precalculus

[vast ravine]

how would i solve this

[uncut mulch]

consider the double angle identities

[vast ravine]

i have the identity sin2(ϴ)=2sinϴcosϴ

[uncut mulch]

yes

wait

poor use of parentehses

sin**(2ϴ)**

[vast ravine]

so am i multiplying by 2 or squaring pi/12

?

[uncut mulch]

does 2theta mean 2*theta or theta^2

[vast ravine]

that's my question?

i don't know

[uncut mulch]

i'm asking you a basic algebra concept

[vast ravine]

oh

typically multiplication

[uncut mulch]

yes

exactly

[vast ravine]

but when the 2 is to the right of the sin it's squaring?

[uncut mulch]

if the 2 is in the exponent position / superscript

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

the double angle identity for sine uses the latter

[obsidian monolith]

ramonov:

[vast ravine]

so the trig identity Sin2ϴ would mean in this equation sin (2* pi/12)

[uncut mulch]

expression not equation but yes

[vast ravine]

so the 2 + 6 would reduce

making it pi/6

sin(pi/6)

[uncut mulch]

wdym 2 + 6

[vast ravine]

oh i mean if you multiply the 2& pi/12 the 2 and 12 would reduce

to 1 X pi/6

[uncut mulch]

yes

[vast ravine]

in the back of my book it says the answer is 1/2

[uncut mulch]

pi/6 is a special angle

[vast ravine]

but isn't pi/6 ≠ 1/2

[uncut mulch]

and you should know the value of trig functions evaluateed at that angle

=sin(pi/6)

[vast ravine]

ohhh

oic

[uncut mulch]

the identity is NOT
$$2x = 2\sin(x)\cos(x)$$

[vast ravine]

30*

[obsidian monolith]

ramonov:

[vast ravine]

so the identity is saying look at the sin of 2* the angle theta

right?

i think i was confused because i was pluggin in degrees in my calculator instead of radians

thanks for clearing that up for me

[uncut mulch]

yes

[vast ravine]

thank you so much

[velvet blade]

It's supposed to be x^3k right?

not x^3^k?

[dim jungle]

x^3^k because for f^3(x) it was x^27 and 27=3^3 not 3*3

[velvet blade]

Okayy

[vast ravine]

i have here verify the identity sin2ϴ/1-cos2ϴ=cotϴ

i'd use my double angles and cot=tan to flip the fraction

and transform into 1 - cos²ϴ-sin²ϴ/2sinϴcosϴ = tanϴ

at which point where do i go

i think i'm forgetting some identities i need

[velvet blade]

@vast ravine I didn't understand your question. Can you elaborate?

i'd use my double angles and cot=tan to flip the fraction
when is cot = tan?

at pi/4 right?

Ok I seem to be of no help

[vast ravine]

since cot = 1/tan

sin2theta/1-cos2theta = cottheta

[velvet blade]

yeah

[vast ravine]

therefore tan theta = 1-cos2theta/sin2theta

right?

[velvet blade]

wait

cot(x) = cos(x)/ sin(x)

okay

[vast ravine]

right

[velvet blade]

I don't think what you typed is the right identity

[vast ravine]

but also tan = 1/cot

[velvet blade]

yeah

[vast ravine]

so if i wanted tan instead of the cot

i'd just flip the fraction

[velvet blade]

tan (x) = sin(x)/cos(x)

yeah

Are you having problems with compound angles?

[vast ravine]

i'm confused about the steps to verify this identtiy

it's about double angles and identities

[velvet blade]

tan (2x) = sin(2x)/cos(2x)

[vast ravine]

is the problem

[velvet blade]

Okay give me a min

[vast ravine]

the way i'm approaching it is flip the fraction to get tan, then using double angles?

after expanding sin2theta and cos2theta i'm lost

[sick steppe]

how are you expanding cos2theta?

cos^2 - sin^2 ?

[ancient rampart]

why dont you turn the bottom into 2*sin^2 pheta

[sick steppe]

2 people are already helping

[velvet blade]

Wait yeah, that's right

I had my sign conventions wrong

[ancient rampart]

ok was just trying to help cuz i have a question and dont want to interrupt

[velvet blade]

@ancient rampart you did better job than me

[vast ravine]

hmm

i realized there's a way to use 1-2sin^2theta

instead of flipping the fraction i just start on the left side

[velvet blade]

wait you don't need to do any of that

[sick steppe]

you're typically suppose to start on 1 side and work to the other

[vast ravine]

so i have 2sinϴcosϴ/1-1-2sin^2ϴ

[velvet blade]

Just expand the identities

[sick steppe]

brackets for the love of go please

[vast ravine]

lol

(2sinϴcosϴ)/1-(1-2sin²ϴ)

[sick steppe]

$\frac{2\sin{\theta}\cos{\theta}}{1-(1-2\sin^2{\theta})}$

[obsidian monolith]

moshill1:

[velvet blade]

Wait sorry

[sick steppe]

see the error you made propr?

[vast ravine]

didn't i write that here?

so i have 2sinϴcosϴ/1-1-2sin^2ϴ
@vast ravine

[sick steppe]

you distributed the - sign wrong

[velvet blade]

What's -(-1)?@vast ravine

[vast ravine]

oh i'm on a mac

i'm using the characters for theta and exponents

maybe it doesn't translate

[velvet blade]

? No, not the characters

your sign convention

Just use any letter instead of theta for ease of typing

[sick steppe]

you're saying 1-(1-2sin^2(x)) = 1-1-2sin^2(x)

[vast ravine]

you're saying 1-(1-2sin^2(x)) = 1-1-2sin^2(x)
@sick steppe i see i need paranthesis to distribute

[sick steppe]

yeah, so after they're distributed, what is it?

[vast ravine]

so i'd be left with -1+2sin^2(x) on the bottom?

[sick steppe]

1-1... yeah

[vast ravine]

derp

so after the 1-1 cancel id be left with -2sin^2(x)

the 2 would cancel each other on the top and bottom

1 sin would cancel each other

leaving cos(x)/-sin(x)..?

[ancient rampart]

no it will be +sin(x)

1-(1-2sin^2(x)) the negative sign will distribute

to both terms in the parentheses

[vast ravine]

oic

[ancient rampart]

so 1 will become -1 and -2sin^2(x) becomes 2sin^2(x)

[vast ravine]

and therefore cos(x)/sin(x)=cot

[ancient rampart]

yep

[vast ravine]

which verifies the identity

[ancient rampart]

ok now i will ask my question

[vast ravine]

thanks for your help guys

@ancient rampart is this directed at me?

[ancient rampart]

no

[vast ravine]

you are in QI, between 0 and pi/2

[ancient rampart]

just asking for help

[vast ravine]

cos x = .4

cos is the x value

[ancient rampart]

ye and by pythagorean theorem i have sinx is root(21)/5 and tanx is root(21)/2

and now i just use half angle identities right

[vast ravine]

i'm working on that stuff right now maybe someone else can take over 😄

[ancient rampart]

nvm i figured it out :P

[vast ravine]

what is it?

[ancient rampart]

well for that specific one u just apply half angle identities cuz u know cosx

so u just plug in cosx for all 3 identities

[vast ravine]

all these trig identities make my brain go crazy

i hate seeing all the same variables repeated again and again

i have here cos⁴(x)-sin⁴(x)=cos(2x) . verify the identity

there's the double angle formula for cos²x-sin²x

but doesn't that leave an extra cos²x+sin²x

that isn't accounted for?

this is what the back of the book says

how can the answer be cos2x if there's still cos²x+sin²x?

[ancient rampart]

because cos²x+sin²x = 1

Pythagorean identity

[nova ravine]

when proving identities, use what you have from the formula sheet.

[vast ravine]

because cos²x+sin²x = 1
@ancient rampart thank you

i need a bigger formula sheet lol

[idle moat]

how do you find the inverse of a sine function?

[gentle vigil]

First you define an interval in which sine is bijective

Then you define a function f that satisfy f(sinx) = x in the interval

[velvet blade]

how do I prove this?

[void zealot]

Is anyone familiar with R^2 values of a graph

[sick steppe]

@void zealot more specific please

[void zealot]

It has to do with how a line fits with a set of point on a scatter plot

This might help https://www.investopedia.com/terms/r/r-squared.asp

[gentle vigil]

It says it's the variance

[void zealot]

Could you explain?

[gentle vigil]

Variance is a quantity that is a measure of difference between two data set

[void zealot]

ok that makes sense Thanks for your help

[plain turtle]

so i didnt take notes for these can someone help?

https://cdn.discordapp.com/attachments/758826217595731989/779177494003318804/Screenshot_36.png

[mystic umbra]

you can graph it

or you can use calculus

your choice

what do you want to do?

@plain turtle

[plain turtle]

@mystic umbra calculus

[mystic umbra]

take the derivative of the function

then do =0

you need to find the absolute maximum of the function

all maximums and minimums have a derivative= 0

write down the possible values for x

it helps to use a graphing calculator

[vast ravine]

how many degrees do i add to 45 to get in Q4?

is there a specific ratio i need

[mystic umbra]

yes

270 <= Q4 <= 360

270-45=225

so you need at least 225

that puts you on the line Q3 Q4

[vast ravine]

so if i had 55.00 degrees and needed q4 should i subtract from 360 then?

[mystic umbra]

no that is not going to work all the time

that is not the formula

270-55= on the edge of Q3 Q4

360-55 on the edge of Q1 Q4

unless 55>90

[quaint mason]

how do u solve for the side from first base to the point where they all meet

[astral mantle]

the center line is 403 feet

the lines from home plat to first base and home plate to third base are 90 feet

use some LAWS

if you know what im sayin

[vast ravine]

can you guys help with this one

2sin²(2ϴ)+7sin(2ϴ)+3=0

i think it's quadratic but i can't figure out how to factor it

[astral mantle]

try factouring out sin(2theta)

[quaint mason]

LAWS?

[astral mantle]

like law of cosines lol

[quaint mason]

oh thats what i did

a^2=b^2+c^2-2(b)(c)cosA

right?

[astral mantle]

depends on how you labelled the diagram

what was your answer?

[quaint mason]

lemme check

and for that

i used 90 for b and 430 for c

[astral mantle]

the answers probably like 300 or near there

i didnt do it just a guess

300 something probably

[quaint mason]

hm

i got like

284.96

unless my calculation was off?

[astral mantle]

$cos(45) = \frac{(403)^2+(90)^2-a^2}{2(403)(90)}$

[obsidian monolith]

Star_:

[astral mantle]

solve for a and should be the answer

[quaint mason]

cant i do um

a^2=90^2+420^2-((2(90)(430)cos(45))

after solving left side

then square root both sides

[astral mantle]

what i did is just a manipulated version of the law of cosines

[quaint mason]

it would still give same answers right?

[astral mantle]

i got like 345.46

that seems like the right answer

[quaint mason]

i got 336.50

oh shoot thats why

ok i got 345.28 now xD

[astral mantle]

close enough

[quaint mason]

ty star

[idle moat]

how do you find sin(pi/2 - theta) and cos(pi/2 - theta)?

[quaint mason]

sain^-1 both side

to get theta by itself

sin^-1

[quaint mason]

are u not supposed to do cos^-1(8839/240000? I also tried tan already

confused on what to do

[willow bear]

have you made a diagram?

i have a strong feeling you're trying to go into this blind

[quaint mason]

So

Yeah i did kinda

The distance from where hes standing to where the mountain is

Is 240000 meters

The height is 8839 meters

So i draw the angle of elevation from where hes standing to the peak of the mountain's height

And im tryna find that angle

And for b, i did 170000 instead bcuz ur just subtracting since hes closer by 70km

I made 240000 and 8839 meters as 2 different sides. Was I suppose to find the hypotenuse then use inverse cos from there instead?

[viscid thistle]

This is a weird question.

So his 240km away from the peak.

And then declares the entire mountain's height is 8.839km?

[willow bear]

@quaint mason can you show me your diagram

[viscid thistle]

This is what I'm imagining but I might be wrong.

[willow bear]

[viscid thistle]

Situation A is possible, but it would probably require more assumptions.
Situation B makes the dude look nooby as hell (since he didn't climb the mountain), but It'd be easier?

[willow bear]

no ok like

irl 240 km is far enough that itd be well over the horizon unless you were like a kilometer up from sea level at least

so we're assuming a flat earth here just for simplicity

[viscid thistle]

Oh okay, so he's further away from the mountain.

Like he's not even close to it.. lol that makes so much more sense haha.

I thought he was gonna climb the mountain or something.

So more like this?

[willow bear]

well the angle of elevation would surely be arcsin(8.839/240)

this diagram makes it near obvious that arccos just gives the wrong angle

[viscid thistle]

Yeah, you have hypotenuse and opposite. And we're looking for that angle where the dude is.

Let θ be "the angle where the dude is to the mountain peak".
Then sin(θ) = o/h
So to find θ, we can use arcsin(o/h) = θ.

btw, I came here to ask a precalc question. I was going through a calculus book and at the start there were some questions which were like "You should know how to do" questions, and I got stuck on one sadly.

$$\text{Factor }3x^\frac{3}{2}-9x^\frac{1}{2}+6x^{-\frac{1}{2}}$$

[obsidian monolith]

ryаn:

[viscid thistle]

My first step was to factor out the 3, which gave me this:

$$3\left(x^\frac{3}{2}-3x^\frac{1}{2}+2x^{-\frac{1}{2}}\right)$$

[obsidian monolith]

ryаn:

[viscid thistle]

But I have no idea what to do after that.

Just a buncho squareroots.

[willow bear]

i mean you could factor out x^(-1/2) from this and be left with 3/sqrt(x) * (x^2 - 3x + 2)

[viscid thistle]

Ah right, that looks good, I'll give that a go, thanks!

[quaint mason]

Sorry for late response @willow bear, basically the height goes up 8839, and the horizontal line connected to it is 240000.

[willow bear]

show don't tell

[quaint mason]

Ok

[willow bear]

i want to see your diagram with the angle of elevation clearly marked

[quaint mason]

Aighty

Um

Ignore the work above, idk what else was expected.

I mean thats the diagram for the angle of elevation

[willow bear]

,rccw

[willow bear]

ok

so...

[quaint mason]

?

Rccw?

[willow bear]

i had the bot rotate your image

[quaint mason]

Ohh

Oops

[willow bear]

anyway

where did you get it from that cos(θ) would be 8.839/240

[quaint mason]

I forgot that cos was adjacent over hypotenuse. Got mixed up w tan(x)

So my 2nd approach wouldve been

To solve for the hypotenuse

[willow bear]

why tbh

just take arctan(8.839/240)

[quaint mason]

Wym

[willow bear]

and be done

[quaint mason]

I did arctan and got a very small number and wasnt sure if that was right the first time around

It was like 2. Something degrees

Cant remember exactly

[willow bear]

two-ish degrees sounds about right

[quaint mason]

Oh.

[willow bear]

think about it - 240 km is like what, 30 times the height of mt everest?

the triangle is gonna be really thin

[quaint mason]

Yeah

If i did it by finding hypotenuse first and used cos^-1

Would i still get same result?

[willow bear]

sure but you might get fucked over by loss of precision

[quaint mason]

Ow, tyvm btw Ann

[bitter sandal]

I need help with B. #2

[viscid thistle]

that's very subjective

try to make some cases

[willow bear]

wym "fibonacci-like sequence" lmfao

[bitter sandal]

I've been trying but I have to get some sleep. Though my #1 would be correct if my observation is that the quotient is the sum of the two previous terms of it, right?

[uncut mulch]

start with any two integers ig

[bitter sandal]

thanks this helped! i can conclude that the sum of the first 10 ten terms = 7th term of that sequence

[proven marten]

question 4 (no calc in both ways)

guessed one root, x+1

x+2

[willow bear]

x+1 isnt a root its a factor

[proven marten]

my bad

okay so question correction

its 2 + 3x - x^3

is there a way to find the turning points without differentiating? (because my teacher hasnt covered any calc yet)

[blissful ridge]

2+3x-x³=(2-x)(x+1)²

So we have a double root at x=-1

So that's your first turning point

<@&268886789983436800>

[kindred sluice]

what

is that not allowed here

[severe verge]

it is not allowed

[blissful ridge]

No, it's not

[kindred sluice]

i just looked up math discord server

ok thanks

[blissful ridge]

And read the goddamn rules

[kindred sluice]

i dont have time for that

🌚

[blissful ridge]

@proven marten

[opal cosmos]

Hey guys

Im doing sum and difference identities in precalc right now, and when I was doing hw, i saw there was pattern in answers depending on what trig fx

for cos and sin My answers were always +-sqrt 6 +-sqrt2/4

And tan, +-1 +-sqrt 3 /+-1 +-sqrt 3

Is there some sort of ways to predict +or - on each number by looking at question?

[idle moat]

how do you sketch the reference angle?

[willow bear]

is there a way to find the turning points without differentiating?
no

[fresh bluff]

how can you find all values of y at a given x if the function passes through that x multiple times?

[viscid thistle]

How would I use mathematical induction to prove

Khan academy got me all messed up on this

[patent beacon]

Base case, which I assume you can do.

Then the inductive step.
Assume 3 + 6 + ... + 3n = 3n(n+1)/2
Prove 3 + 6 + ... + 3(n+1) = 3(n+1)(n+2)/2

That make sense?

[viscid thistle]

Not really

I got the base case proven, just unsure how to prove the inductive step

[patent beacon]

Like, the inductive step as I layed it out doesn't make sense?

Or it does, but you just can't prove it?

[viscid thistle]

Where did you get 3(n+1) from?

This is where I'm at rn

No clue if its even remotely correct but

[blissful ridge]

You see the second last term second be 3(n-1) and not n

[viscid thistle]

The what?

the 3n^2+9n?

[blissful ridge]

$1+3+6+..........3(n-1)+3n$

[obsidian monolith]

The Godfather:

[viscid thistle]

I'm lost

[blissful ridge]

If you are summing upto 3(n-1) term

Then

$\underbrace{1+3+6+......+3(n-1)}_{\frac{3n(n-1)}{2}} + 3n$

[viscid thistle]

o

$\underbrace{1+2+3+......+3(n-1)}_{\frac{3n(n-2)}{2} + 3n$

Do I just leave it and it makes it?

[obsidian monolith]

KEVIN:

$\underbrace{1+2+3+......+3(n-1)_{\frac{3n(n-2)}{2} + 3n$
```Compile error! Output:

! File ended while scanning use of \underbrace.
<inserted text>
\par
<*> 703195331877404792.tex

I suspect you have forgotten a }', causing me to read past where you wanted me to stop. I'll try to recover; but if the error is serious, you'd better type E' or `X' now and fix your file.

[blissful ridge]

Bot is slow

[viscid thistle]

oh

[obsidian monolith]

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

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

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

[viscid thistle]

Okay so

Oh so 3(n-1) is going to be the term before 3n

[obsidian monolith]

The Godfather:

[blissful ridge]

Yes

[viscid thistle]

So this would be my proof?

I have the base case proven too just not in picture

[blissful ridge]

You still have to write the step where you assume the formula to be correct

The induction hypothesis step

[viscid thistle]

Oh I have to write that?

Okay

[viscid thistle]

Which is this?

[blissful ridge]

Yes

[viscid thistle]

So this would be my full answer

[blissful ridge]

If you go this route, then write the formula is true upto (n-1)th term

This is called strong induction

[viscid thistle]

ok thanks

[patent beacon]

Your prove line is wrong

When we do these questions, we are discussing some proposition P(n) (which means "the proposition is true for n")
.

Base case:
P(0) is true

(or not necessarily 0)

Inductive step:
Assuming P(n)
Prove P(n+1)

[heady field]

would a local extrema be an absolute minimum or absolute maximum?

[sick steppe]

Not necessarily

[heady field]

how many would this graph have?

i am doing test corrections and it says it has 1

[quick mirage]

if f(c) >= f(x) of all x near c then its a local extrema

assuming obviously x = c in f(c)

[heady field]

what is c

[quick mirage]

c is a value

any local extrema value when plugged into f(x) in this case

[heady field]

oh

so it would be then

would that also be an absolute minimum?

wait nvm that's a stupid question

it's not

[heady field]

so i got (x^2+1)(x-2), and i think that's right, but how would i make it so it passes through the point (3,5)?

[sick steppe]

@heady field co-efficient of k at the very front scales the shape

[heady field]

yeah but isn't that too change the y intercept?

[sick steppe]

Yeah

but you dont care about the y-int

[heady field]

how would i find out what k is?

oh would i plug in 5 = k(3)?

nope that's def not it

wait no it is

i divided the wrong thing

thanks that's the last thing i was struggling with 😄

[sick steppe]

bad notation as a heads up

you implied k is a function

[heady field]

oh my bad

[sick steppe]

it's fine

[velvet blade]

Is this a good proof to show inverse mapping of F_(a,b) exists?

[plain turtle]

My fellow math peoples...

Can you help me with my work? I need answers with work so I know what to do for similar questions

I urgently need it, someone plz save me ;-;

I'm gonna need screenshots of work

[velvet blade]

If $f^5 = I$, express $f^-1$ as a positive power of $f$.

What did he mean in that sentence?

[obsidian monolith]

lazypawtato:

[sick steppe]

$f^{-1}(f(x)) = I$

[obsidian monolith]

moshill1:

[sick steppe]

@velvet blade this is my guess?

Like what power do you raise f to get the inverse of f

[velvet blade]

I wrote $(f^{-1})^5$

[sick steppe]

put powers in {}

and my interpretation is that that's wrong

[obsidian monolith]

lazypawtato:

[velvet blade]

He asked to show it in positive power right?

wait, positive power of f

[sick steppe]

yeah

so by definition: $f^{-1}(f) = I$ right?

[obsidian monolith]

moshill1:

[toxic bridge]

how would I solve this? I just need a refresher

[patent beacon]

Just the slope of the secant line

Between those two points

[toxic bridge]

just plug in -5 for x, solve and do the same for 6?

[velvet blade]

@sick steppe Is I^-1 = I?

[sick steppe]

I think so?

[velvet blade]

Okay, But I still am not sure what the question was

Let me send a screenshot

[sick steppe]

$f^{-1}(f) = I$

[obsidian monolith]

moshill1:

[velvet blade]

Yes I got that

But I can't determine what f is if only f^5 is given right??

I'm sorry for late reply, I'm doing the problem below

[sick steppe]

@velvet blade you dont need to know what f is

f^2 * f^3 = f^5

as a hint

[velvet blade]

wait oh yes

bruh I'm having difficulty figuring how to write it down

Wait

I got it

F^5(f^(-1))= I (f^-1)

f^(-1)=f^4

Damn thanks a lot for sticking with me dude

[mystic umbra]

@plain turtle I replied to you yesterday with the answer

try this https://www.mathsisfun.com/calculus/maxima-minima.html

detailed explanation

that only shows the local maxima

you want the absolute maximum of the function which could be the local maxima

sometimes however, you have more than one local maxima so they can't all be the absolute maximum of that function

there can be only one absolute maximum of a function if a function has an absolute maximum

some functions have 1 or more local maxima but no absolute maximum because of the function running off to inf

[willow bear]

@velvet blade note that in your problem with f^5 = id if you write out all the powers of f in order they'll follow a cycle of length 5

and there is nothing really special about the number 5, this sort of thing will happen any time some power of your function ends up as the identity

[velvet blade]

Ohh I didn't see it that way

I have a different question now, wait let me send the screenshot

How does $\tau_n\sigma$ leave n fixed?

[obsidian monolith]

lazypawtato:

[velvet blade]

whoa that worked well

[willow bear]

$\tau_n(\sigma(n)) = \tau_n(k) = n$

[obsidian monolith]

Ann:

[velvet blade]

Okay

[willow bear]

i assume J_n refers to {1,2,...,n}

[velvet blade]

Yes

it's chapter on permutations

[willow bear]

uh huh yeah fun stuff

[velvet blade]

I also liked mappings

Hey so if it's viewed as a permutation of $J_{n-1}$ does that mean that only other numbers are associated differently but not n in this case?

[obsidian monolith]

lazypawtato:

[velvet blade]

@willow bear Just a small doubt

[willow bear]

it says nothing of where the first n-1 points go, only that they don't go to n since n is fixed

[velvet blade]

Okayy

[viscid thistle]

How would I manually calculate something like:
$$2^{0.123}\text{ or }2^\pi$$
I imagine the first could be done like:
$$2^{\frac{123}{1000}}$$

[obsidian monolith]

ryаn:

[viscid thistle]

And then a bunch of squareroots or something.

$$a^b = e^{b ln a}$$

[willow bear]

\ln(a)

anyway

these are rarely calculated manually to any number of decimal places

why exactly do you need to do so

[viscid thistle]

Just out of curiosity

Since I realized I couldn't do them manually.

I guess you could use binomial theorem too maybe.

But it seems e^(b * ln(a)) is the standard way.

[willow bear]

eh

[viscid thistle]

Yea, I feel pre 'eh' about it too.

Apparently a slide rule can do it lol.

Not that I intend on buying one for this.

[willow bear]

might as well just fire up a calculator

[velvet blade]

Do transpositions have a unique order you need to follow?

I am unable to figure out what the number in the subscripts mean in a transposition of a permutation

They're different in every problem's solution

Does tau_1 mean the interchange of first number or the interchange of the permutation of number 1?

Okay nevermind I got it

[viscid thistle]

lol yea agreed

[willow bear]

@velvet blade they're just names & are context-dependent

[umbral current]

@velvet blade What do you research in pre algebra

[velvet blade]

@umbral current I self study maths and have self doubts

[umbral current]

Understood

[velvet blade]

What about you?

[umbral current]

I am majoring in math

at my local university

and have plenty of self doubts as well

[velvet blade]

Oh what have you learned recently?

[umbral current]

I am just starting Calculus 1

[velvet blade]

Okay

[umbral current]

I am a newbie

[velvet blade]

In maths major?

I thought majors have real analysis

[umbral current]

yes

they do

but I haven't gotten to that point yet

nowhere near it

[velvet blade]

Okayy

I'm studying mechanical engineering 3rd year

[left fable]

Real analysis will give you eye bags and dark circles around your eyes

[willow bear]

jokes on you i already had eyebags

[left fable]

lol. Yah, I never caught much sleep when I learned real analysis, hence my eye bags. Perhaps it’s a mathematicians trait whether before or after analysis 🤷🏽‍♂️.

[velvet blade]

I'm already feeling mentally miserable

I don't want physical misery

[tough yoke]

could anyone help me out here with the solutions for part 3?

[sick steppe]

,rotate

[sick steppe]

part 3 is an optimization question @tough yoke

[viscid thistle]

How would I do the induction proof for this sequence?

Would I be using (n+1)^3?

[lunar axle]

could someone verify if this is right for me? or is it 12/25 instead

[viscid thistle]

Should be positive 24/25 as the angle is in quadrant 3

Since sin and cos would both be negative

[sick steppe]

@viscid thistle test the base case, assume true for n=k, prove true for n=k+1

[lunar axle]

thank u sm!!

are any of these incorrect?

would this be positive as well??

[obsidian monolith]

AstragaleDB9:

Yes that it positive. We have $sin(2\theta) = 2sin(\theta)cos(\theta) = 2 * frac{-3\5} * frac{-4\5} = frac{24\25}$
```Compile error! Output:

! Undefined control sequence.
l.54 ... = 2sin(\theta)cos(\theta) = 2 * frac{-3\5
} * frac{-4\5} = frac{24\25}$
The control sequence at the end of the top line
of your error message was never \def'ed. If you have
misspelled it (e.g., \hobx'), type I' and the correct
spelling (e.g., `I\hbox'). Otherwise just continue,
and I'll forget about whatever was undefined.

AstragaleDB9:
Compile Error! Click the reaction for details. (You may edit your message)

[willow bear]

that's now how you type fractions in latex

$\frac{-3}{5}$

[obsidian monolith]

Ann:

[willow bear]

in general, \frac{num}{denom}

[obsidian monolith]

AstragaleDB9:

AstragaleDB9:

AstragaleDB9:

[willow bear]

also, \sin and \cos exist

and \cdot for the multiplication sign

[neon escarp]

what is dy/dx of e^(x+y)=y

[sick steppe]

@neon escarp implicitly differentiate and then isolate for dy/dx

[granite crescent]

i think this is the right spot to ask-

can someone help me with #5??

[sick steppe]

@granite crescent did you find the derivative?

[granite crescent]

yes i did

[sick steppe]

Ok what did you get?

[granite crescent]

it’s y’=-2sinxcosx

[sick steppe]

yeah, so what do you next?

[granite crescent]

you find the slope but that’s where i’m kinda stuck

[sick steppe]

ok how do you find slope given the derivative?

[granite crescent]

you put in the point given

[sick steppe]

yeah

[granite crescent]

but like i get stuck here

[sick steppe]

evaluate the trig

[granite crescent]

idk if i’m doing it right

[sick steppe]

no

where did the denominators go?

[granite crescent]

isn’t it /4 ??

[sick steppe]

yeah

$-2\frac{\sqrt{3}}{2} \cdot \frac{1}{2} = \frac{-2\sqrt{3}}{4}$

[obsidian monolith]

moshill1:

[granite crescent]

wont the 4 =2 after ?

cuz reduction

[sick steppe]

yeah

so why did you write -2sqrt(3)?

[granite crescent]

i wrote - sqrt(3)/2

[sick steppe]

In the pic of your work you have -2sqrt(3)

anyway

yeah it's -sqrt(3)/2

[granite crescent]

yes okay thank youuu

can i get help for the sixth question too?

i know the slope which is “2”

the derivative is y’ =e^x

so then you do 2=e^x

but idk where to go from there

[gaunt mason]

The tangent will contain the point (x0, y0) on the curve. The slope of the tangent is the value of the derivative at x0. You have the requirement that the slope has to be 2.

So 2 = e^(x_0)

To find x_0 you make an ln operation on both sides of the equation.

ln 2 = x_0 by the rules of the logarithms

@granite crescent

[sick steppe]

@granite crescent solve for x from 2= e^x

[granite crescent]

x=0.69 ?

[sick steppe]

ln(2), yes

[granite crescent]

but then y=1.99 or 2

so would the point be (0.69,1.99) ?

and the slope =2 ?

[gaunt mason]

You don't usually approximate these values with decimals. You put it as ln2

Which will give you the point (ln 2, 2)

And the slope is 2 as per requirement. So you have a point and a slope and you need to find the equation that satisfies this. That will be the equation of the tangent line, that is your solution.

[quaint mason]

how do u solve for this problem?

[sour hemlock]

let's see what you have tried so far

[viscid thistle]

@quaint mason

[quaint mason]

oops

one sec

so i wasnt too sure, but i started off by adding the 2 values that makes up the horizontal leg which was 85 feet

and then i wasnt so sure about how to figure out the angle, but could u use laws of sines afterwards?

[viscid thistle]

I mean

There's a way where we don't use one piece of information

[quaint mason]

i was tryna do

[viscid thistle]

If you consider the supplementary angle of 95.5°, the one inside the smaller triangle

You can just use a trig ratio

[quaint mason]

60/sin(95.5)=

erm

[viscid thistle]

What?

[quaint mason]

trig ratio??

[viscid thistle]

The bigger triangle isn't a right triangle

You can't use trig ratios

Yeah

soh cah toa

[quaint mason]

but u said to use trig ratio

[viscid thistle]

Read carefully

[viscid thistle]

.

the one inside the smaller triangle

The smaller triangle is indeed a right triangle

Let me post a pic to clarify this

[quaint mason]

wait but

how do u know that it is a right triangle, i get that the angle is getting smaller but

it doesnt say that its a right triangle

[viscid thistle]

This angle marked is just 180-95.5

And as we know the smaller triangle is a right triangle

Because of the height

[quaint mason]

the height determines that its a right triangle?

but i thought that

[viscid thistle]

You ain't gonna measure the height in any form but in a vertical line

[quaint mason]

the angle determines that its right triangle

90 degree angle

[viscid thistle]

I feel like you know what i mean, but you don't know how to express it into words

Yes, if we have the height of a building, which is just a vertical line, it is assumed that it forms 90° with the horizontal

[quaint mason]

ohh

ye

[viscid thistle]

It does even tell you explicitely at the second question

[quaint mason]

yeah i gotcha now

so to solve for height, i would have to solve for the hypotenuse

and then h^2=c^2-85^2

erm

ye

[viscid thistle]

I mean

[viscid thistle]

.

You have one angle, and you have the opposite side

And you want the adj side

Consider the appropiate trig ratio

[quaint mason]

tangent

[viscid thistle]

Yes

[quaint mason]

lemme try it

i got a very small height

[viscid thistle]

That's weird

Let me see what you got

[quaint mason]

so its tan(84.5)=85/h?

[viscid thistle]

Wait

[quaint mason]

or is it suppose to be tan(84.5)=25/h

[viscid thistle]

Isn't it 25

Ofc

Why would it be 85?

Wait

,calc 180-95.5

[obsidian monolith]

Result:

84.5

[viscid thistle]

It's all the way around btw

Remember tan is opp/adj

Not adj/opp

[quaint mason]

so its h/25?

[viscid thistle]

Yeah

But do you get why

I don't wanna have you saying "yes" without having understood it

[quaint mason]

yeah, u basically took the line which is 180 degrees, subtract 95.5 to get the opposite angle, then ur using that small angle to solve for the height. it gives u the horizontal value

you have the angle

and use tangent because of o/a

[viscid thistle]

Yes

[mighty iron]

,calc 5²

[obsidian monolith]

The following error occured while calculating:
Error: Syntax error in part "²" (char 2)

[mighty iron]

,calc5^2

[viscid thistle]

@mighty iron #bots

[mighty iron]

oh

srry

[viscid thistle]

All g

[quaint mason]

tan(84.5)=h/25 then multiply 25 both sides to get h alone

thats it?

and tyvm btw

[viscid thistle]

Yes

Yw

[little plank]

Are you free in this channel

Well I just need help with the f (x) =3 [x-4]

[viscid thistle]

,rccw

[little plank]

You see the box its blank that's because i don't understand

[little plank]

Which formula to use

I don't know how to put my answer together

The vertex is 0,4
Domain -infinity sign , infinity sign
Range [0,infinity sign ]

But my teacher didnt explain how to put the equation togther

[sick steppe]

what do you need to answer?

[little plank]

Well yes but i don't just want the answer i wana know how to do this

Like the formula when to use it

[sick steppe]

formula for what?

i asked what do you need to answer cause idk what the question is

[little plank]

The blank

And thr formula above it

[lapis sphinx]

Write each equation in transformation form it says

[little plank]

But i don't know how

[sick steppe]

I know how to write transformation form, but the question doesnt give enough info.

since you need the base function defined

[little plank]

I have the vertex domain and range

[sick steppe]

g(x) = |x| means f(x) = ag(k(x-h)) + c

[little plank]

Okay can you help me solve it

Or better yet what's the k x HC and represent

[sick steppe]

k is horizontal dilation, h is horizontal translation, c is vertical translation

but like i said, not enough info unless you're gonna define the base function yourself

[little plank]

Oh what info would I need to in order to put it all together

[sick steppe]

g(x) = |x| means f(x) = ag(k(x-h)) + c

"define the base function"

said 3 times now 🙃

[little plank]

Thanks

So another words k&h RX including the other X and C is y

Are x

Ok thanks for your help

[torn pebble]

can someone explain how the 13 became neg

[limber compass]

dk how to do this?

[viscid thistle]

@limber compass this channel is occupied, please proceed with a free one.

Not even more than a minute passed between the last question asked...

@torn pebble okay so it's negative due to the derivative of (1/(x+1))

Let me show you why

[limber compass]

okay my bad.

[viscid thistle]

$$13\cdot \dv{x}\left[\frac{1}{x+1}\right]$$ $$13\cdot \dv{x}\left[(x+1)^{\color{green}{-1}}\right]$$ $$13 \cdot{\color{green}{-1}} \cdot \underbrace{\overbrace{(x+1)^{-1-1}}^{(x+1)^{-2}}}_{a^{-n}=\frac{1}{a^n}} \cdot \dv{x}[x+1]$$ $$-13\cdot \frac{1}{(x+1)^2} \cdot \dv{x}[x+1]$$ $$-13 \cdot \frac{\dv{x}[x+1]}{(x+1)²}$$

@torn pebble

[obsidian monolith]

Al𝟛dium:

[torn pebble]

oh wow okay thanks a lot

that makes a lot of sense

so would you be able to use the quotient rule to solve (13/x+1)

[viscid thistle]

I mean sure

As long as you apply it appropiately, you get to the same thing

And write 13/(x+1) not (13/x+1), they aren't the same but i guessed what you meant by context

[hexed ermine]

Aledium god at latex

[velvet blade]

Can I ask about precalc proofs in proofs and logic channel if this channel is occupied?

[viscid thistle]

Lol

[torn pebble]

okay thanks!

[viscid thistle]

Yw

[viscid thistle]

I mean uh i guess?

That's a good question to ask

[velvet blade]

lol thanks, If it's free here I can ask it here

[viscid thistle]

Maybe it is more appropiate here, but i don't think it gets to a point where it is innapropiate there

[velvet blade]

oh okay

[hexed otter]

guys can someone help me in mathematical induction ;-; having a hard time

[hard hedge]

I’m having trouble trying to set this up

nvm I got it

can someone help me with this problem please

#4

[viscid thistle]

a 6m long ladder is leaning against a wall at a height of h and angle of theta.

the ladder slips down the wall at a constant rate of 0.05m/s. find the rate at which the angle is changing when the height is 2.5m

how do i do that

https://cdn.discordapp.com/attachments/419072946787844097/779986668513918976/unknown.png

[astral mantle]

alright write out h in terms of 6 and theta

then you can differentiate both sides

and solve

for dθ

@viscid thistle still need help?

[viscid thistle]

i solved it just now

sorry

[astral mantle]

np

[patent lance]

Hey

so Im tryna do this problem

but ig im doing it wrong

theres a video showing how to format it

but I dont know waht Im doing differnelty

Im doing 5000((1.06/12))^60

but that seems to not be the answer

[quaint mason]

Can someone help draw out these triangles for me? kinda confused on how its suppose to look like

[quick mirage]

im not the best drawer but i can try lol

[quaint mason]

ty :)

[quick mirage]

@quaint mason x and y would be d1 and d2 (depending on which one is larger / smaller).

If you ever see something like 34 degrees east of north or 49 degrees west of north, just think of the angle starting at north and just moving 34 degrees east or 49 degrees west

Also notice the right angles, you can solve for the triangle's angles that way

And the rest is just applying trigonometric laws

[quaint mason]

ty

[patent lance]

Im hvaing some trouble

with

I cant figure out how to format this

thought itd be 8.50(1.05)^10

I figured it out

im slow

[quaint mason]

i thought that these had no solution at first but, thats bcuz i tried using the law of sines

i then tried to use the law of cosines but i dont have the value of c...

well the value for side c

how would i solve for side c?

[sick steppe]

@quaint mason pretty sure you do law of sines to find B first

[quaint mason]

thats what i tried doin so

i did um

11/sin136=4/sinB right?

and then i divide both sides by 4

[sick steppe]

yeah

[quaint mason]

to get sinB=whatever value

then i did sin^-1 both sides

[sick steppe]

you'll have 1/sinB but yeah

[quaint mason]

and

it said domain too big

[sick steppe]

You didnt take the reciprocal before taking sin^-1

[quaint mason]

wait what

wait

why is it still 1/sinB

[sick steppe]

You divided by 4, so you're left with 1/sinB

[quaint mason]

i divided both sides by 4

[sick steppe]

yeah

the sinB is the denominator

[quaint mason]

shouldnt that get sinx alone

wait

then

ugh

ok

so then

[sick steppe]

$\frac{\sin{136}}{11} = \frac{\sin{B}}{4}$

[obsidian monolith]

moshill1:

[quaint mason]

(11/sin136)/4=1/sinB

wait

u can flip em?

[sick steppe]

Yeah...?

[quaint mason]

i thought it was

side/sin(angle of the side)

would it still be the same?

[sick steppe]

as long as they're all consistent you can have sin(A)/a or a/sin(A)

[quaint mason]

so i can multiply both sides by 4

and then do sin^-1 both sides

[sick steppe]

yes, if you write law of sines the way i did

[quaint mason]

tyvm moshi!

[sick steppe]

np

[slate forge]

I'm confused on how to determine what is included in the terminal on the unit circle

Even when I get certain things right (say pi/3 is on the unit circle), the answer will include things like 15pi/3

idk how to A) come to these answers if they're not on the unit circle and B) determine if they fit the interval

can someone offer their assistance?

[sick steppe]

all angles are on the unit circle @slate forge

[opaque olive]

any hint about the second part

that the value of the limits are same?

is that all? -_-

[merry basin]

Hello, how would I go about proving that the limit of x + a, as x approaches infinity = infinity by definition?

[slate forge]

@sick steppe how do I know the specific angles when it only identifies a handful

[sick steppe]

if the number is in the interval it's in the interval

[slate forge]

Ik that but I need the angles that cot(x)=1 equals, for example

I can give examples of what the other problems r looking for when I get home from work 😭

[fringe imp]

hellO! I am having trouble on a problem if anyone is willing to help me out

https://gyazo.com/896e5f2983171b676f147a2ae4391b54

[sick steppe]

@slate forge cot(x) =1 means tan(x) =1 means x = pi/4, pi + pi/4

[fringe imp]

ive done previous problems like this in terms of solving the equation and finding points on the unit circle where these points would be true, but these equations are usually something like cos(2x) = -1/2

nver have i had a problem where sin(2x)=-sinx not sure where to start on this one

[sick steppe]

double angle @fringe imp

[fringe imp]

how would the double angle formulas apply here

do i start somwhere with sin(2x)+sinx=0?

[slate forge]

@sick steppe thank you, but what are the steps to solving any problem like these? I have like 80 of them and they use cot, tan, sec, cos, sin, etc.

[sick steppe]

@fringe imp so you can cancel the sinx?

@slate forge just get it in terms of trig = number, then cast rule

[fringe imp]

yeah move it to the other side

not sure what you mean by cancel though

[sick steppe]

$2\sin{x}\cos{x} = -\sin{x}$

[obsidian monolith]

moshill1:

[fringe imp]

oh okay yuse one of the double angle formulas to change the sin2x

[sick steppe]

yes...

that's why i said double angle

[fringe imp]

yeah right i ws just brain farting

then it would be sinxcosx=0 yeah?

[sick steppe]

what

no

the sine is canceled

[slate forge]

@sick steppe thanks for ur help

[fringe imp]

wait im brainfarting even harder now

[slate forge]

I'm gonna try it when I get home

but I'm afraid that I am going to simply fail ashvsbshs

[merry basin]

I don’t know how to approach this quesfion about limits of seauences

Any hinter towards any of these points would be great

[fringe imp]

@sick steppe okay i know what you mean by cancel now

i was working through it

and i ended up factoring out a sinx of x

leaving

sinx(2cosx+1)=0

$sin{x}(2cos{x}+1)=0$

[sick steppe]

that works too... or just cancel the sinx right away

[fringe imp]

haha

[obsidian monolith]

danmyles:

[fringe imp]

idk what you mean by cancel though

[sick steppe]

$2\sin{x}\cos{x} = -\sin{x} \implies 2\cos{x} = -1$

[obsidian monolith]

moshill1:

[fringe imp]

yeah that just

doesnt make sense in my brain

for some reason

[sick steppe]

weird

[hexed otter]

what do u do if there's a square outside of sigma equation??

helpppp

[viscid thistle]

@hexed otter i think that means you have to evaluate the sum and then square the result

[sick steppe]

,rotate

[sick steppe]

= (1+2+3+...+n)^2

[astral mountain]

Can i use L'hospital when i get undefined/0?

or how else am I supposed to solve such limit

we just consider undefined as a number or

[sick steppe]

what's Bgtan(x)?

[hexed otter]

wait so like uhh

[hexed otter]

oh thanks!

[sick steppe]

the big square just means square the whole sum

[hexed otter]

:oo ok thanks now i can work on it

[sick steppe]

yeah

[astral mountain]

Bgtan = arctan I guess

[sick steppe]

well cot = 1/tan

so the limit = 1/(tan(x)arctan(x))

[astral mountain]

oh ok

thanks

[patent beacon]

@astral mountain
Consider left and right handed limits instead

[slate forge]

this is the answer I got, the cast rule wasn't helping me

but it's wrong

this is the answer they were looking for

idk how to arrive to that

another example

any advice? @sick steppe

[cobalt swallow]

@slate forge when is cos( 3theta /2) = 1/2 ??

[viscid thistle]

You can always use a u-substitution so you can get a simple cos(u) = some value)

and then when you do that, you solve for u, and then use the original substitution to solve for x (or theta)

[viscid thistle]

Hey

Can someone help

I can’t understand this question

Because I heard that you can use any for X and Y but when I use different one they come out differently

[undone glen]

the function f can also be written in the form f(x) = ln ax/ln b

write down the value of a and b
write down the equation of the asymptote

<@&286206848099549185>

[fossil crown]

From

#❓how-to-get-help

Rule 4

If your question has not been answered for a minimum of 15 minutes, you may use the Helpers tag once. Please do not try to bump your question using this ping unnecessarily. Do not abuse this ping. Do not individually ping users with the Helpers tag without their express permission.

[undone glen]

my bad i forgot i guess

[echo wagon]

What is the function?

[undone glen]

i figured it out but thanks