden locust Yet another great video. )) Keep it up!

[sudden locust]

@chrome sinew Its Cuberbatch when hes playing Alan Turing (The Imitation Game)

Great movie!

@upper karma Thanks buddy ❤

[chrome sinew]

I know. I have watched that movie. It is the best biographical movie I've ever seen. Love it. But I said that because B. Cumberbatch is not a mathematician himself ;-) never mind @sudden locust

His acting is so good people take him as real the real person

The Theory of everything and American Sniper came out the same year otherwise he probably could have won the oscar

[upper karma]

what does the zero with a line through it mean?

[dark sparrow]

that's not a zero with a line through it

that's a greek letter

specifically, theta

$\theta$

[somber coyote]

Ann:

[sudden locust]

@chrome sinew One of my favourites aswell! Awesome movie... Only seen it once, so I should rewatch it sometime soon!

[upper karma]

ok thanks

[ornate tusk]

how would i find the exact value of acos(-1/3) by hand

[keen aspen]

@buoyant spruce what do all the angles add up to in a triangle

[buoyant spruce]

180

[keen aspen]

yep

now add them all up and set equal to 180

and solve for x

[buoyant spruce]

ok thank you

[sly marlin]

Either there's extra information or that problem has no solution

[night karma]

D

[upper karma]

what would the third one be

[fossil lotus]

There are three similar triangles in that diagram @upper karma

[upper karma]

oh i see what you mean

[fossil lotus]

You got it?

[upper karma]

ye

ty

[languid radish]

How do I solve sin x> -1/2

sin x> sin(7pi/6)

Is where I’m at

[median crown]

Well

Is the problem bounded?

0 to 2pi?

[languid radish]

I mean
The question says:
“Solve sin x > x1/2 or find the domain of 1/sqrt(1+2sin x)

The textbook says draw the graph from -pi/2 to 3pi/2

I don’t get why this and why not 0 to 2pi

@median crown

[fossil lotus]

What does your textbook say about how to solve trigonometric inequalities

[languid radish]

It brings it up only in one solved example

So I’m a bit confused

It’s explanation is only half a line
“Draw a graph”

That’s literally it

@fossil lotus

[fossil lotus]

Hm

Based on that, I'm guessing they want you to figure out when sin(x) = -1/2, and then look at the graph of sin(x) to determine in what intervals sin(x) is greater than -1/2

[languid radish]

Yes I did solve sin x= -1/2

That gives n*pi + (-1)^n 7pi/6

I found some YT videos but I’d rather take the help of discord

[fossil lotus]

What are you having trouble with then

[languid radish]

With what?

The problem,or the videos?

I solved for n*pi + (-1)^n 7pi/6

x > n*pi + (-1)^n 7pi/6

[fossil lotus]

What's your question

[languid radish]

Solve for sin x> -1/2

[fossil lotus]

Brb

[languid radish]

OR:Find the domain of 1/sqrt(1+2 sin x)

[languid radish]

Anyone there?

[fossil lotus]

@languid radish Since the example in your textbook mentioned drawing a graph, I think that’s what you have to do

You know where sin(x) = -1/2

So between any two points where sin(x) = -1/2, y is either going to be greater than or less than -1/2

And you can figure out which one it is by looking at the graph of y=sin(x)

[languid radish]

Yeah
But is there any other way
Without a graph

@fossil lotus

[fossil lotus]

Probably

I mean, you could just test points

Or look at the derivative and see whether the function is increasing or decreasing at the points where y=-1/2

I’ve never had to do inequalities involving trigonometric functions though so idk what the normal way to solve them is

[languid radish]

Okie

Thanks

[fossil lotus]

Np

[formal raptor]

trig help?

[silent plank]

where are you stuck?

[formal raptor]

Trig is just extremely confusing

I just need help putting the problem together and solving it

[silent plank]

have you attempted to draw a diagram?

[wind crypt]

This. Diagrams are really useful for trigonometry word problems.

[languid radish]

How do we solve:
Find the period of f(x)= sin(x+cos x)

[dark sparrow]

what have you tried so far

[languid radish]

sin(x+t + cos(x+t) ) = sin(x + cos x)

I have no idea how to proceed

I feel so dum dum

If x=0?

[dark sparrow]

well remember that you want this to hold for all x

[languid radish]

Yeah

[dark sparrow]

so yeah sure

what happens if x = 0

you get

sin(t + cos(t)) = sin(1)

[languid radish]

sin(t + cos t)= sin(0+cos(0))

Yeah

[dark sparrow]

so t + cos(t) = 1 + 2kπ or t + cos(t) + 1 = π + 2kπ

yknow it looks like sin(x + cos(x)) should be 2π-periodic.

just from eyeballing.

maybe its period is smaller

[languid radish]

oh yeah I forgot about this thing for some reason
How’re you getting it from just eyeballing lol

[dark sparrow]

well i tried 2π

f(x+2π) = f(x) as it turns out

[languid radish]

Yeah

But how’d guess from eyeballing lol

[dark sparrow]

i mean

it looks like it should be 2pi-periodic

like

cos is

and so's sin

so maybe

and yknow if you plug it in

you get that it works

[languid radish]

Okie doke

Thanks

[upper karma]

I need help

Trigonometry problem

Trigonometry function

Its in Portuguese tho

It wants me to calculate the fundamental period of the function

Hope I translated that correctly

G(x) = sen(3x+(pi symbol)/4)

How do I calculate the fundamental period from that

I know I gotta use the 3 somewhere

And I know that the function repeats itself at 2pi

Please ping me if you know the answer

[upper karma]

@upper karma i assume you mean sin?

and to find the period of the function you should use this: iirc

you know that b is 3

Oh my god thank you

Just in time

Youre a life saver

no problem

[median crown]

No problem

[inland canyon]

hello, i have a vector problem which i haven't been able to solve for about 2 hours, perhaps someone could help. https://i.imgur.com/qAaGxgA.png

[errant nebula]

Can anyone help me with proofs (starting geometry)

[dusky tiger]

can anyone help me with Proving Angles Congruent?

[devout shell]

we have to see the givens first

[formal raptor]

Anyone able to help with some trig

[dark sparrow]

[formal raptor]

I have this question

[dark sparrow]

uh huh...

[silent plank]

where are you stuck?

[formal raptor]

I’m not sure how I’m supposed to find the larger acute angle. Like it’s really confusing to me

[silent plank]

do you understand what the larger acute angle is?
can you find the smaller acute angle?

(in a right triangle)

||a right triangle has a right angle and 2 acute angles||
||the larger acute angle is the larger of the two acute angles||

[formal raptor]

Yea I understand that and I’ve solved a smaller acute angle problem before

Is that how I would solve this one?

[silent plank]

what's the angle sum of a triangle?
what's the angle sum of the 2 acute angles?

[formal raptor]

I think it’s 180?