geometry-and-trigonometry

[dark sparrow]

$\frac{\theta}{2} = \frac{\pi}{3}$

[somber coyote]

Ann:

[dark sparrow]

from here, they multiplied both sides by 2

do you understand how that line was arrived at? (you should, unless you lied in response to my previous question)

@oak minnow

[oak minnow]

yea

im reading

what do you mean by (you should, unless you lied in response to my previous question)

[dark sparrow]

i mean exactly what i said

[oak minnow]

what?

multiply by 2

[dark sparrow]

i asked you to point out the first thing you do not understand. if what you pointed at really is the first thing you don't understand, then it must be the case that you understand everything that came before it.

[oak minnow]

im lost

why 2pi/3 though

[dark sparrow]

$\frac{\theta}{2} = \frac{\pi}{3} \ 2 \cdot \frac{\theta}{2} = 2 \cdot \frac{\pi}{3}$

[somber coyote]

Ann:

[dark sparrow]

that's why

[oak minnow]

but where did 3 come from in the first place

[dark sparrow]

... you should've said you didn't understand how the text arrived at θ/2 = π/3 ffs

[oak minnow]

ok, so answer it

I don't understand how the text arrived at θ/2 = π/3

[dark sparrow]

ok finally we're getting somewhere, at least we're in the clear as to what needs explaining.

okay so the line just before that is $\cos(\theta/2) = \frac{10}{20}$

[somber coyote]

Ann:

[dark sparrow]

in other words, $\cos(\theta/2) = \frac12$

[somber coyote]

Ann:

[oak minnow]

yeah 10/20 reduces to 1/2

[dark sparrow]

so

what angle has 1/2 as its cosine?

it's one of those "easy" angles whose sin and cos values have simple-ish expressions

[oak minnow]

cos30

wait

nvm

[dark sparrow]

if you name this angle in degrees you'll need to convert it to radians afterwards

also, if you're doing it in degrees, make it abundantly clear you're doing that

i.e. put the degree symbol

[oak minnow]

can i times it by 180/pi

[dark sparrow]

you can't "times" it, because "times" is not a verb

[oak minnow]

multiply

[dark sparrow]

if you're asking whether you can multiply your angle by 180/pi, then of course you can! nothing is stopping you. you can also multiply it by 10, by -5sqrt(37) or by 58.2234.

which is to say, you can do anything you want to anything as long as you know what the result will mean in your context

[oak minnow]

no what i meant was can multiply that 180/pi to get my answer

[dark sparrow]

will multiplying your angle in degrees by 180/pi give you the same angle but in radians?

[oak minnow]

i hope so

because i dont know how to convert to radians

ok going back to your question

o what angle has 1/2 as its cosine? it's one of those "easy" angles whose sin and cos values have simple-ish expressions

[dark sparrow]

180 degrees = pi radians

[oak minnow]

ah im confused

[dark sparrow]

okay before we go back to the text

we need to clear up your confusion regarding radians

are you still here?

[oak minnow]

ye

s

@dark sparrow

[dark sparrow]

ok

[oak minnow]

if 7 is my radius

[dark sparrow]

wai T

sgsdjkg

[oak minnow]

and 6 is the chord length

[dark sparrow]

are you throwing another problem in already

holy shit

[oak minnow]

no

well yes

[dark sparrow]

okay no

[oak minnow]

i want to ask something

would i have to multiply by 6

[dark sparrow]

multiply what by 6

can you state the entire problem you were going to ask about

[oak minnow]

in the example its cos(theta/2)=10/20

[dark sparrow]

ok what the hell

you're going to confuse both yourself and me here

are you trying to sidetrack the conversation to another problem, unrelated to what we were discussing just now?

[oak minnow]

no its completely related to this

did you not see my question i posted

I want to know how to do it

[dark sparrow]

do you want me to just give you a formula and nothing else

[oak minnow]

formula would he nice but i also want to know how to do it,

[dark sparrow]

how does knowing how to do it differ from knowing the formula?

knowing a formula implies knowing how to apply it and what each variable involved in it represents

[oak minnow]

ok go on tell me what to do and not do with this problem

[dark sparrow]

honestly no i've just about had it with your inability to respond to simple requests

i don't want to help you and have no real obligation to

[oak minnow]

likewise,

i would of preferred someone else to help

[upper karma]

@dark sparrow was wondering if you could help me...

I put the answer D, but it's C...the thing is

if A = CD/2

then the ratios should be the same, since they're similar

why isn't ED = 5/2?

[dark sparrow]

$AD = \frac12CD$, but the scaling factor between triangles $AED$ and $ABC$ is not $\frac12$, because $AD \neq \frac12 AC$

[somber coyote]

Ann:

[upper karma]

OH!

And how are we supposed to notice that it is 1/3?

[dark sparrow]

DC = 2AD

AC = AD + DC

[minor arch]

https://i.imgur.com/1pkxQB5.png

For problem 94. I get the answer

\frac{2\sqrt{7}}{7}

[dark sparrow]

that is correct

[minor arch]

Im shocked you actually read what I was trying to type.

https://i.imgur.com/1WViOY7.png

But yeah. Is it odd then that when I input it into the calculator, the value that's given is different from the value of csc?

Or do the different trigonometric functions always supposed to different numerical values?

...

Dumb question.

Of course not.

I was losing my mind for nothing. Obviously they wont have the same value when switched.

[hallow smelt]

does this equal Cot and Tan at the same time?

[dark sparrow]

what?

[hallow smelt]

I don't dispute that if you do the additive formula for sin over cosine you get cot

but when I use the tan subtraction formula I get tan(x)/1

0-Tan(x)/ 1+0

[dark sparrow]

you what

tan(π/2) isn't 0

[hallow smelt]

oh wait ya its undefined

[dark sparrow]

at best tan(π/2) is unsigned infinity

[hallow smelt]

okay thanks that explains it

guess you don't actually have to memorize the tangent additive and subtraction formula

if you can just sin cosin it

[dark sparrow]

you don't have to, but they are handy occasionally

[slate pewter]

i can know the radius and the diameter

becuse the circumfrnce is 60

so the raidus is 10

becuse 10 * 2 is 20

20 * 3 is 60

in the quistion they said list took the PI is 3

not 3.14159

and becuse we can know the radius

we can know all the area

we can make it

the radius is 10

10 sqared is 100

100 * 3 is 300

but i dont know how to do the rest

[copper valve]

since a circle has 360°, and a right angle is 1/4 of 360°, the area of this region will be 1/4 of the area of a circle with the same radius

oh woops

thought it was right

it's 120°

but same procedure

[slate pewter]

if it right angle its easy

[copper valve]

you divide

120°/360°

to find how much of a circle you have

[slate pewter]

= 3

[copper valve]

not

no*

1/3

[slate pewter]

oh yes

so do you mean

the 300 is also 1/3

so the answer is 100

[copper valve]

what's the formula for area of a circle

[slate pewter]

Pi X sqared

[copper valve]

what's X

[slate pewter]

any numper

you sqare the numper and multiply it by PI

[copper valve]

no, it has a certain meaning

with respect to the circle

X is the radius

[slate pewter]

yes

i did say it wrong

but i know how to find the area

of the circle

for a given radius or diameter

[copper valve]

what's the radius here

[slate pewter]

raidus need to be a 10

becuse the circumfrnce is 60

[copper valve]

right

[slate pewter]

10 * 2 = 20

20 * 3 is 60

[copper valve]

what are you multiplying for

[slate pewter]

nothing

[copper valve]

ok

[slate pewter]

i am saying is the circumfrnce fourmela is 2PiR

this is what i mean

[copper valve]

oh yeah

ok

[slate pewter]

they gived the circumfrnce

we can know the raidus

[copper valve]

then yeah

[slate pewter]

so radius is 10

10 sqared is 100

100 * 3 is 300

[copper valve]

yea

[slate pewter]

the area of all the circle is 300

you said

120 from 360

is 1/3

do you say is the area also need to be

300 1/3

[copper valve]

yes

[slate pewter]

so the answer is 100

100cm sqared

[copper valve]

yea

[slate pewter]

ok thanks

[copper valve]

np

[slate pewter]

thank you see ya

[viral needle]

This is wholesome :D

[slate pewter]

i didnt understand what you mean

@viral needle

wholesome

[viral needle]

Good luck with any others @slate pewter

[slate pewter]

i make things bigger

XD

[devout shell]

what command gives that?

[slate pewter]

yes kawwwwwai

its cuuuuteeee

[devout shell]

[viral needle]

[craggy ridge]

can anyone explain to me why sin b is positive when sin theta is negative?

[upper karma]

Related acute angles are always positive

They refer to the angle formed by the terminal side and the x-axis

@craggy ridge

[craggy ridge]

ohhh okay thank u!

[cloud void]

Simple and easiest way to calculate TG without a calculator?

[warm forge]

1. a triangle, if the numbers are simple
2. maclaurin series expansion

[hallow smelt]

I'm having trouble understanding the reasoning here

[sleek imp]

They defined s as x+y then plugged into the equation with s expanded out with t and got the values

[waxen vale]

Could someone help.

Im aware I have to use 1/2absinC
So I rearranged the formula to 1/2 ab sin 110 = 85

But it's an isosceles triangle so I dunno how it works

[devout shell]

This is actually a nice way to apply some geometry

So if this triangle is isosceles, what is true about the angles at the triangles base?

@waxen vale

Law of Sines: $\frac{A}{\sin(A)}=\frac{B}{\sin(B)}=\frac{C}{\sin(C)}$

[somber coyote]

⚡Amphy⚡:

[waxen vale]

They would be equal, so both being 35

[devout shell]

Let's look at AC and AB, you can apply the law of sines here since you know the angles and what else so you know about AC and AB?

wait huh...that tells us nothing lol

[waxen vale]

yeah

[devout shell]

take a new approach

so we are given the area

so drop a line segment from point A, that will split the triangle in half

[waxen vale]

I tried that

[devout shell]

huh, let me think a bit more about this then lol

[waxen vale]

Oh wait

I think I got it

You do 1/2 * a * a * sin (110) = 85

a = b since it's an isosceles

so you'd get a^2 = 170/sin(110)

Anyway, thanks for taking your time to help.

[devout shell]

that's good then lol

[raw prism]

<@&286206848099549185> can someone please do number 2. I tried it many times but can’t seem to get the right answer

[devout shell]

,rotate

[waxen gorge]

,rotate 90

[devout shell]

what identities have you applied so far?

[raw prism]

got to sin 2x/cos 2x = tan2x

[devout shell]

upon first glance the 3cos(2x) will cancel

[raw prism]

but it equals out to 3tan2x -3

i think i got it

tan 2x = 3

but i still cant do the second part to it

[dire rampart]

how did u get 3

[devout shell]

you can use the trig ratios to figure out angles and then can use those angles to find the length of AC and AD, then x=AC-AD

[empty trout]

@devout shell Cant I just used the pythagorean theory to find line AD then I use it again to find line AC then I just subtract

[devout shell]

HAHA you're right lol

please use that method then

Sometimes I don't see the easier method lol

[empty trout]

thx

I was asking cause my friend did what you did

and I was confused

[devout shell]

you can use Law of Sine and Cosines to solve this

[upper karma]

I have a question regarding vertices, vectors, and graphics

How would I take a position and size vector and find all of the vertices necessary to make a rectangular prism?

I've heard of this, but I'm not completely sure it's the solution:

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

[heady juniper]

How does one solve Sin(apple) = Sin(potato)

i think there was a general method

where you considered if it was in the same quadrant

or different quadrant

and u added some kpi or 2kpi to it

but ive forgotten what that mehod was

[dire rampart]

apple =potatoe +2kpi

[plucky marlin]

[dark sparrow]

nope that's not a complete description

sin(x) = sin(y) iff x = y + 2kπ OR x + y = (2k+1)π

[hallow smelt]

is there a reason they do that extra step at the end?

[dark sparrow]

not really? they could've just as easily factored out say 1/8 instead

[plucky marlin]

i dont get the x + y = (2k+1)π part tho

@dark sparrow can u explain maybe ?

[dark sparrow]

x = (2k+1)π - y

= 2kπ + (π - y)

[dire rampart]

oh yea tru

[hallow smelt]

is there a reason they don't just stop at -.322+Pi(K) and find a positive angle?

because for Tan(X)= - Root 3 they just stopped at -Pi/2 + Pi(K)

[muted sparrow]

Is anyone aware of an algorithm for classifying a volume formed by set operations of simple primitives (half-spaces, quadrics) as bounded/unbounded?

[dire rampart]

@knotty matrix stick to one channel

[knotty matrix]

sorry

[upper karma]

I need help verifying the identities

[valid pike]

#❓how-to-get-help

[rain charm]

@upper karma

[surreal bolt]

In a 2D coordinate plane, the point (x, y) is part of a lattice L iff x and y are both integers. Pick any three points A, B, and C in the lattice L. Show it is impossible for angle ABC to have a measure of 30 degrees nor a measure of 72 degrees. Show further that the cos^2 (mABC) must be rational (thus again proving 72 degrees cannot be the measure of angle ABC).

... for fun if anyone is interested. 😃

[dark sparrow]

the second thing you ask to show is trivial

[surreal bolt]

Is it?

I'd like to know if so.

[dark sparrow]

$\cos(\angle ABC) = \frac{AB^2 + BC^2 - AC^2}{2 \cdot AB \cdot BC}$

[somber coyote]

Ann:

[dark sparrow]

AB, BC and AC are each the square root of an integer

[surreal bolt]

okay fair enough. You're using law of cosines basically?

[dark sparrow]

that is exactly what i am using

[surreal bolt]

Okay. cool.

I'll keep that in mind.

Thank you 😃

[dark sparrow]

to prove that no triangle with vertices in Z^2 can have 30° as an angle it suffices to show the stronger statement that no triangle with vertices in Q^2 can have 30° as an angle

[surreal bolt]

Hmm is that well-known already?

[dark sparrow]

it's not hard to prove

[surreal bolt]

OK and well a similar problem:

Let the XY-plane be tiled / tessellated with equilateral triangles. Let L be the set of all vertices. Pick any three points A, B, and C in L. Show it is impossible for angle ABC to have a measure of 45 degrees.

That's all I have today.

[dark sparrow]

right

this is most easily done by identifying the plane with C

so that the lattice becomes the set of all numbers of the form $a + b \omega$, where $a, b \in \bbZ$ and $\omega = \exp(2i\pi/3)$

[somber coyote]

Ann:

[dark sparrow]

actually this has a name

the eisenstein integers

if one can show no number in L has pi/4 as its argument, then the required result follows

[surreal bolt]

Ah cool. I have not heard of that.

I used ... geometry and trig.

Would your solution be considered "complex analysis" or just geometry using the complex plane?

[dark sparrow]

it's not complex analysis by any measure

[surreal bolt]

But yeah. ABC dilated about point B by a factor of 2 will land A', B' and C' all on a rectangular grid of 1 by root(3). And some work with a general angle with a bit of law of cosines solves it nicely.

wikipedia-ing Eisenstein's integers.

Thank you.

[young hemlock]

can someone help

[dark sparrow]

what's holding you up here?

[mortal socket]

ez

[young hemlock]

uh

[dark sparrow]

@mortal socket please don't give out the answer.

[young hemlock]

i dont know how to find the perimeter and area of this specific shape

[mortal socket]

get the perimeter of the rectangle

[dark sparrow]

this shape has been broken into three pieces for you

[mortal socket]

and get the permiter of half circle

[dark sparrow]

and its perimeter into four

[mortal socket]

it got law

[young hemlock]

into 4?

[dark sparrow]

yes

[young hemlock]

ok i see it

so do i work out the perimeter and area of the different pieces?

[mortal socket]

it is 2 pi r /2

[dark sparrow]

yes, do it piece by piece.

[young hemlock]

ok

[mortal socket]

yah

[dark sparrow]

that's how you find the perimeter and area of any shape that can be broken down into simpler pieces.

[young hemlock]

@dark sparrow so if i want to find the perimeter of the semi circle on the right i do pi times 8 plus 2 times 8 right

[dark sparrow]

what

the diameter (which is 8 anyway, not 2*8) isn't part of the shape's perimeter

[young hemlock]

so how would i calculate the semicircle on the right

[dark sparrow]

if i asked you to find the circumference of a full circle of diameter 8 cm, would you be able to do it?

[young hemlock]

yes

[dark sparrow]

okay

so what about a semicircle of diameter 8 cm, what is the length of its arc?

@young hemlock

[young hemlock]

uh

[dark sparrow]

don't overthink it

it's almost right there in the name

[young hemlock]

half times pi times 8 plus 8

[dark sparrow]

no!

no! i asked you for just the arc, but you keep including the diameter!

but the diameter isn't part of your shape's border!

[young hemlock]

pi times r

[dark sparrow]

4π. the length of that semicircle arc is 4π.

1/2 * π * 8.

[young hemlock]

oh

[dark sparrow]

you kept including that +8

[young hemlock]

i have a question

Why do we not include the diameter?

[dark sparrow]

look at the shape

[young hemlock]

You said break it up into pieces

[dark sparrow]

the diameter of that semicircle is not part of the border of the whole shape.

[young hemlock]

But I'm finding a seperate shape

[dark sparrow]

okay look

for AREA, it's fine to imagine like cutting your shape up into pieces

but for PERIMETER, you'll want to cut only the border

because the perimeter of a shape is the length of its border

[young hemlock]

I see

[dark sparrow]

this should be common sense

[vast garnet]

Consider the triangle ABC where $AC = 110$, $BC = 102$, $\angle ACB = 60$.
Note that segment AC is parallel to the north. What is the bearing of A from B?

\begin{align*}
&AB^2 = 110^2 + 102^2 - 2\times 110 \times 102 \times \cos(60) \implies AB = 106.2261737991\
&\frac{\sin(\angle CAB)}{102} = \frac{\sin(60)}{106.2261737991} \implies \angle CAB = 56.26044354
\end{align*}

Thus, the bearing of A from B is
$$360 - (180 - CAB) = 236.26044354$$

Could someone tell me if this is correct pls?

[somber coyote]

CabbageGuy:

[heady thunder]

How do i calculate the Points A and B, when P and alpha is given?

[flint compass]

Why is b=2pi/period? I’ve seen how you derive it, but why is it like that?

[flint compass]

<@&286206848099549185>

[upper karma]

You can re-arrange that to get this

$Period = \frac{2 \pi}{b}$

[somber coyote]

Starly:

[upper karma]

Because b is some constant that is multiplied by input x, that means when b > 1, the period will be less than 2*pi

You could say that the period is inversely proportional to constant b

[flint compass]

But what could the ratio be thought of? When b=2pi/period then could it be though of as 2pi is one revolution and the period of how long it takes?

[gritty flare]

I think it would be that modification of input a squishes a function

Thus lessening the period

,w graph x^2 and 4x^2

[somber coyote]

Display results online and refine query

Requested by Solarkoid#3482

[gritty flare]

You see 4x^2 is (2x)^2 which is x multiplied by 2 so f(2x)

And is a little squished

[flint compass]

I still don't understand it. I'm not really looking for a complex Why is the period 2pi/b?
explanation. Just some way to visualize it or think about it

I think it would be best if I could think of B as a ratio. I've been trying to think of it as 1 revolution of the circle per period but then when I try to think of period as P=2pi/B I can't seem to wrap my head around what type of ratio it could be though as.

<@&286206848099549185>

[umbral snow]

e^(it) has a period of 2π. This is of course because
e^(it) = cos(t) + isin(t)

@flint compass

Or are you just talking about the period of sin(bt)?

[flint compass]

@umbral snow, Sine.

[umbral snow]

sin(t) has a period of 2π
sin(t/b) is a horizontal stretch by b, multiplying the period

So sin(t/b) has a period 2πb

[flint compass]

Why does sin(t/b) stretch the period?

I mean I understand that dividing t/b makes it longer but I don't really understand why that happens.

[umbral snow]

You have to plug in a number that is twice as large, in order to get sin(t/2) to get to the same place as sin(t)

sin(4/2) = sin(2)
So getting to t = 2 in sin(t)
Is getting to t = 4 in sin(t/2)

[flint compass]

Is there another way to think about the period? Like 2pi/B where 2pi is one revolution and B is something?

[upper moat]

So I have 2 problems of a certain “genre” that I can’t solve for the life of me.....
How would I solve these, and all problems of their kind in general?

1. Given a set of coordinates (x_1, y_1) to (x_n, y_n), at which coordinates should you place 2 “node” points such that the sum of the squared distances of each point 1 to n to the nearest node point is minimized? then, what is this minimized sum?

2. Given a set of coordinates like above (with y-coordinates all > 0), and m “node” points with m < n, place all node points on the line y = 0. Each point 1 to n and its nearest node point defines the diagonal corners of a rectangle. What is the minimum sum of all the rectangles?

Sorry if that was kinda unclear - ping me and I’ll clarify anything

Computation is allowed for these problems

[upper karma]

Heya

Trying to solve this but, honestly, don't know where to begin

The topic being circles and lines secants and tangents.

Perhaps something like.. (360 / 2) - 118 = (-11x+7)

[acoustic vector]

Central angle and the degree a sector is are the same

so label the central angle of -11x + 7 as c degrees

c and 118 make a straight angle

so 118 + c = 180

c = 62

So then -11x + 7 = 62

-11x = 55

x = -5

[minor arch]

https://i.imgur.com/ICsIN4c.png

Can anyone explain #47 for me?

book never explains how I should find the equation

[devout shell]

Thinking back to the unit circle and the relationships the exist with x,y and sine and cosine, what is the exact value of sin(30)? cos(30)? and how is that related to x and y?

[minor arch]

I haven't actually made it to the unit circle yet. Guess its just odd placement.

Unit Circle is in chap 6.

this is chapter 5.

I do know the values of sin(30 tho. and cos(3)

[devout shell]

oh...well you know about how x=rcos(theta) and y=rsin(theta)?

ok that is very good then

[minor arch]

sin(30) = 1/2
cos(30) = root(3)/2

[devout shell]

so on a circle of radius 1, x=cos(30) and y=sin(30) from the relationship I just told you

so now you have a coordinate that is on that line

$(\frac{\sqrt{3}}{2}, \frac{1}{2})$ is a point on that line

[somber coyote]

⚡Amphy⚡:

[devout shell]

the problem also says the line passes through the origin

so can you see how to find a line with this info?

[minor arch]

slope would equal

y = mx

so it'd be . .

$(\frac{1}{2}}

(I failed

[devout shell]

missed a dollar sign at the end

just add that in

[minor arch]

(dont know the rest of the rules for the syntax. where can I see it?

$(\frac{1}{2}}$

[somber coyote]

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

[devout shell]

you used } instead of ) at the end

[supple abyss]

$\frac{1}{2}$

[somber coyote]

CaptainLightning:

[minor arch]

But yeah, 1/2 = m(root(3)/2)

No?

[devout shell]

yes, so then m= what?

[minor arch]

1/root(3)

rationalize it

root(3)/3

[devout shell]

so the line's equation is...

[minor arch]

That kinda makes sense.

I sorta get it, but I'll come backk to this question when I'm done with unit circles.

[devout shell]

$y=\frac{\sqrt{3}}{3}x$

[somber coyote]

⚡Amphy⚡:

[minor arch]

Yeah makes perfect sense.

I imagine I could do the same with

60 degree angles.

[devout shell]

I mean I personally would have written it as $y=\frac{x}{\sqrt{3}}$ to reduce the clutter

[minor arch]

or rather, the 60 degree angle that forms.

[somber coyote]

⚡Amphy⚡:

[devout shell]

yes, that is the same idea you can use for angles that you know that exact values for

[minor arch]

Yeah. I don't like to rationalize it unless its required for a problem. makes using the identities

a pain in the ass

since I sometimes forget the rationalization changes the outcomes.

[cloud void]

4sin(a)=2sin(60

A?

[devout shell]

Well, what is sin(60)?

[mortal socket]

how did we know the sin, cos, tan of the 90

[plucky marlin]

using unit circle ?

tan(90) is weird and disgusting

[mortal socket]

tan 90 isn't possible

but i don't understand yet

can you send a picture of drawin or explanation @plucky marlin

[plucky marlin]

behold

the unit circle

[mortal socket]

this is the legendary unit circle

[plucky marlin]

it sure is

[mortal socket]

[plucky marlin]

ik

uwu

[viral needle]

Forgot how glorious it was

[steady sleet]

$\sin(\hphantom00^\circ)=\frac{\sqrt0}2=\cos(90^\circ)$\~\$\sin(30^\circ)=\frac{\sqrt1}2=\cos(60^\circ)$\~\$\sin(45^\circ)=\frac{\sqrt2}2=\cos(45^\circ)$\~\$\sin(60^\circ)=\frac{\sqrt3}2=\cos(30^\circ)$\~\$\sin(90^\circ)=\frac{\sqrt4}2=\cos(\hphantom00^\circ)$

@dark sparrow berhaps pin this and unit circle

[somber coyote]

SAckalope:

[plucky marlin]

[idle bloom]

Oh that's an cool pattern

[viral needle]

[upper karma]

Hello

I have a stupid question.

Say you have 2 points AB.

You connect them, forming a line segment

It just so happens, those two points are on opposite ends of a circle

Does that mean that AB can be

both a diameter

and an arc

?

[flint compass]

If Asin(x) can be visualized as A changing the circle’s size then what can sin(Bx) be visualized as?

[astral hornet]

Well, b is the frequency

[flint compass]

So could it be visualized with a circle?

[upper karma]

Is the secant technically a chord in the parts inside the circle?

[sinful nexus]

@upper karma If you think about it geometrically... Think of the graph of f(x)=cos(x)... One of the points it intersects the x-axis at is pi/2 (90 degrees).

Another way to think about it is with the unit circle... When the angle from the x-axis is 90 degrees and you let (x, y) be (cos(x), sin(x)), then at 90 degrees, the x-coordinate (cos(x)) is 0.

I hope that makes sense.

[upper karma]

I am having trouble understanding how to graph tan and cot

y=tan4x

my period is pi/4 right?

Have really hard trig related question in #help-5 thank you for any help.

[sinful nexus]

@upper karma EDITED: Correct, f(x)=tan(x) has a period of pi. So g(x)=tan(4x) has a period of pi/4 because you can fit tan(4x) four times over the same period (pi) as tan(x). So a period of pi/4 compresses tan(x) by a factor of 4.

I hope that makes sense. Sorry I'm on mobile and don't know how to use the TeX bot yet >.<

@upper karma No worries, I'm glad you understand!

@upper karma Sorry I can't look at it atm as I'm at work, hopefully someone else can help :)

[minor arch]

https://i.imgur.com/9BzjSmE.png

Can anyone explain to me why#148? I imagine that the answer is that the larger the cosign, means the less deduction there will be from the product?

But I know that has to be wrong. It's too simple of an answer.

[upper karma]

Still needing a trigonometry wizard

[dark sparrow]

repost your q

[upper karma]

in here?

!TRIG!
Hello, sorry this question is kind of hard.
Given the conic section: 5x^​2 + 6xy + 5y^​2 - 8 = 0

A) Discriminant: ellipse -64
B)Angle of rotation: 45 or pi/4
C)equations for x' and y'
x=x'cos(pi/4)-y'sin(pi/4)
y=y'sin(pi/4)+ycos'(pi/4)

D) Now I need the standard equation which I will attach where I'm stuck.

https://cdn.discordapp.com/attachments/486844829209198613/573333274982875148/20190501_221742.jpg

[upper karma]

nvm

[plucky solar]

does anybody have any ideas for proving this:

sin(2*alpha) + sin(2*beta) + sin(2*gamma) = 4*sin(alpha)*sin(beta)*sin(gamma)

alpha, beta and gamma are angles in a triangle ( so alpha + beta + gamma = 180deg )

[dire rampart]

hmmm

maybe take sin(a+b+y)=sin180=0, expand lhs and simplify

[plucky solar]

finally figured it out

[dire rampart]

epic

[upper karma]

lol

Why can't one just...sum 6+2

Which would give the diameter

and divide by 2...

[gaunt vine]

i need help

in math

[plucky marlin]

[dark sparrow]

@upper karma because you do not know whether that horizontal line is a diameter

[steep mantle]

How do you go from sin(5t+x) = 0 to
5t+x = pi?

@dark sparrow

😄

[dark sparrow]

needs more context. in general the solutions of sin(x) = 0 are x = kπ where k runs over the integers

[umbral snow]

Likely looking for the first non-zero solution?

[steep mantle]

ill post the whole chegg solution

:c

[covert rune]

@upper karma How do you know 6+2 is the diameter?

[upper karma]

@covert rune it just looks like itd be more or less

🤷

[covert rune]

Length of C is unknown though

Visual references aren't supposed to be measured for reference, you need to do the math

[neon void]

how do i find C when i only know B and b

C = 71.5 degrees and b is 594 meters

[dark sparrow]

context?

[neon void]

in a triangle

can i use sine rule in a non 90 degree triangle

ive got c

aswell

[dark sparrow]

wait

so what do you have

B, b and c?

[neon void]

yep

[dark sparrow]

3 elements, great

then C can be found using the law of sines, which applies to any triangle

[neon void]

ok even when its non 90 degree

so this is my situation atm

i found arcsin(3/4) or something

[pulsar delta]

@everyone can someone help me take my math test

[devout shell]

Lol nope

That’s against the rules anyways

[dire rampart]

[floral stratus]

and u shouldnot do @ everyone for tht

[daring burrow]

How do you prove that the graph of a linear equation is a line? (using high school maths)

[plucky marlin]

equation of a line looks like a line

[dark sparrow]

you need a definition for straight line then lol

[daring burrow]

i meant mathematically

a straight line is a series of points in any one direction and the opposite direction ig..

[iron nest]

I believe a straight line is a line with fixed gradient?

In which case, since the derivative of any linear function is a constant, which is fixed, it is a straight line?

[dark sparrow]

oh god no

[plucky marlin]

[astral totem]

Just a noob asking for help in #help-1

[mossy vine]

@upper karma AB isn't necessarily the diameter

and in fact probably isn't

[upper karma]

@upper karma oh, cool

Do you know of a theorem which is; 4r^2 = a^2 + ... + d^2

It works with that specific case

@upper karma no hahaha it works here

So like

so for example

Set y to be below c

Then y, by point theorem, is said to be y(3) = 6(12)

so y = 4

And a theorem says

4r^2 = 2^2 + 6^2 + 3^2 + 4^2

4 + 36 + 9 + 16

,calc 4+36+9+16

[somber coyote]

Result:

65

[upper karma]

,calc 65/4

[somber coyote]

Result:

16.25

[upper karma]

,calc sqrt(16.25)

[somber coyote]

Result:

4.0311288741493

[upper karma]

And 4.03 is the radius of the circle

🤔 unit

wdym

unit youtube bruhhhhh

oh

10months in and I'm learning the radius of a circle

😂 😂 😂 😂

thats what we are doing school wise

lol

its a theorem but Ive not been able to find it

youtube power bruh

[eternal flare]

ow do I know when to change cos^2 x-sin^2 x to cos2x, and vise versa?

How do I know when to multiple by conjugate?

How do I know when to change the '2' in a 2-sin^2 x to get to 2sin^2 x + 2cos^2 x - 2sin^2 x to get 2cos^2 x?

I know the simple steps like change them all to sin/cos, factor, etc. But how do I know when to do these other things?

How do I know when to use what when I solve trig ids

[upper karma]

to know how, you need to know why.

Why do you multiply by the conjugate?

[eternal flare]

I really have no idea

My teacher didn't explain, she just said do it

She said it was to make a 1-cosx in the denominator to a 1-cos^2 x to then change to a sin^2 x

[dark sparrow]

can you show the problem this is from

[eternal flare]

It's from a variety of problems, but mainly this

They're my friend's

In the first question right side, he changed 3 to 3sin^2x+3cos^2x and -sin^2x to 1-cos^2x

When and how do I know when to do stuff like that

[dark sparrow]

honestly there is no One Rule

[eternal flare]

Then how would u know when to do it?

Is there like a list of steps?

[dark sparrow]

experiment lmao

there is rarely only one way to solve a problem

[eternal flare]

Oh ok

Thanks

[acoustic peak]

I have a quadratic formula and I need to find it's vertex.

I've forgotten how parabolas work..

$$2x^{2}+16x$$

[somber coyote]

Deceive:

[compact shuttle]

Vertex of a quadratic formula (y=ax^2+bx+c) is at x=-b/(2a)

In this case a=2 and b=16

[acoustic peak]

Okay

So the vetrex of this one would be -16/4?

[umbral snow]

@acoustic peak
You can also factor it as
2x(x + 8)

It has a root at x = 0, and a root at x = -8, so the vertex is in the middle at x = -4

[acoustic peak]

Don't you mean 2x(x+16)?

[umbral snow]

Sorry, all better now

[acoustic peak]

It's cool

We were both dumb

Let's pretend it never happened.

[umbral snow]

What happened?

[acoustic peak]

Idk

Anyways, the y point?

[umbral snow]

Plug x = -4 into y = 2x² + 16x

[acoustic peak]

Oh yeah

I don't know why I have an issue with that

Remembering that I can plug it in

[umbral snow]

You can also do all of this by completing the square, which is worth knowing

[acoustic peak]

Completing the square?

[umbral snow]

You can write
2x² + 16x = 2(x + 4)² - 32

That vertex form tells where the vertex is, at (-4, 32)

[acoustic peak]

Hmm..

I get where the 32 comes from, but not why it's negative.

Nor why it's subtracted from the quadratic equation.

[umbral snow]

Sorry, the vertex is at (-4, -32)

,calc 2(16) + 16(-4)

[somber coyote]

Result:

-32

[acoustic peak]

Could you explain how you got the roots?

Or did you just look at Desmos?

[umbral snow]

@acoustic peak
Factored it
2x(x + 8)
2(x - 0)(x - (-8))
Root at 0, root at -8

[acoustic peak]

That's some next level factoring.

[delicate perch]

hey everyone!

i'm probably an idiot for not knowing how to do this

but can someone help me with this?

just ping me, i'll be here

[devout shell]

@delicate perch do you know where the center is?

[delicate perch]

(5,7)

it shows

[devout shell]

radius?

[delicate perch]

isn't the radius 2?

[devout shell]

yes, because the radius is the length from the center of the circle to any point on the circle

can you show me the options you can choose from for a and b?

[delicate perch]

sure

[devout shell]

oh man lol

[delicate perch]

haha

[devout shell]

let's work through this then

[delicate perch]

alright sounds good

[devout shell]

so there are two options really

we have the extreme point at 7 and then there's the center which is at 5

[delicate perch]

yes

[devout shell]

the value of a...well we see that we have a point x

that point x is actually the sum of two smaller distances

can you see what those distances are?

[delicate perch]

?

[devout shell]

try this, so drop a vertical line from point x to the x axis

[delicate perch]

i don't think i can do that on the software

[devout shell]

just imagine it in your head

[delicate perch]

okay

[devout shell]

so we have this line that is x away from (0,0)

[delicate perch]

yes

[devout shell]

if you drop a vertical line from the center point you see that it is how far away from (0,0)?

[delicate perch]

yes

[devout shell]

how far away is it?

(what's the x coordinate of the center point)

[delicate perch]

5 units, correct? because it falls on the x-axis

[devout shell]

perfect

now is 5 greater or smaller than our distance x?

[delicate perch]

smaller

[devout shell]

so how much more do we need to add to 5 to get to the distance x?

you should be able to see it now, how much more horizontal distance does it take to get from 5 to x?

[delicate perch]

x-5? that's one of the answers

[devout shell]

perfect

you see that 5+a=x correct?

[delicate perch]

yes

[devout shell]

you can see how we got that?

[delicate perch]

ahh

yes

thank you

ik that this is probably a pretty simple problem but i got intimidated as soon as i looked at it

[devout shell]

try the length of b then

tell me what you get

[delicate perch]

okay

i'm not sure if this is correct but y-7?

[devout shell]

that is right though, but you see why that would be correct?

[delicate perch]

i think so

because if we take a horizontal line from the center, we arrive at the point (0,7).

[devout shell]

perfect

that is correct

[delicate perch]

oh

that's great

for the pythagorean theorem i'm assuming it would be (x-5)² + (y-7)² = 2²

[devout shell]

it would be that

[delicate perch]

okay! thank you so much

[devout shell]

notice that's actually the equation for the circle

[delicate perch]

it was probably a waste of time for you considering the simplicity of the problem, but looking at the answer choices startled me

yes, i do see that

[upper karma]

anyone can help?

[dire rampart]

angel

[upper karma]

Never mind the crappy grammar

The source of translation isn't exactly that proficient lmao.

[dire rampart]

do they mean which expression equals 3 when u plug in 72 for.theta?

[upper karma]

Not sure, maybe.

Keep in mind that's a no calculator for some reason

[dire rampart]

aight in that case it's either B or D

cuz 5 times 72 gives a nice number

,w 5*72

[somber coyote]

Display results online and refine query

Requested by the one n only#7133

[dire rampart]

now just plug in and check which one gives u what u need

[bitter onyx]

can anyone help me with a trig problem?

struggling on it a lot

Use the sum and difference identities to evaluate exactly. Find 2 angles that add or subtract to yield the given angle. Then use cos(A+B) or cos(A-B)
cos105°

[spark stag]

alright, so

there are some common angles which you should know the cosine of

like cos30°, cos45°, cos90°, etc.

can you think of a way to add two "common angles" together to get 105°?

(or subtract)

[bitter onyx]

i mean cos60+cos45 no?

[spark stag]

right

well, no

cos(105) = cos(60+45)

but thats not the same thing as cos(60) + cos(45)

do you know the cosine sum identity?

[lime flame]

<@&286206848099549185>

[chrome fiber]

what have you tried

[lime flame]

treated as a trapezoid with a hole

then as a rectangle then took out the square and triangle

how do u justify this?

[chrome fiber]

bad quality

[lime flame]

Assume angles are right and all sides are equal

r is the ratio of the area and perieter

Idk how to solve it

[chrome fiber]

its not 1?

since all side lengths are 1 you can easily find the area of the figure by adding the number of small squares

and perimeter is fairly simple as well

[lime flame]

Is there a faster method though?

As opposed to just simply counting the perimter and area

[solemn holly]

Can someone give me a hand with how am i suposed to calculate "transformation with a rotation angle of 45°"?

[dark sparrow]

it's a rotation by 45°

[cloud merlin]

You can use some basic trig

Wait a bit let me just do it now

[dark sparrow]

@solemn holly if the rotation were about the origin, would you be able to calculate that

[cloud merlin]

Ok I think I got it

Let me quickly check it

Ok so in degree mode this is the formula

[dark sparrow]

[cloud merlin]

I think in the end you multiple by the original y over x

Yea that’s it I’m pretty sure

Ok this is the full one

Again all of these work with 45 degrees but you can change it

Remember if you are using a calculator you have to use degrees

[devout shell]

You used 45 rad? Lol

[cloud merlin]

No degree

But this won’t work with that problem I just realized

[devout shell]

Nah, just use gradians instead

[cloud merlin]

Cuz it’s assuming the center point is 0,0

Give me some time I’ll make a new one

[solemn holly]

@dark sparrow probably not; this isn't something teached in schools here but it's needed for a scholarship i'm interested

I tried to look on Khan Academy video but i simply didn't understand it

And what i found on the internet where a bunch of formulas

So i'm asking here for help, including links to study

@cloud merlin i really appreciate your help

If it's best for you, you can send me a link for a material instead of explaining it all the way, i won't mind

And, thanks

[cloud merlin]

Ok this might work

Let me check

[supple abyss]

the fuck is going on in here

[cloud merlin]

Who knows

[solemn holly]

With a two dimensional surface, if we take (2, 1) as the center point and consider
a transformation with a rotation angle of 45◦, then point (3, 3) is transformed
into point ?

Is going on

Btw

Why the sin-¹?

[cloud merlin]

Inverse sin

Helps to know range angle

[supple abyss]

arcsin

[cloud merlin]

Nvm

I already have the idea I should have the answer at least tmr tho

[solemn holly]

https://www.quora.com/What-point-is-the-point-3-3-transformed-into-if-with-a-two-dimensional-surface-we-take-2-1-as-the-center-point-and-consider-a-transformation-with-a-rotation-angle-of-45

I found this

"The line from (0,0) to (1,2) has a slope of 2. The angle made with the x-axis is tan^-1 (2)."

What is "a slope of 2"?

Sorry, math isn't my forte and english isn't my first language

[dark sparrow]

what is your first language

[solemn holly]

Portuguese

[cloud merlin]

Ok quick question

Would you say that 2.8 is close to 3?

Cuz if it you do then this is probably the formula

[devout shell]

,rotate

[cloud merlin]

Yea

[dark sparrow]

god honestly

why even use arcsin

and arccos

[spark stag]

@solemn holly i believe its called declive in portuguese?

[cloud merlin]

Cuz then you get the side length

[spark stag]

its a value that represents how fast an output changes as an input does

for example, for the line y=3x+1, the slope is 3

[dark sparrow]

$x' = x \cos(\theta) - y \sin(\theta) \ y' = y \sin(\theta) + x \cos(\theta)$

[somber coyote]

Ann:

[cloud merlin]

I have homework to do so I’ll be back later

[dark sparrow]

this is for rotations around the origin

[spark stag]

for $y = \frac{-1}{2} x + 9$, the slope is $\frac{-1}{2}$

[somber coyote]

Namington:

[cloud merlin]

Wait is this a bit?

Bot

[spark stag]

yes im a bot

[devout shell]

Texit is a bot

[supple abyss]

is what a bot

[spark stag]

beep boop

[cloud merlin]

xD

[solemn holly]

@spark stag ohhh, thanks, that helped a lot

[devout shell]

beep boop

[cloud merlin]

Ok is there like a way I can show you the formula

Like explain or something

[devout shell]

you already showed us the formula on the white board though

[cloud merlin]

Do you guys understand?

[solemn holly]

Well, i'll have to study acrsin

[cloud merlin]

I don’t even know if it works fully though

[devout shell]

no because I don't even know what the original equation is lol

[solemn holly]

Because i've never seen it before

[cloud merlin]

Oh well study it it’s cool

[solemn holly]

And did you deduct the formula?

[cloud merlin]

Are you talking about inverse sin cos and tan?

Yea I did

Deduct means make right?

[solemn holly]

Yes

[cloud merlin]

Ok

[solemn holly]

That's impressive

Btw, let me give you the awnser

[cloud merlin]

Thanks but like I said idk of it works fully

Cuz it gives you like 2.89 or something but it should be 3

[dark sparrow]

no

deduce

deduce != deduct

deduct = subtract

[cloud merlin]

Anyways gtg do some hw

[solemn holly]

2 - 1/√2, 1 + 3/√2

Is the answer

Thanks for your help

[cloud merlin]

but whats the question?

like for example is it giving you a cordinate?

wait is the center point of rotation 2,1?

[solemn holly]

Yes, the center point of rotation is 2,1

With a two dimensional surface, if we take (2, 1) as the center point and consider
a transformation with a rotation angle of 45◦, then point (3, 3) is transformed
into point ________.

The question is "wich point will (3,3) be after the 45° rotation?"

[cloud merlin]

yea it should be transformed to 2.8 then something

im pretty sure

or 2.9

2.9 is y

x is 1.683

wait how does the other calculation work?

ans im pretty sure that x cant be more than 3

[cloud merlin]

yea because 2 plus the square root of 2 to the power of 2 and 1 is 3

meaning anything to the side of it is less

and 45 is less

[hollow crow]

I have k how do i get x?

2 lines are parallel

[plucky marlin]

180 - k = x

[hollow crow]

K. Thanks

[hallow smelt]

whats the point of polar coordinates?

[dark sparrow]

some things, particularly those having to do with circles or ellipses or highly symmetric curves, admit easier or more generalizable descriptions in polar than in cartesian

[gilded acorn]

So I had this question on a test and i didn't notice the arc congruence. So I tried to solve using law of cosines

(The part on the bottom is what I solved first to find the radius)

I can't find any errors, but the answer is 31, not 36.39. Am I doing something wrong or is the question just unsolvable this way?

[umbral snow]

@gilded acorn
What is the question?

[gilded acorn]

@umbral snow Finding the length of AB

[serene viper]

Question:
How can the terminal arm of *theta* lie on two different quadrants?
I checked the answer for the question, and the answer was quadrant 2 & 3. The given angle was cos(theta) = -5/12

[somber coyote]

Nichterwitz:

[solemn holly]

With a two dimensional surface, if we take (2, 1) as the center point and consider
a transformation with a rotation angle of 45◦, then point (3, 3) is transformed
into point __.

I used the point (2,1) as if it was (0,0) and (3,3) as if was (1,2) wich (using the formula) gave me a result

how do i transform that result in to the one it would be if done with the original points? or did i simply messed up?

The correct answer is that but with a
"2 - (what i found) on the x" and "1 + (what i found) in the y"

can someone please give me a help with this mess?

[minor arch]

https://i.imgur.com/5xuzhYO.png

How in the world did they get c?

oh wait. pi is 3.14

[keen aspen]

Ya

Gotta do some approximations

[upper karma]

https://i.imgur.com/8HJ5dTB.png

Can someone explain to me how they are arriving at those answers? When I attempt to simplify the complex fraction inside the square root, it comes out to sqrt(3/5)/2, not any where close to their answers...

[gritty siren]

cos(2t) = 1-2sin²(t)

[vocal swift]

hey guys

If I have a circle and 2 points on it

is it certain that the perpendicular line from the radius to the line between these 2 points will bisect the distance?

[upper karma]

its trivial

[upper karma]

so we have $OA$ = $OB$, right?

[somber coyote]

nydyгd:

[upper karma]

since both are radii

[vocal swift]

yeah

[upper karma]

and also OP is the common side

[vocal swift]

yeah?

[upper karma]

and its perpendicular to the chord

So by the SAS test of congruency we have...

$AP$ = $PB$

[somber coyote]

nydyгd:

[upper karma]

[vocal swift]

wait an isosceles triangle vertex bisector is perpendicular to the facing edge right?

[upper karma]

you seem confused

its given to be perpendicular, right?

[vocal swift]

no

[upper karma]

thats the given condition

[vocal swift]

I am trying to prove this condition

[upper karma]

read the question you asked a minute earlier

[vocal swift]

yeah sorry for the confusion

it goes both ways

[upper karma]

no probs

[vocal swift]

if it bisects the chord then it is perpendicular and vice versa

[upper karma]

so the triangles are congruent and your theorem is proved

[upper karma]

Geometry is so hard smh

[robust breach]

A contractor needs to put carpet in the hallway of a house, the diagram of the house is below. All of the sides of the figure are 4 feet long, except for the two longer sides that are each 8 feet long. All angles in the figure are right angles. What is the area of the hallway?

would it be 56?

[dark sparrow]

56? how'd you get that?

[robust breach]

I did this last night and its a day late so I cant really tell you how, I probably was really stupid and added the sides all up.

[dark sparrow]

that will not get you the area