precalculus

[red tree]

Ok what is the definition of domain that you currently have

[silk thunder]

base must be greater than zero

And must not equal 1

[red tree]

Errr, I said definition

like what is a function? What does domain of a function mean?

[silk thunder]

Domain means the x values possible

you mean what does domain mean right?

[red tree]

That is what I mean

Ok thank you, that's more or less what I expected (although it's a weak definition)

but that's on your teacher

Ok so you seem to already know how to start this problem

[silk thunder]

the domain of a function is the set of inputs accepted by the function

This is a better definition

[red tree]

yes

[silk thunder]

\sqrt{log(x+1)}

sorry i am new

i dont know how to use the symbols

[red tree]

$-\sqrt{\log(x+1)}$

[silk thunder]

Ye thank you and please add a negative sign before the square root

[obsidian monolith]

Icy001

[silk thunder]

Exactly thank you , now how do i find the domain of this function

[red tree]

Since you know the definition, you can at least try something with it

Try some inputs and see if the function accepts them

This will help your intuition, and maybe you can derive the solution from there

[silk thunder]

i tried but this one is a bit challenging

[red tree]

I literally mean try some numerical inputs

Try a large number, then a small number, then a negative number

[silk thunder]

i did

negative number didnt work

How does that help me solve it ?

[red tree]

Okay now suppose x is an arbitrary number, try to think about exactly the conditions on x in order for the expression to be defined

[silk thunder]

x should be greater the 0

[red tree]

Sounds good to me

[silk thunder]

but because it is in square root the result must also be positive

Now thats the problem

[red tree]

Suppose x > 0

Then it follows that log(x + 1) > what?

[silk thunder]

0?

[red tree]

ok

Then it follows sqrt(log(x+1)) is defined

Now suppose x < 0. Then it follows that log(x+1) < 0, and so sqrt(log(x+1)) isn't defined

What about x equals 0

[silk thunder]

log(1)

=0

Defined

[red tree]

cool

Problem complete

[silk thunder]

Thanks but do you know a way to solve it algebraically

[red tree]

Logically is the best way

Here's a cleaner logical solution

$\sqrt \square$ is defined iff $\square\geq 0$

[obsidian monolith]

Icy001

[red tree]

Therefore $\sqrt{\log(x+1)}$ is defined iff $\log(x+1)\geq 0$

[obsidian monolith]

Icy001

[red tree]

Now $\log(x+1)\geq 0$ iff $x+1\geq 1$, iff $x\geq 0$

[obsidian monolith]

Icy001

[red tree]

[silk thunder]

problem is for exam i need to write solution

[red tree]

You don't like writing words in your solutions?

[silk thunder]

Let me check

[sick steppe]

Are you currently writing an exam?

[silk thunder]

No

how can i use my phone in exam ? If it is not allowed in exam halls

[sick steppe]

Take-home exams...

[silk thunder]

i meant school exams …

anyway thanks

[rocky finch]

Could someone help me please and explain where the 1/5 went too 🧐

Like under the 3x and under the full radical

[uncut mulch]

prod to sum law for logs

ln(k) is just another constant

[rocky finch]

Aaaaah so it's in the C, I was assuming that but I wasn't completely sure, thank you very much

[unkempt girder]

hello i was wonder what $H/to0/frac{/sqrt{47+H}{H}

wait

$H/to0/frac{/sqrt{47+H}{H}

cool

$\lim_{x \to 0} \frac{\sqrt{47+h}-5}{H}

$\lim_{x \to 0} \frac{\sqrt{47+h}-5}{H}$

[obsidian monolith]

andmam

[unkempt girder]

YES

ok so i got 2 as my solution i just want to make sure im correct

also X is H sorry

$\lim_{H \to 0} \frac{\sqrt{47+h}-5}{H}$

[obsidian monolith]

andmam

[unkempt girder]

mathway told me it is undefined soooooooooooooooooooooooooo

but i am stilll confident in my answer

[upper tundra]

How did you get 2?

[unkempt girder]

ill send my work

actually https://www.youtube.com/watch?v=8cKN-J118NM this video should explain how i got 2 they are very simmilar problems

and my work is messy

[upper tundra]

You had 47 here

[unkempt girder]

yes but the method is the same

in the video and my problem

[upper tundra]

The limit in the video is a finite value because after rationalisation of numerator, we are just left with h which can be cancelled with h in the denominator

[unkempt girder]

could you not cancel h in my problem?

[upper tundra]

No

[unkempt girder]

oh

[upper tundra]

And first of all, the "h" in the numerator and "H" in the denominator was meant to be same right?

[unkempt girder]

yeah thats my bad

oh i think i understand

alright thanks

[lone vault]

Hi, I'd like a little help with the following equality.

I know the sides equals, but I'd like to know why (this is one of the steps in order to solve exponential equation)

Any idea how to get it? 😅

[vivid void]

pretty much just multiply the left by $\frac{2+\sqrt{3}}{2+\sqrt{3}}$

[obsidian monolith]

quantum

[vivid void]

it’s a difference of squares

[red tree]

Alternatively, whenever I see the claim that a = 1/b, this is equivalent to the claim that ab=1, so you can just check ab=1 in this case

Wait, I just now read the original question

[vivid void]

oof

[red tree]

I know the sides equals, but I'd like to know why (this is one of the steps in order to solve exponential equation)

Hmmmm....

[vivid void]

pretty sure that what i said was enough

[red tree]

What you said is the procedure

What I said is like, using the definition of 1/b

[vivid void]

i assumed they meant, how to get the left side in the form of the right side

[red tree]

It's possible they wanted to ask how (what the steps are), not why

[median sun]

[median sun]

For the most part yes

Normal is perpendicular to a face or side.

You wouldn’t call these normal to each other

[lone vault]

Exactly what I needed! Thank you so much ^^

[viscid thistle]

hello i'm having trouble with some questions, can anyone help me with that? it's section 5.2

[outer briar]

@median sun thankyou sir/ma'am

[vapid plaza]

For 41 you can multiply (1+sinθ) to both the top and bottom of the fraction in LH

The result should follow after simplifying

[vapid plaza]

Alternatively, you can square both sides after first showing that both the LH and RH are positive

[vapid plaza]

(this is necessary because otherwise it could be that one side is the negative of the other)

[tight compass]

What question?, what are you trying to do?

[cedar bramble]

what is 3 plzz

[tight compass]

What have you tried? Have you tried using sign analysis test(testing your inequality for intervals created by your roots).

[cedar bramble]

i sovled it

[tight compass]

Nice.

[viscid thistle]

41,47 and 49

[ivory rover]

If the following function h(x)=x^3-3x^2+3x has a maximum value in the interval [0,3], then the coordinates of the point at which the function has a maximum value are:

can anyone help me?

[sterile valley]

find x-coordinates at which h'(x) = 0
choose the one which lies in interval [0, 3]
check whether it is maximum point by substituting this x value in h''(x), h''(x) should be < 0 for maxima

[ivory rover]

(check whether it is maximum point by substituting this x value in h''(x), h''(x) should be < 0 for maxima) it is coming out equaling 0 not <0

[sterile valley]

then it is an inflection point, not maxima

oh wait I think I get the question now

x^3 - 3x^2 + 3x is an increasing function in interval [0, 3] (because h'(x) is >= 0 for x in [0, 3])

so maximum at that interval is just h(3)

[ivory rover]

so the coordinates at maximum value will be (3,9) right?

[inland birch]

which is right? when simplifying 2/√2
-rationalize denominator, multiply by √2/√2 = (2√2)/4 = √2/2
or
-split numerator into two root 2 values and cross out one from the numerator and denominator
so like this 2=√2*√2, that divided by √2. (√2*√2)/√2 = cancel one and the answer is one √2

I'm working on trig identities and i keep seeing both ways so whats up?

[uncut mulch]

you didn't apply the first method properly

sqrt(2) * sqrt(2) isn't 4

if done properly, both approaches are valid

[inland birch]

ahhh i see i completely went pass that part lol thank you

[plain bolt]

how would i find 6^x=4^(x+1)

[tight compass]

Trying using log base 4 on both side to see what you get.

[plain bolt]

huh

[plain bolt]

whats the forumla / rule for that?

[tight compass]

log_4(6^x) = log_4(4^(x+1)).

if you know the power property of logs you can continue from there.

xlog_4(6) = x+1 implies x(log_4(6) - 1) = 1 thus x = 1/(log_4(6) - 1).

[solid sundial]

Would like help with solving for variables w, x, y, z with c_i being constants:

[vivid void]

take ln of both sides for all lines

[solid sundial]

I have tried this but am not able to express any variable in terms of constants

Could it be possible that this is involves differentials, or probability density functions? Or something else apart from simple algebra? Would be glad to hear any thoughts

[past meadow]

once you take the ln of every equation its just 4 linear equations in ln(x), ln(y), ln(z) and ln(w) right?

so standard methods for solving linear systems should work

oh wait

im a dumbfuck

ignore that

[solid sundial]

Don't say that lol, its a tough one I've been stuck on it all day

[past meadow]

wait yeah it is still pretty easy after taking ln. renaming ln(c_1) to a, ln(c_2) to b etc, after taking the log of every equation, we achieve a/w=lnx, b/z=lnx, c/w=lny, d/z=lny. Then a/w=b/z and c/w=d/z which is a linear system we can solve for z and w.

(edited to fix typo)

[solid sundial]

I'm not able to solve this for z and w, can you help?

[past meadow]

az-bw=0, cz-dw=0. This can be solved just like any two variable system of equations. For instance, multiplying the first equation by c/a gives cz-bc/a*w=0. Then take the second equation from the first.

[solid sundial]

maybe I am missing something, the best I can get from this is (bw)/a - cw = 0

[inland birch]

when you rationalize a negative denominator, like -(1/√3). is the rationalization also negative?
so -(1/√3) multiplied by -(√3/√3) or is it just positive √3/√3

[late mica]

Well, what's a negative number * a negative number? The same rule applies.

[inland birch]

i dont think i explained it right, basically i want to know if the (√3/√3) on its own will that be positive or negative

so if original √3 is negative will the rationalize part also be negative or not

[late mica]

Well, -(1/√3) * -(√3/√3) = √3/3
As for what will the rationalization be, I'm assuming you mean when you rationalize the fraction will you multiply it by a negative or a positive? For this case I would stick with * √3/√3 - because you're multiplying by 1, which effectively changes nothing. You can multiply by 1, add and subtract the same number, those kinds of tricks.

[inland birch]

yeah but -1/-1 is still multiplying by one no?

hang on let me take a pic

[late mica]

That is true.

[inland birch]

Alright so which is the one

[late mica]

The bottom one - because you're multiplying by 1.

[inland birch]

ok thanks for the help

[fossil flare]

Does anyone know the answer to 8?

[full pagoda]

what have you tried?

well actually that looks a lot like a test/quiz

[fossil flare]

Idk what to do

[fossil flare]

It's a practice for my finals

[full pagoda]

take the derivative twice?

[fossil flare]

Do you have a working out for clarity?

[full pagoda]

i'm not gonna do it for you, no

but if this is a practice for your finals, you should know how to do it

or at least where to start

[stuck lark]

@fossil flare pls dont make comments like that

[fossil flare]

I wouldn't have asked if I knew...

[fossil flare]

Calm down.

[full pagoda]

this server isn't about getting your work spoonfed to you, if you want to learn a new concept that you need for your final exam, i suggest google

[uncut mulch]

have you tried anything so far?

[fossil flare]

I worked it out don't worry

[sharp mist]

I’m struggling with derivatives

What is the difference from a tangent line and a derivative?

Or are they totally unrelated

If so, I’ve got a long way to go

[red tree]

A tangent line is a line which is a certain set of points, and a derivative is a number

[mild swan]

@sharp mist ^^

A derivative can be used to find a tangent line

[sharp mist]

I think I get it better

Thank you guys

[solid sundial]

Thank you to those who helped, through help I believe this may be an acceptable solution so I am posting it for others. If you have any thoughts/corrections, just @ me.

https://colab.research.google.com/drive/1caVlspyAMT8ngutMIJqw3a-e66Izhb4h?usp=sharing

[topaz swift]

quick question

is https://media.discordapp.net/attachments/844681108473118750/923120690356428880/549047761518198795.png

this a valid perspective?

[woeful wing]

i believe this is just a (poorly written) interpretation of the fundamental theorem of calculus

[topaz swift]

oh

I see that now

[queen brook]

Tldr: could someone please check these two courses and compare their content and tell if i'm ready to take calc1 if i've taken the second course?

I was confused whether to take this course: https://www.udemy.com/course/trig-by-krista-king/

or this one: https://www.udemy.com/course/trigonometry-the-unit-circle-angles-right-triangles/

I went for the second one cause it had more content + it was specially for trig, since trig is the important pillar of calc (I think) i thought it would be wise to take the second one!

So can I now start learning calc 1?

in the second course it doesnt have analytical geometry and matrices sections so are those needed for calc 1 or i can just skip them if I've taken the trig course already?

Any advice is appreciated

[viscid thistle]

So i've been given a function f(x)=x^2+bx+c, so there are to unknown coefficients. Minima x=2, and zero point x=1

How do i find the coefficents?

[full pagoda]

what is true at a minimum?

[viscid thistle]

a=/0

idk

[full pagoda]

perhaps something about the derivative?

[viscid thistle]

ye

[full pagoda]

what about it?

[viscid thistle]

ummm

This is harder than what i expected 😦

[full pagoda]

,w graph x^2

[obsidian monolith]

Results provided by WolframAlpha

Click here to refine your query online
Upgrade to WolframAlpha Pro!

Requested by a disappointing son#0540

[full pagoda]

what is the slope at the minimum

[viscid thistle]

x=2

[full pagoda]

?

look at the function i just graphed

what is the slope at the minimum

[viscid thistle]

0

[full pagoda]

right

so you know at x=2 in your original equation, the slope is 0

so what should you do?

[viscid thistle]

so it will be (x-2)?

[full pagoda]

what

[viscid thistle]

I haven't taken math in english 😦

so it it's hard

srry

[full pagoda]

i'll use notation

should be universal

f'(2)=0

you have f(x)=x^2+bx+c

[viscid thistle]

yes

[full pagoda]

find f'(x)

[viscid thistle]

2x+b

[full pagoda]

so f'(x)=2x+b

now use f'(2)=0

solve for b

[viscid thistle]

b=-4

[full pagoda]

now go back to f(x)=x^2+bx+c, zero point at x=1

so f(1)=0

you have b

solve for c

[viscid thistle]

4

[full pagoda]

are you sure?

[viscid thistle]

jk

😉

c=3?

[full pagoda]

yep

[viscid thistle]

broOOO

i knew this already

it's just my brain goes stupido

Thank you sir : )

[full pagoda]

👍

[cold kestrel]

what is precalc

[sick steppe]

pre-calculus

[cold kestrel]

what is this so called pre-calculus

[willow bear]

a weird mix of algebra, combinatorics and trigonometry taught by US schools, nominally intended as preparation for calculus but generally serving no such purpose in practice.

[cold kestrel]

i think its unnesecary

if you wanna study calculus

screw it

just do the thing

[vivid void]

calculus just needs algebra 2 and trigonometry

what more could precalculus cover

[tight compass]

precalculus usually covers what Ann said and basic stuff on vectors, matrices.

Tho precalculus is usually called algebra and trig here.

[sullen zealot]

Haven't learned matrices

I should probably do that

[muted owl]

from what i know, they're used with multivariable calculus

[tight compass]

Just learn linear algebra before multivariable calculus.

[sick steppe]

not really.

used much more in LinAl specifically.

[muted owl]

ah okay

[sullen zealot]

well i've already learned multivariable calculus

and im around algebra 2

[viscid thistle]

I am given this graph, function of this graph is a cubic polynomial function.

I found that, the minima is (3, 1) and maxima (1, ?)

I need to find the function

[vivid void]

there’s no minima or maxima

[viscid thistle]

😮

[uncut mulch]

you should mention local

[vivid void]

relative extrema sure

absolute extrema no

[viscid thistle]

f me these words are just insane

[vivid void]

local means exactly what you think it means

[uncut mulch]

you could consider this as the graph of a certain cubic that was translated 1 unit up

[viscid thistle]

ye

How would i be able to find the function?

[vivid void]

you would need to use a system of equations

you want to find something in the form of f(x) = ax^3+bx^2+cx+d

you know d = 1

[sick steppe]

.

[viscid thistle]

mhm

[vivid void]

$ax_1^3+bx_1^2+cx_1+1 = y_1 \\ ax_2^3+bx_2^2+cx_2+1 = y_2 \\ ax_3^3+bx_3^2+cx_3+1 = y_3$

[obsidian monolith]

quantum

[vivid void]

choose three different x values and their corresponding y values

then you will have a system to solve

@viscid thistle

[viscid thistle]

(3,1)

[vivid void]

don’t use (0,1) if you couldn’t tell

[viscid thistle]

OK that's the only one i found

[vivid void]

well

you don’t have anymore coordinates where both x and y are integers

so this part is up to you

[viscid thistle]

I still don't know what to do

[vivid void]

find two more x values and their corresponding y values

it’s just that both of them won’t be integers now

realized i made a typo

$ax_1^3+bx_1^2+cx_1+1 = y_1 \\ ax_2^3+bx_2^2+cx_2+1 = y_2 \\ ax_3^3+bx_3^2+cx_3+1 = y_3$

[obsidian monolith]

quantum

[vivid void]

just find two more (x,y) pairs you can work with @viscid thistle

[viscid thistle]

I don't know how to 😩

[vivid void]

it’s the same way you found (3,1)

[viscid thistle]

but there aren't anymore i can see

am i blind

[vivid void]

there are

they just aren’t both integers now

so you have to estimate

[viscid thistle]

could you give hint

https://tenor.com/view/tom-and-jerry-tom-falling-asleep-too-sleepy-sleepy-gif-12075429

[vivid void]

i’ve literally told you everything you need to know

look at the graph

estimate what the numbers would be

[viscid thistle]

(4,2)?

[vivid void]

what would the approximate value of f(4) be

it’s more like (4,2.5)

[viscid thistle]

oh

lol

is that even allowed

[vivid void]

well yeah

why wouldn’t it be

[viscid thistle]

Idk, sounds it just sounds wrong

and one last one

(1,2.2)

[vivid void]

maybe (2,1.75)

[viscid thistle]

hhhh- i hate desimalcscdsjd

[tulip talon]

Should I talk about Analytic geometry here or in #geometry-and-trigonometry ?

[vivid void]

$ax_1^3+bx_1^2+cx_1+1 = y_1,\ (x_1,y_1) = (3,1) \\ ax_2^3+bx_2^2+cx_2+1 = y_2,\ (x_2,y_2) = (4,2.5) \\ ax_3^3+bx_3^2+cx_3+1 = y_3,\ (x_3,y_3) = (2,1.75)$

[obsidian monolith]

quantum

[vivid void]

@tulip talon here is fine

since analytic geometry is just a complicated sounding way of saying coordinate geometry

[vivid void]

@viscid thistle

[viscid thistle]

wow

would it be like (x-3)(x-4)(x-2)?

[vivid void]

no clue

probably not since i don’t see how you’d get decimals from that

just plug in the x and y values into what i just sent

[tulip talon]

is there a known formula for representing the area of a triangle defined by the two axis and a line equation ax+bx+c = o or y =mx +n?

[vivid void]

and solve the system

[tulip talon]

the area of the triangle in function of n and m from y = mx + n

(or a,b and c for that matter)

I couldn't find such a formula online

Have y'all heard of one?

[vivid void]

are you basically asking about the area of the triangle formed by the x and y intercept of a linear function

probably could have phrased that a bit better

[tulip talon]

surely could

I don't know english mathematical terms that well

[vivid void]

no clue why you have the equations set equal to 0

[vivid void]

.

[vivid void]

should be approximately correct then

[viscid thistle]

yesssssssssssssssssssssssssssssss

https://tenor.com/view/help_urself-vereena-harukihq-gif-23792104

ur the teddy bear

[vivid void]

[hushed sphinx]

I think you're supposed to read off the graph that f(0)=1, f'(1)=0, f'(3)=0, f(3)=1. That gives four linear equations to determine the three coefficients of the polynomial.

[vivid void]

[hushed sphinx]

Hmm, no, that won't work.

[vivid void]

this is precalculus

lol

i purposely didn’t bring that up

[hushed sphinx]

It would be underdetermined anyway, neither of those data says anything about how far away from y=1 the graph gets.

[red tree]

Hmm

Slightly different approach: f(x) - 1 has a root at 0 and a double root at 3, hence f(x) = ax(x-3)^2 + 1 for some a

Maybe a can be determined from the fact that f'(1) = 0

[vivid void]

how can you know it’s a double root

[red tree]

A couple of ways exist to see that

One is a sign analysis of the curve to the left and right of the root

another is to use the fact that $(x-a)^2\mid f\iff f(a)=0\text{ and }f'(a)=0$

[obsidian monolith]

Icy001

[hushed sphinx]

Unless I'm mistaken f(x) = ax(x-3)^2+1 gives f'(1)=0 no matter what a is.

[vivid void]

neither of these really help me see that honestly

[red tree]

Dam that is unfortunate

so we will need to use the fact that f(1) = 2.something

[vivid void]

probably because i don’t understand either of them lol

[sharp mist]

Is there a common standard for the curriculum that is taught in precalc?

[sick steppe]

no.

Curriculum is state/province/whatever specific usually

Just look up your place's curriculum.

[sharp mist]

Thank you! That did work

[daring lark]

how would you do this problem? i got sin= square root of 24/5 bc i used the pythagorean theorem but it doesn’t sound right lol

[sharp mist]

1^2=1
1/5^=1/25

1-1/25=24/25

The other leg is sin t

You messed up in the process of squaring 1/5

@daring lark

[sharp mist]

Whoops

[daring lark]

wait nvm

lol

ty!

[viscid thistle]

Yo it isn't √25 but it is √24

[vivid void]

[mild swan]

^^

[vivid void]

25+1 = 25

[daring lark]

tyyy

[restive bridge]

https://cdn.discordapp.com/attachments/163780366329577474/924156183026229288/wPdVyYO.png

Can anyone break it down to me how they got 5n + 5n + 5n?

I used this method and it worked for the first line.

By would it actually be 20n??

so 5n + 5n + 5n + 5n

[obsidian monolith]

abc42

[viscid thistle]

nvm i calculated the tangent incorrectly

[restive bridge]

@topaz swift
That's a calculus question

[topaz swift]

oh

wait what even is precalc

sorry

[restive bridge]

Pre-calculus covers: Exponential Functions, Logs, Systems of Equations, Sequences, Probability/Counting

[topaz swift]

oh ok cool

[sharp mist]

And trig sometimes

[tepid cloak]

welp I already found y which is B(2,0) what do I do next

[uncut mulch]

is that supposed to be for part a) or b)

once you have: B(2,0)
you have the respective y value as requested

[tepid cloak]

is the y the same for both parallel and perpendicular or the y value is different from each other?

[uncut mulch]

different

otherwise you're implying that parallel lines and perpendicular lines are identical

in case the question is clear enough
a) is parallel to the line whose inclination is 135°. find y
a) is perpendicular to the line whose inclination is 135°. find y

[tepid cloak]

oh okay I get it now thankss!

[agile gazelle]

0.7.15 a

I was thinking I should just literally draw a regular pentagon inside the unit circle and find the solutions geometrically

But it felt like cheating

How would you guys do it?

[obsidian monolith]

abc42

[hushed sphinx]

In practice? I'd remember the golden ratio has something to do with it, fool around with a calculator until I discover that cos(pi/5)=phi/2, and use that to construct an exact representation of e^(i·pi/5) that I can then prove is correct by direct calculation on a clean sheet of paper.

[sullen raptor]

bruh what

how did u find

$\cos(\frac{\pi}{5}) = \frac{1+\sqrt{5}}{4}$

[obsidian monolith]

abe

[hushed sphinx]

As I said, literally fooling around with a calculator. Throwing numbers related to phi at the inverse trig functions until I hit one that gave a nice whole number of degrees :-)

[viscid thistle]

aye I know this is a weird question, but does anyone have a good source for why we preform a different process to factor quadratics with a non one leading coefficent versus when the leading coefficent is one

[hushed sphinx]

Do we?

[sick steppe]

We don't

[viscid thistle]

thanks for the response, seems to be something that is relatively obscure since there are so many different approaches to factoring quadratics

guess I will continue just using what I have been taught

[copper vigil]

Because in a non monic quadratic there are two extra unknowns

(x+a)(x+b) vs (ax+b)(cx+d)

So when you expand (x+a)(x+b) you get x^2 + (a+b)x + ab

But when you expand (ax+b)(cx+d) you get (ac)x^2 + (ad+bc)x + bd

[agile gazelle]

That sounds even more like cheating lol

[hushed sphinx]

It just said the formula shouldn't include trig functions, not anything about how you discover it.

[agile gazelle]

But like

Is there a way without knowing beforehand that the 5th roots of 1 are uniformly spread across the unit circle?

[hushed sphinx]

If you don't know that, the relation between part a and b of the exercise is meaningless anyway.

[agile gazelle]

Does there have to be a relation?

[hushed sphinx]

There happens to be one, and the fact that the two questions are part of the same exercise number strongly suggest you're expected to make use of it. ¯_(ツ)_/¯

[agile gazelle]

Ok ig

Thanks for the insight!

[viscid thistle]

thank you, I was looking for an explanation using a general approach like that

[tepid cloak]

welp how do I do this

[willow bear]

presumably you have been taught how to find the area of a polygon defined by its vertices

is that the case? y/n

@tepid cloak

[tepid cloak]

no ://

oh wait yeah

I think the area is 0

but im not sure

I did A = 1/2 (1 + 5 - 4/3 - (-1 + 4 + 5/3)

nvm I get it now

[chrome glen]

so i got this and im confused for part C

when i use the whole sigma notation in my calculator it gets a diff answer to when i do the whole S=a(1-r^n)/(1-r)

im just wondering where the 1.03 that ive circled comes from and why it needs to be there

[wide vale]

This is weird

Oh wait

Doesn't the formula work when the starting index is 0

Yeah my bad here

Oh I think I found the answer for your question

$5000\sum_{n = 1}^{20}{1.03}^n = 5000(\sum_{n = 0}^{20}{1.03}^n - 1)$

[obsidian monolith]

Touch Our Beans

[wide vale]

$= 5000( \frac{1 - 1.03^{21}}{1 - 1.03}- 1)$

[obsidian monolith]

Touch Our Beans

[wide vale]

$ = 5000(\frac{1 - 1.03^{21} - 1 + 1.03}{1 - 1.03}) = 5000(\frac{1.03 - 1.03^{21}}{1 - 1.03})$

$5000(\frac{1 - 1.03^{21} - 1 + 1.03}{1 - 1.03}) = 5000(\frac{1.03 - 1.03^{21}}{1 - 1.03})$

[obsidian monolith]

Touch Our Beans

[wide vale]

Now you just factor out a 1.03 here @chrome glen

[chrome glen]

wait lemme just read all that thank you

[chrome glen]

ok thank you could you tell my why in terms of the interest the regular formula doesnt work 😭

[wide vale]

You mean why $1.03 + 1.03^2 + ... + 1.03^20 \ne \frac{1 - 1.03^20}{1 - 1.03}$?

[obsidian monolith]

Touch Our Beans

[chrome glen]

yes

[wide vale]

Well the formula for geometric sum is $1 + r + ... + r^n = \frac{1 - r^{n + 1}}{1 - r}$

[obsidian monolith]

Touch Our Beans

[wide vale]

So you need 1 in the front

And the exponent of r in the formula is n+1, not n

[chrome glen]

wait hold on a minute whats ur geometric sequence formula

[wide vale]

That's why this equation failed

[wide vale]

Wdym

[wide vale]

I used this one only

[chrome glen]

the one where its just getting the terms not summing them

[wide vale]

Oh

[wide vale]

Well yeah you factor out the 5000, which is the value of a

So I looked at $1 + 1.03 + ... + 1.03^{20}$ separately

[obsidian monolith]

Touch Our Beans

[chrome glen]

wait is the 1 supposed to be there for it

[wide vale]

There was no 1 there, so I added 1 and subtracted 1 to have it

[wide vale]

To apply this

[chrome glen]

ahh ok yes

[chrome glen]

thank you

[wide vale]

You're welcome

[rapid temple]

Someone can help me here?

This is what i tried to do here:
I expressed z as polar cordinate, and it's adjcent
and then tried to find the the multiplications for all the solutions in image

My final solution that i got is: 0.49 i

<@&286206848099549185>

[grave tartan]

shouldnt you have k=3 and k=4 here too

since we have 5Ø

[rapid temple]

ohh fuckkkk

its k=3 i guess

since its from k=0 to k=3

[grave tartan]

thats 4 solutions right

[rapid temple]

I was confused with r and his 3 🤦‍♂️

[grave tartan]

yeah me too i had to make sure you werent cube rooting

also I think you do need k=4

you get a series of -1, 3, 7, 11, and 15

the next one k=5 would be 19 but that is just coterminal with -1

which we know

[rapid temple]

but k=3 its a full rotation isnt it?

[grave tartan]

if it was a full rotation we would only need to write up to k=2 tho

see what you are doing isnt cube rooting $4e^{i5\theta}$

[obsidian monolith]

vrang

[grave tartan]

its splitting it up into $4$ and $e^{i5\theta}$

[obsidian monolith]

vrang

[grave tartan]

cube rooting the 4

and fifth rooting the $e^{i50}$

[obsidian monolith]

vrang

[grave tartan]

so we should have 5 angles

in order to make a full rotation

0, 1, 2, 3, 4

with the 5 being coterminal with 0 so we dont need to do further than 4

but also I haven't done this in a while so idk why you added all the angles but subtracted the one for k=0 in your second to last step

[rapid temple]

or yea now that i draw it, i see this is 4

[tepid cloak]

can someone check if my answer is correct

[uncut mulch]

show work

[tepid cloak]

oh wait my angle b and c is wrong

i've used the slope formula then the angle between two lines formula

[tepid cloak]

welp

[violet narwhal]

I've just started learning calc and I've got to find the minimum for y=sin3t, I've currently gotten "t=cos^-1(0)/3=0.5235" which im certain is the maximum, but I'm just confused on how I can now use this to get the min

[sick steppe]

Don't need calc for that

[violet narwhal]

It's required to be done w differential calc

[sick steppe]

Ok so you have t=(2n-1)pi/6

From solving cos(3t)=0

[violet narwhal]

Yeah I got that

but Im just confused on how that gave me the Min

[sick steppe]

Pick an n value st you get -1

Since you know you expect to get -1

[willow bear]

or

you could additionally calculate y'' and check at which extreme points it's positive to get your minimum points

[thorny geyser]

if youre certain thats the max then just use the amplitude to get the min

[storm spoke]

dude do i skip pre calc

[turbid arrow]

need help w what to do w the bottom part of the fraction

managed to get this far

[sick steppe]

cot isn't sin/cos

[turbid arrow]

mb just realized

yeah it's cos/sin

[sick steppe]

so you have $\frac{\cos^2(x)}{\sin(x)(1-\sin(x))}$

[obsidian monolith]

Mosh

[turbid arrow]

can we simplfy the 1-sinx into cscx?

[uncut mulch]

no

(because 1-sin(x) isn't csc(x))

[turbid arrow]

i got sinx for the bottom

[uncut mulch]

what do you have elsewhere

[turbid arrow]

$\frac{\cos^2(x)}{\sin(x)-sin^2(x)}$

[obsidian monolith]

balloony

[turbid arrow]

this is what i have so far

[uncut mulch]

expanding the denominator doesn't really help you

instead consider applying conjugates/difference of two squares/pythagorean trig identities

[vague crow]

keep the unexpanded denominator

if its proofs you can add a form of zero and use a pythagorean identity

but basically you want to get the top which is

1-sin^2(x)

lemme write it down rq

you use difference of squares

then you can factor out the quantity (1-sin(x)) from top and bottom

and by splitting the fraction into its own parts we are left with this which we can then simplify to csc x + 1

@turbid arrow

[grave tartan]

depends

on how strong your math is or the courses that come before precalc

Personally I think its good practice and if you want to skip you should skip earlier classes

but if precalc is skippable then it really matters on if you think you will be okay in calc

[stoic flint]

henlo!

these are the two questions and their solutions are given i am confused that why in the question 7 we only took one equations but in question 8 we took the 2 also as g(x)

why is that so?

[trim hemlock]

because in question 7, the area of the slice of your solid (perpendicular to the axis of revolution) is simply pi*x^2, i.e pi(1+y). However, in question 8, the area is the bigger "circle" (2^2 *pi) subtracted by the smaller "circle" (pi * (1+y)), and in summary the area is now 2^2 * pi - pi(1+y)

[stoic flint]

ahhhh i got it aight aight

[vivid void]

are you aware that this is the precalulus channel

[stoic flint]

yo is that not pre cal?

[vivid void]

no

integrals are calculus

[stoic flint]

ooof i am sorry

[storm spoke]

they are?

im pretty sure they are pre calc

[errant laurel]

they introduce integrals in calculus. not in precalc or honors precalc

[turbid arrow]

not sure what to do after the last step

[muted owl]

you could pull out a $cos^2 = 1 - sin^2$

[obsidian monolith]

Atie

[muted owl]

and after that just simplify

[vague crow]

or if you keep them together just factor out a cosx from top and bottom (cancel them out) and then you're left with $1/cos(x)sin(x)$

[obsidian monolith]

WulfZono

[vague crow]

then you can split that up into 1/cos(x) * 1/sin(x)

[turbid arrow]

got it!

i got this far w this question but confused on what to do after the last part

[vague crow]

so go ahead and replace the term in the initial part (1-cos^2(x)) with sin^2(x)

then use the reciprocal identity on csc(x) making it 1/sinx

and multiply sin^2(x)/sin(x) leaving us with sinx=sinx

@turbid arrow

[turbid arrow]

so like this?

[vague crow]

sorry for the late response but yes

[tepid cloak]

does anyone know any checking method for the perimeter of a triangle given it vertices or checking method if the distance is correct using distance formula

[uncut mulch]

wdym by checking method

[tepid cloak]

to know that the answer is correct

like for this solution

lets say the perimeter is correct but I want to show that it is correct by means of checking

[vapid plaza]

Do you expect a slick oneliner that spits out the perimeter?

I don’t think it exists

[uncut mulch]

double check your work?

and/or do it again if you're not sure

[tepid cloak]

can someone check if my answer is correct

directed distance = -15.21 units?

or d = 6.8 units?

[willow bear]

it is impossible for us to verify this without someone going through the numbers themself or you posting your own work for verification

[tepid cloak]

oh sorry just a sec

sorry for the messy solution

[willow bear]

your handwriting makes it hard to distinguish 1 and |

this is bad, especially in math

[willow bear]

the second line in this image can be misread as $\frac{14(3)-3(-5)+71}{\sqrt{4^2+(-3)^2}}$

[obsidian monolith]

Ann

[willow bear]

aside from this, this work you have presented seems correct

[tepid cloak]

yeah I was writing it in a hurry and its just a draft ill probably rewrite it

okay thanks

[willow bear]

you should make it a habit to always write numbers in such a way that they cannot be confused for other symbols

[tepid cloak]

yes maam

welp

[gilded sapphire]

get the equation of the line with
$y - y_1 = m(x - x_1)$
assume the point B is $2\sqrt{10}$ dustance from A put B's co-ords in the line's equation get a relation between x and y and then apply distance formula.

[obsidian monolith]

cosh²(x) - sinh²(x) = 1

[vivid void]

there’s an even and an odd exponent there

what you have is basically x^3-2x^0

remember that an even function satisfies f(x) = f(-x)

and an odd function satisfies f(-x) = -f(x)

[vivid void]

@solid fractal not sure if you saw this

not sure why

c is the answer

notice that the function has even and odd exponents

that means it’s neither

[glossy pebble]

This is just a note that can help, but also if they were to be equal, then you just have to compare f(-x) and -f(x) and see what is making it not odd

[low slate]

kind of confused on how to do this

[willow bear]

@low slate do you still need help with this?

[low slate]

No I’m good ty

U can just add or subtract 2pi to get ur 2 angles

[grave delta]

hi everyone

do you have a pdf or somethin? im planning to take my master's and I want to recall everything from the start

[willow bear]

a pdf of what

[vast totem]

Just start reading precalculus 6th edition blitzer book

[tepid cloak]

helpp

[mild swan]

Adding onto what others have said:

A function f(x) is odd if and only if f(-x) = -f(x).
A function f(x) is even if and only if f(-x) = f(x).

[mild swan]

I'd try graphing the points on the triangle and try drawing it out to begin with

[flint urchin]

could someone help me complete one of these, i am quite confused T-T

[vivid void]

@flint urchin the domain of a function is every number you can input that will make the function defined, the range of a function is, well, the range of the output values

[flint urchin]

what about part c?

what is that awhat am i supposed to do on that part?

[vivid void]

probably say what it approaches near those invalid x values

if i had to guess

[flint urchin]

could u give me an example of what to say for 5?

if the domain is (-infinity, 1)U(1,infinity)

[vivid void]

it becomes infinite near the excluded x values

[flint urchin]

can i say that

for all of them?

in part c?

[vivid void]

i would say yes

i didn’t say it approaches infinity though

because for some of them it kinda approaches $\pm\infty$

[obsidian monolith]

quantum

[flint urchin]

oh

[vivid void]

which means the function doesn’t exist there

so i just said infinite instead

[flint urchin]

oh

so its either approuches negative or positive infinnity for them?

[vivid void]

which doesn’t imply positive or negative

no

[flint urchin]

oh

[vivid void]

look at 5

[flint urchin]

do i need to imply if they are or not?

[vivid void]

wait i have a better answer for the first two

you could say, the function approaches two different values near the excluded x values, so the function is undefined there

[tepid cloak]

okay

[low slate]

how is part d incorrect?

[sharp mist]

how did you calculate that?

[low slate]

i thought sin= csc(90-theta)

[mellow tapir]

$\sin(\theta) = \frac{1}{\csc(\theta)}$

[obsidian monolith]

Yottachad

[mellow tapir]

on the other hand:

$\cos(\theta) = \sin(90-\theta)$ and vice versa

[obsidian monolith]

Yottachad

[lavish edge]

Guys for the special inequalities part, first they say ap is less than mod theta but then say ap squared≤theta squared. How does that work.

Can anyone please explain how this is correct? Or if the proof given is wrong and there's some other kind of proof. Thanks

[mellow tapir]

it says "we picture theta as a nonzero angle" seemingly to just visualize it

[lavish edge]

yes

[mellow tapir]

and then at the top it says "any angle theta in radians"

so the author is including theta in the final example

cuz of course when theta = 0 AP = 0

so by that if you include theta = 0 you could say less than or equal to (as its equal to at 0)

[lavish edge]

but then at the beginning why did he say AP<mod theta

shudnt it be <equal to

?

[mellow tapir]

wdym by mod theta?

[lavish edge]

absolute value of theta

[mellow tapir]

absolute value?

oh

yeah its b/c he stablishes he says theta > 0

in that case AP would ALWAYS be < theta

|theta|

[lavish edge]

whats b/c

[mellow tapir]

because

this is pretty much the synopsis:

if you don't include theta = 0 AP < |theta|

if you do include theta = 0, AP <= |theta|

[lavish edge]

so then when he says the squared part he includes 0?

[mellow tapir]

yes

[lavish edge]

oh ok one more thing

[mellow tapir]

that can be shown at the top b/c it says "for any angle theta in radians"

not "for all theta not equal to 0 in radians)

[lavish edge]

the terms on the left hand side of the equation( that entire sentence) what does that mean

[mellow tapir]

where

[lavish edge]

right there

in the pic

[mellow tapir]

well left hand side of what equation

[lavish edge]

sin^2theta+(1-costheta)^2= AP^2

[mellow tapir]

yeah

ur asking how he got that?

[lavish edge]

they say in the left hand side of the equation

since they are positive ,"each is smaller than their sum and is therefore less than or equal to theta ^2

[mellow tapir]

yeaj

[lavish edge]

how can it be equal to theta squared

if both are positive

[mellow tapir]

well if theta is 0

sin^2(0) = 0^2 = 0

[lavish edge]

yea

[mellow tapir]

(1-cos(0))^2 = (1-1)^2 = 0

so we have 0 + 0 = 0 (the theta)

[lavish edge]

but 0 isnt pos

itive

right

[mellow tapir]

yeah ur right

[lavish edge]

so

[mellow tapir]

he just said that for the purpose of not having to bother with negatives

it should say positive or 0

[lavish edge]

oh ok

thanks

alright so to visualize it he took theta>0 but in fact it can be equal to 0 if we dont visualize it?

[mellow tapir]

yers

at least that is what i think

[lavish edge]

ok

[sharp mist]

What happens if you put complex numbers into trig functions?
Or, if its easier could you send a video?

[winter geyser]

using eulers equation i think

you have to let x be a complex no

[vivid void]

cosine gives real values for imaginary inputs

i know that much

notice i said imaginary and not complex

[winter geyser]

yes

[winter geyser]

https://www.youtube.com/watch?v=RCG_5op1FOo this might help

[sharp mist]

Thank you

[sharp mist]

According to that video, at the end he mentions a triangle with a cosx>1

And he talks about how it would act kind of

Would that shape be a polygon?

[winter geyser]

I think the angle has to be taken in context of complex plane and extend it to 3d, I think it will still be a triangle but in complex plane. I do not fully know to be honest, but there probably is better way to explain the triangle w.r.t to complex plane.

[amber nimbus]

hi can anyone explain to me trig identities? I'm having a hard time proving/verifying them

here's a sample question from our practice questions

[uncut mulch]

you could start by expressing cot(-x) using cos and sin

[timber obsidian]

you might need to know the fact sin(-x) = -sin(x) and cos(-x) = cos(-x).

[timber obsidian]

say a =-x and maybe just start here

[lean owl]

One message removed from a suspended account.

[vivid void]

it’s not very bad

in fact calculus isn’t even very bad

[lean owl]

One message removed from a suspended account.

[sick steppe]

greek letters are just... other letters

like a and b and x and y

[timber obsidian]

calculus is really nice

the more you do it, the more you’ll like it

[timber obsidian]

i was about to say he’s prolly talking about pi and sigma in calculus for example sinx/x can be represented by an infinite series which you can then do a neat little thing with fourier transformation to get the whole value from negative to positive infinite as pi, because sinx/x is even. but he hasn’t done most of the stuff yet so you’re right, it’s judge like any other letter

[sick steppe]

That doesn't come up in high school.

bar pi

Like... this is differential calculus we're talking about, not calc2 or calc3 or whatever

[lean owl]

One message removed from a suspended account.

One message removed from a suspended account.

[amber nimbus]

ohhh rightt. thank u!!

[warm forum]

u can just write it out ;-;

[shut folio]

what?

[icy yarrow]

Are these all correct?

[timber obsidian]

yeahhh

[timber obsidian]

it doesn’t ask you to solve it, it asks you to list the steps in order

[timber obsidian]

the rest are right. albeit, i’ve not bothered with the calculations

[coral token]

From the upper equation we have:

[obsidian monolith]

Menez

[coral token]

Then:

[obsidian monolith]

Menez

[coral token]

,wolf -4.5(\cos(2t)) - 3(\sin(2t)) = 3(-\sin(2t)) + 2(\cos(2t))

[obsidian monolith]

Results provided by WolframAlpha

Click here to refine your query online
Upgrade to WolframAlpha Pro!

Requested by Menez#3747

[coral token]

However, the answer is only 3pi/4 and 7pi/4. Does anyone know why is it wrong? (0-2pi range)

[willow bear]

is it not possible from here to get the values of sin(2t) and cos(2t)?

you overcomplicated this massively

[coral token]

Where?

[willow bear]

from your original equations...

-3cos(2t) - 2sin(2t) = 2
-3sin(2t) + 2cos(2t) = 3

this is a linear system in cos(2t) and sin(2t)

[coral token]

yeah

[willow bear]

ok so why not treat it as such

[coral token]

The problem is that idk why the answer I got was wrong

Did I do something wrong?

[willow bear]

dunno about wrong but you certainly made your own life more difficult

an arithmetic fuckup could have done you in

,w -3x-2y=2, 2x-3y=3

[obsidian monolith]

Results provided by WolframAlpha

Click here to refine your query online
Upgrade to WolframAlpha Pro!

Requested by Ann#0413

[willow bear]

cos(2t) = 0 and sin(2t) = -1

[coral token]

ty

[tepid cloak]

can someone help me with this pls

[uncut mulch]

have you drawn a diagram yet?

[tepid cloak]

i've already drawn it

[willow bear]

show your diagram

[willow bear]

ok, and have you calculated the coordinates of the other endpoints of the median and altitude?

[tepid cloak]

i've only calculated the slopes of the altitude and median so far

[pale solstice]

yo does anyone know how to solve this? im lost

[viscid thistle]

Hi, help me with this one please.

[rare totem]

@viscid thistle identities are always true... try to prove R.H.S = L. H. S by simplyfing

[opaque idol]

Try to find counterexamples for equations - those will be the conditional equations

[rustic swallow]

I got hi

gotchu

a+b+z

use pyhtaroean thorem

🧠

[winter geyser]

idk how to prove it, but largest triangle will from edge point of parabola, (-2,0), (2,0) and (4,0) [triangle coordinates) intuitively this will be the largest triangle.

[marsh cipher]

@pale solstice area(∆) = 0.5•PR•height

Here if you maximize PR and height then you'll get maximum area

That happens at co ordinates Buzzy mentioned above

[pale solstice]

i sat with it all night yesterday, im pretty sure i solved it, thanks for the help tho, @marsh cipher you too

[tropic wolf]

i need help please ive been stuck on a problem forever imma lose my mind

[viscid thistle]

Alright, go on into a help channel and I'll assist you there.

[tropic wolf]

where do you do that

[viscid thistle]

#help-12

[viscid thistle]

hey, may i ask what is the formula to find the rate of the maturity value

[viscid thistle]

idk how to derive

[torpid hemlock]

by maturity value you meant maturity value of compound interest?

[viscid thistle]

simple interest

[hidden dew]

what operation is this?

[sick steppe]

a/b = 1/(b/a)

[uncut mulch]

divide numerator and denominator by x

[hidden dew]

gotcha

[hidden dew]

does this fact have a name?

inverse of a fraction, something like that?

[sick steppe]

I guess just division by a fraction

not everything needs a name tho

[hidden dew]

I guess I was hoping that it would have a name so I could google it and find practice problems to make that and other related identities stick in my brain

I can't stop stumbling over this kind of stuff while attempting to learn calc 1

[uncut mulch]

definition of division,

[lavish edge]

so in my book it says

a point P that intersects the circle has coordinates (x,y)

where x=rcostheta and "sin theta"

so when we look at the origin and the line extending from the origin at 0 degrees

does this apply for this oo

too

you might not understand this so

this

x is what im talking about

if it intersects the circle

will it also be bcos0, bsin0\

xcos0 xsin0 sorry not xcos0

!help

[vapid plaza]

First of all, a point doesn’t intersect a circle. A point lies on a circle.

[lavish edge]

ok

then

[vapid plaza]

Not sure what you’re stuck on

[lavish edge]

look

it say a point P has coordinates x,y right

if it lies on the circle

[vapid plaza]

Then?

[lavish edge]

yea so

if we look at initial side x

[vapid plaza]

What point on initial side x?

[lavish edge]

just x

not any point

look at the sid

side

[vapid plaza]

So you’re talking about the line segment.

What’s the problem?

[lavish edge]

yea

now it lies on 0 degrees right]

just x

[vapid plaza]

It’s inclination would be 0 degrees, yes.

[lavish edge]

wdym by inclination

[vapid plaza]

The angle that line x makes with the x-axis

But that’s not the focus

What is troubling you?

[lavish edge]

look so if the point lies on the circle it wud be acos theta, a sin theta . If somehow the initial side goes up to the circle and lies on it will it be xcos0,xsin0

[vapid plaza]

“Initial side goes up to the circle and lies on it”?

[lavish edge]

yes

if x extends till it touches the circle

like how r is touching the circle

[vapid plaza]

So you mean just a bit further to the right of segment x?

As in the figure

[lavish edge]

yes

[vapid plaza]

That point would be (r.0)

[lavish edge]

yes but

does it follow the rule

[vapid plaza]

It does

Try

[lavish edge]

xcos theta, xsin theta

yes but theta is already given

[vapid plaza]

Theta is 0

[lavish edge]

say its 30

[vapid plaza]

Oh nvm

Continue

[lavish edge]

yes but theta is already given

so how is theta 0

[vapid plaza]

?
Theta is given only the name theta.
If you sub in theta=0 for
(r cos(theta), r sin(theta))
You would get (r,0)

[lavish edge]

yes but

how can we have theta for r and theta for x

look at the figure

theta is given

say its 30

now how can theta be 0 for x

[vapid plaza]

The actual theta in the figure is just an example of the rule

[lavish edge]

yes ik

[vapid plaza]

in fact theta can be anything
Not just “30”

But 30 is used for the figure for demonstration

[lavish edge]

but can theta be 2 diff things

can it be 30 as well as 0 for the same figure

[vapid plaza]

Hmm

Let me try to say sth

Suppose a circle of radius r centred at the origin. Then all points on the circle will be of the form
(r cos(theta), r sin(theta))

the circle is defined to be all the points “generated” from all the different thetas

[lavish edge]

yes

look basically what im trying to say is

theres a theta in the figure

say its something like 30 ok

now the point P formed where it lies on the circle

wud have the coordinates x=rcos theta and y=rsin theta

now if we look at the initial side x

it would have the coordinates x,0

and x,0 is the same thing as xcos0,xsin0

but this initial side doesnt always lie on the circle

so howcome

[vapid plaza]

If theta = 30, we get a point on the circle.
If theta = 0, we get another point on the circle.

[lavish edge]

but what if theta is 30 and 0 both

for side x only its 0

and its 30 if we look at the full figure

[lavish edge]

why does this work

[vapid plaza]

You keep saying “initial side”. You mean the blue segment lying on the x-axis?