precalculus

[rigid sun]

I’m unsure

[lethal oracle]

<@&286206848099549185> is there anyone available that can help me with this?

[native timber]

if you have $z \in \mathbb{C}$ with $r = \mid z \mid$ and $\theta = \arg(z)$, $z^n = r^n\left(\cos\left({n\theta}\right) + \text{i}\sin\left({n\theta}\right)\right)$

[lethal oracle]

Am I correct that my r is 2 and theta 0?

[native timber]

i don't know what your question is

[lethal oracle]

Find the nth roots of the complex numbers for the specified value of n

[obsidian monolith]

Jiramide:

[lethal oracle]

-2i, n=6

[native timber]

we can't determine what r and theta is without the actual complex number

in that case, r = 2 and theta = 3pi/2 rads (or 270 deg)

[lethal oracle]

Oh alright that’s where I went wrong

[native timber]

oops being dumb again

[lethal oracle]

Do I only have a theta of 0 when the coefficient of i is 0?

[native timber]

yes

if the complex number is solely real then theta is 0 or 180 deg

[lethal oracle]

Does the fact that it’s -2i change anything

[native timber]

yes

if you draw a graph where the x-axis is the real number line and the y-axis is the imaginary number line, where would -2i lie?

it's 2 units below origin right

[lethal oracle]

Ya

[native timber]

therefore theta would be 270 deg

[lethal oracle]

Doesn’t that give my the same solutions as theta being pi/2

[native timber]

while cosine does give out the same outputs for the input pi/2, sin does not

[lethal oracle]

I think that’s why I’m not getting the correct outputs

What should my arguments be?

Is it not pi/12+(4pi/12)k

[native timber]

i dont know where you managed to get those numbers

[lethal oracle]

Alright let me take a step back

Pi/2 is the theta

[native timber]

3pi/2

pi/2 points upwards, 3pi/2 points downwards

hold on lemme do this for myself

[lethal oracle]

Right I forgot it’s -2i

Let me find the roots again

Ya I’m not getting them

[native timber]

yep i just did it for myself and i don't seem to be getting your numbers

[lethal oracle]

What argument did you get

[native timber]

$m = 2, theta = \frac{3\pi}{2} \implies \sqrt[6]{2}\cdot\left(\cos\left(\frac{3\pi}{12}\right) + i\sin\left(\frac{3\pi}{12}\right)\right)$

[lethal oracle]

Shouldn’t it be 3pi/12+(2pi/6)k

[obsidian monolith]

Jiramide:

[lethal oracle]

For each argument

[native timber]

by argument are you talking about the angle

[lethal oracle]

Yes

[native timber]

why would it be (2pi/6)

the period of sine and cosine is 2pi

it's 3pi/12 + 2pi*k

[lethal oracle]

I thought I divide the 2pi by 6

Along with the original angle

That’s what I’ve been doing and I was getting the correct roots

[lethal oracle]

He wants roots given like this

[uncut mulch]

just an issue with your base angle

[lethal oracle]

This isn’t for that problem I’m just giving an example

So the base angle is 3pi/2

?

[uncut mulch]

yes

[native timber]

i don't recall having to divide 2pi*k by n as well

wouldn't that just divide the unit circle into nths

[uncut mulch]

yes.

it would

[lethal oracle]

My notes say to divide 2pi(k) by n

Is that the right way though ramonov?

[uncut mulch]

yes

[native timber]

hmm i wonder where i got confused

[lethal oracle]

Ya I got the correct roots now

And would the boxed portion be the right way to present the roots?

[native timber]

oh i get why you divide the periodicity by n as well

mb

[signal mountain]

Hi, I'm trying to solve part two of this problem. I tried solving for m=9/20(m_0) but i end up with the wrong answer, can anyone help?

[uncut mulch]

what did you get for k?

also you are solving the wrong thing

the original mass is m_0 (NOT 9/10 m_0 which was the mass at t=10)

what is half of m_0?

[signal mountain]

i got k=0.0105

[uncut mulch]

would've preferred an exact value but ok

[signal mountain]

i'm confused about half the mass

[uncut mulch]

what's the original mass?

[signal mountain]

m

[uncut mulch]

no

"m" is the mass at time t

do you understand that "m_0" denotes mass when t=0?

[signal mountain]

ah

i think so

[uncut mulch]

and 2nd part of the question is asking you to solve for t when
m = ?

[signal mountain]

i realise i don't understand this question at all

[uncut mulch]

what is half of the original mass?

[signal mountain]

m_0/2 ?

[uncut mulch]

yes

$ \frac{m_0}{2} = m_0 e^{-kt}$ where $k = \frac{\ln (10) - 2\ln (3)}{10}$

[obsidian monolith]

ramonov:

[uncut mulch]

solve that for t

[signal mountain]

ah, that works. i think i understand now, thanks

[odd helm]

For #39 why does the key do ln7 / ln2

[uncut mulch]

change of base law

your calculator doesn't have a log base 2 or log base 7 button

[odd helm]

Alright thank you

[green zenith]

I need help with triangles

Is it possible to have 2 triangles when you have the given area and 2 sides?

[uncut mulch]

what are your thoughts on it?

[green zenith]

Thats not possible, right?

In the case of both sas and ssa, its not possible because there is a fixed area. One of the triangles is smaller visually than the other one.

[uncut mulch]

triangles that look different can also have the same area

think about the area of a triangle
Area = 1/2 ab sin(theta)

can there be more than one solution for theta in (0,pi)

[green zenith]

So for example, if we have the given ssa that have two triangle, the triangles within it have the same area?

[gentle vigil]

Do you mean 2 sides are equal and an angle (not made by those two sides) are equal for 2 triangles?

They do not mean triangles are congruent

[green zenith]

No. What i meant is in a SSA triangle where one of the sides is opposite the angle, and the altitude<opposite<adjacent. This creates two triangles and would those be equal in area?

[gentle vigil]

They won't

Need not be*

[green zenith]

Okay that is what i thought

But another thing confuses me is if we had the given area and 2 sides

[gentle vigil]

So question you have to ask is, 2 sides and area fixes the other side?

Or fixes the angle made by them?

If yes, there is just one such triangle

If no, there can be multiple

[green zenith]

Oh

Okay

So earlier we had this one for example, we have area of 20 and 10 and 8 as sides

So do you mean that if we look for the angle, we can have multiple triangle but if we look for the other side, then we can only have one, is that correct?

[gentle vigil]

No

I'm saying that

You know that two sides of triangle is equal

If you knew the angle between those side us also equal

Then you would have said that they both are equal

Or

If you knew that the 3rd side is equal

Then also you would havr said that both triangles are equal

So you have to look for either of them in terms of area

Do you know how do you express area in terms of other sides/angles?

[green zenith]

Oh this is deeper than i thought

We only just started and its just the standard formula k=1/2bh, k=1/2absinC and k=(1/2)((a^2sinBsinC)/sinA)

[gentle vigil]

Yes!

So you know a and b for both triangles

And areas

Does that give you a unique value of C?

sinC =2k/ab

What does it say about C?

[green zenith]

It says sinC=1/2

So C=30°

[gentle vigil]

Only 30°?

[green zenith]

30 and 150

[gentle vigil]

Yes!

So does having equal area mean that angles are equal?

[green zenith]

Not necessarily?

[gentle vigil]

Yes!

But what if sides were 5 and 8?

[green zenith]

With area 20?

[gentle vigil]

Yes

[green zenith]

So if those sides were a and b, we jave C=90°

[gentle vigil]

So what's your conclusion about that?

[green zenith]

We just have one possible angle

So just one triangle?

[gentle vigil]

Yup

But for all the other cases, there will be two triangles

[green zenith]

Okay wow

But back on the case of the other one

Something is still bothering me

So does thag mean if you have 2 triangles, we have two different values of c? And still the same b value?

[gentle vigil]

Do you mean B and C?

A,B,C are angles and a,b,c are sides

[green zenith]

Yes c is what i meant

[gentle vigil]

a,b are same. But C is different. So c will also be different

[green zenith]

Ohhh

So i think this is my last question, the final piece of the puzzle, how do you solve for 2 different c's when you have the same a and b and and the value of both sinC is the same?

[gentle vigil]

There will be two solution to this. x and 180-x

You need to find one solution x

Do you know inverse sine function?

[green zenith]

Yes. Im familiar with basic inverse sine.

[gentle vigil]

So your angle is $\sin ^{-1}(\frac{2k}{ab})$

[obsidian monolith]

oscillatingEquilibrium:

[gentle vigil]

And another angle is

$\pi-\sin ^{-1}(\frac{2k}{ab})$

[obsidian monolith]

oscillatingEquilibrium:

[green zenith]

So those are my angle C's?

[gentle vigil]

Yup

[green zenith]

How about my side c's?

If i were to use the law of cosine, i would find one, but what about the other c?

[uncut mulch]

c would depend on which angle you are using

[green zenith]

Oooh

So in law of cosines it would make sense

I get it now

Thanks

[sullen sorrel]

can someone help me with c

[proud jetty]

https://gyazo.com/f6e8b02d16a1503ae8f9c869bccf6676.png

why can we just add cos(pie)cos(x)-sin(pie)sin(x)+ .....?

[queen grail]

Wdym @proud jetty

[proud jetty]

Could you explain me that problem?

Like where does cos(pie)cos(x)-sin(pie)sin(x) comes from?

[queen grail]

You're just applying the sum of angles identity

As in

sin(x+y) = sinxcosy + sinycosx
And
cos(x+y)= cosxcosy - sinxsiny

Where x and y are two angles

[proud jetty]

oh ic

so where does pie come from. wuts value of it

1?

[queen grail]

Do you mean what's cos (pi) ?

[proud jetty]

yea

[queen grail]

cos (pi) is the same as cos (180 degrees) and with an angle greater than 90 degrees we need to use the reference angle, and in this case it's zero. So we'd say cos (180 degrees) = cos (0 degrees). But you have to remember that cosine is negative in the second quadrant (where our angle is) so it becomes cos (180 degrees) = - cos (0 degrees).
Cos of 0 is just 1 so that's negative 1.

[willow bear]

pi, not pie.

[proud jetty]

ah ic.

thx alot

[obtuse kiln]

pee pee poo poo

[proud jetty]

@willow bear toss me sum pie

[obtuse kiln]

oops wrong discord

[willow bear]

?????

[obtuse kiln]

hey cutie

[willow bear]

?????

what do you want

[obtuse kiln]

digits

[willow bear]

what

[obtuse kiln]

like in math

[willow bear]

are you just screwing around rn or what

[obtuse kiln]

feet pics?

[willow bear]

ew.

why would i want to satisfy your foot fetish.

[normal sundial]

[obtuse kiln]

not mine

your fetish

[willow bear]

my what

what the fuck are you on about

[obtuse kiln]

you dont gotta be ashamed

[willow bear]

what the fuck are you on about

[normal sundial]

low effort

get better at trolling tbh

[obtuse kiln]

yeah kinda

[willow bear]

what the fuck are you on about

[obtuse kiln]

idk ive been in here for like 2 years and i never use it

you guys think julitet is a good cat name?

[grizzled orchid]

silence incel

your kind is not permitted here.

[obtuse kiln]

bruh im not an incel

youre an incel

[lost mesa]

your mom's an incel

woaooooooh im comedy gold guys

laugh

[obtuse kiln]

https://youtu.be/8WEhD3TxBDM

[proud jetty]

https://gyazo.com/bf72db2045ae65cbd2a3ffdb4e933790.png

what happened to the (cos^2(x) + sin^2(x))?

[wise kelp]

cos^2(x) + sin^2(x) = 1

[proud jetty]

oh I forgot about that thx

https://gyazo.com/9288d7decbea67578bdfba7bf83870ab.png

why can you just put 1. for csc^2(x)? If you manipulate it don't you need cot^2(x) there?
1=sec^2(x)-cot^2(x)

[wise kelp]

They're simplifying csc^2(x)/csc^2(x) to 1

[proud jetty]

and could you tell me how you end up with 1-2sin^2x?

[wise kelp]

csc(x) = 1/sin(x)

So 1/csc^2(x) = sin^2(x)

[proud jetty]

and could you tell me why 1-2sin^2(x) becomes cos(2x)?

[wise kelp]

That's just an identity

[willow bear]

cos(2x) = cos^2(x) - sin^2(x) = 1 - sin^2(x) - sin^2(x)

[proud jetty]

ah thx alot guys.

[pseudo sonnet]

https://i.imgur.com/Ct1U52V.png

here is what i have so far

am i able to draw that right angle in my drawing

i assumed because his vision is straight at the tree

in relation to the height

<@&286206848099549185>

alright this is what I got so far

can someone verify this

my setup

[pale bison]

looks fine, but you have to taken the height of Ed into account btw

[pseudo sonnet]

so

the total height

is actually

14.6 + 5.5 ?

[pale bison]

probably, i guess the question assumes that Ed has eyes on top of his head, not very scientific but 14.6 + 5.5 should be the height of the tree

[pseudo sonnet]

and for the second part, since the fence is 34 from the tree in all directions

and if the tree were to be cut 3 ft and above from the bottom

the tree would collapse a maximum distance of 20.1 - 3 = 17.1 ft

and since the fences are 34 ft away

it would be safe to cut the tree since they are at a greater distance further away

am I correct?

[pale bison]

sounds good to me

[warm crescent]

Redefine this function

f(x)=|x-2|+|x+2| , x \in [-3,3]

[willow bear]

wdym "redefine"

[warm crescent]

For example refining f(x) =|x| =(x^2)^{1/2} we get
f(x)=x , x>0
=-x , x <0
=0 ,x=0

@willow bear

[willow bear]

uh

so you're asked to write |x-2| + |x+2| as a piecewise function?

$f(x) = \begin{cases} 2x & x \in (2,3] \ 4 & x \in [-2, 2] \ -2x & x \in [-3, 2) \end{cases}$

[obsidian monolith]

Ann:

[warm crescent]

Yes!

How did u do that

I tried this for 1hr

[willow bear]

the boundaries of your pieces are 2 and -2

where the expressions inside the absolute values are equal to 0

[signal mountain]

hi, how do i find the minimum of y=x^2+e^x ? i took the derivative but it seemed tricky to isolate x

[willow bear]

,w 2x + e^x = 0

[obsidian monolith]

Display results online and refine query

Requested by Ann#0413

[willow bear]

hmmmmmmmm

well

do you only need the minimum itself or the x value at which it occurs

[signal mountain]

the answer gives an x and y value

[serene heath]

what are they

maybe they want u to approximate or sumn

[signal mountain]

-0.35, 0.83

the first part was to graph it, so maybe they only approx as you say

[serene heath]

yea mostlikely

cant really solve that exactly

[signal mountain]

ok, thanks

[viscid thistle]

W H A T

apparently this is right

that definitely to me does not look like reduced row echelon form

[abstract dome]

reduced = all digits above and below the leading 1s are 0s

[viscid thistle]

Hm ok that makes sense

thanc

[hazy hull]

https://documentcloud.adobe.com/link/track?uri=urn%3Aaaid%3Ascds%3AUS%3A620abfb2-c85e-49fb-a7f9-90c790b015e3 for #4 is x^2/(3x+1) a composition of those 3 functions

[tardy crescent]

That is not the composition of those 3 functions. Remember to substitute h(x) for x in g(x), and then sub that in for x in f(x)

[hazy hull]

so it would become something like 1/(3x^2+1)

[tardy crescent]

Sorry I was afk, that's closer

You need to substitute the whole expression. So since h(x) = x^2, and g(x) = 1 / (3x + 1), g(h(x)) = 1/(3x^2 + 1). Since f(x) = 1/x, then f(g(h(x)) = 1 / (1 / (3x^2 + 1)) = 3x^2 + 1

[trim fable]

i have a question

how would u prove this

tanx * cotx = sec * cscx

[abstract dome]

wait

[uncut mulch]

uh... those aren't equal...

[abstract dome]

yeah

[trim fable]

we have to

prove

it tho

[abstract dome]

tan(x) * cot(x) = 1

[trim fable]

trig identities

oh..

[uncut mulch]

which would be imposible if they arne't equal

[abstract dome]

but sec(x) * csc (x) = 1 / (cos(x) * sin(x))

[trim fable]

the first was +

[tardy crescent]

Yeah, those aren't equal, 1/ cos (x) * sin (x) is not equal to 1

[trim fable]

tanx + cotx = sec * cscx

my bad

[abstract dome]

oh

[tardy crescent]

oh ok

[abstract dome]

yeah that makes more sense then

[uncut mulch]

zzz

[trim fable]

lol

sorry

[uncut mulch]

write stuff in terms of sin and cos

[abstract dome]

^

[trim fable]

rn i did

sinx/cosx+cosx/sinx

good so far?

[abstract dome]

yes

combine those into one fraction

[trim fable]

ok

so

sinx^2+cosx^2/cosxsinx

good?

[uncut mulch]

parentheses

[trim fable]

?

[abstract dome]

(sin(x))^2

[trim fable]

ye

[abstract dome]

not sin(x^2)

[trim fable]

oh

i meant

[abstract dome]

ye

[uncut mulch]

and around the whole numerator

[trim fable]

sin^2x

[abstract dome]

parentheses just makes it clearer

[uncut mulch]

put parentheses around the angle

sin^2(x)

and put parentheses around the whole numerator when typing in plain text

can you expression be simplified?

[trim fable]

so now

sin^2x+cos^2x/cosxsinx
1/secxcscx

[uncut mulch]

no

[trim fable]

why?

it makes sense

[uncut mulch]

how did your sin and cos change into sec and csc like that?

[trim fable]

oh it should be just

secxcscx?

[uncut mulch]

yes

[trim fable]

oops

[abstract dome]

and there's the proof

[trim fable]

question

u know how

cos was

in the denominator

how it was

sin^2x+cos^2x/cosxsinx

and then

[uncut mulch]

parentheses

[trim fable]

1/cosxsinx

and then

sec=1/cos

so then

u would just

multiply it by the reciprocal?

to bring it to the top?

[abstract dome]

?

[uncut mulch]

1/(cosx sinx)

[abstract dome]

^

[trim fable]

yees

[uncut mulch]

don't ignore the parentheses

[abstract dome]

and that's just (1/cos(x)) * (1/sin(x))

which is sec(x) * csc(x)

[trim fable]

oh..

omgg

[uncut mulch]

by definition
sec(x) = 1/cos(x)

[trim fable]

yesssss

it makes senseeee

[uncut mulch]

and csc(x) = 1/sin(x)

and you can go directly from 1/(sin(x)cos(x)) to csc(x)sec(x) without doing anything extra

[trim fable]

ok

[brisk frigate]

How does one get the value of "X" in a trig function with another number inside?

The one I'm doing is cos(4x)=0

Usually if they ask for "x" from just cos(x)=0 I use my calculator and what i would do is arccos(0) to get the answer

[uncut mulch]

find the general solution for 4x
then divide by 4 to get x

[brisk frigate]

I didnt know about a general solution, I looked it up and it shows this: cos x = 0 is x= (2n+1)(pi)/2

But I'm wondering what value to use for "n"

I suppose since it was 4x then I just do this: x=((2n+1)(pi)/2)/4

[native timber]

if we let y = 4x, how do we solve cos(y) = 0

[brisk frigate]

i think i should learn how to do that first then, i usually just do arccos(0)=y on my calculator

shift cos of 0, and use that as the answer for y

[torn swift]

you want to find when cos(4x)=0?

[brisk frigate]

yeah

but the 4 is messing me up

[torn swift]

oh wait, someone is already helping you

[brisk frigate]

i normally just look at the unitary circle after making cos(x)=0

[torn swift]

but you know when the cosine function is equal to zero?

[brisk frigate]

so idk what to do with the 4

[native timber]

ok maybe replacing it with y might be confusing

[torn swift]

are we restricted to some interval or do we have to have find all solutions?

[native timber]

if we let a = 4x, how do you get a from cos(a) = 0

[brisk frigate]

[0,pi)

[torn swift]

so when does cosine have a value of zero in this interval?

[brisk frigate]

I would just look for when 0 shows on the unitary circle and say that "a" is there

[torn swift]

were on the unit circle does cosine equal zero then?

[brisk frigate]

on pi/2 and

and 3pi/2

if im not mistaken

[torn swift]

for our interval, we are only concerned about pi/4

so we know that cos(4x)=0

and we want x

so apply arccos to both sides

arccos(cos(4x)) = 4x
arccos(0) = pi/2 as you said

so now 4x=pi/2

now solve for x

[brisk frigate]

oh

thats amazing

would that work with anything inside the cos()?

like adding/roots/etc

[torn swift]

sure, arccos(cos(x))=x by the definition of an inverse function

[brisk frigate]

thats so smart

[native timber]

assuming its within the range of arccos yes

[brisk frigate]

i see

[torn swift]

yes, there is that small detail

[native timber]

generally arccos(cos(x)) = q where x = q + 2pi*k for some k

[odd badge]

i.e. arccos(2) is undefined

[torn swift]

but the bigger picture is to see the effect of an inverse function

[odd badge]

if you look at the cos curve it's a good way to see what vals are defined for arccos and what aren't

[brisk frigate]

@native timber is that the general solution the other guy was talking about?

where n is an integer or something like that

[native timber]

yeah if its asking for all solutions then that's the proper way to write it

[torn swift]

general solution is the solution on any interval

[brisk frigate]

ah i get it, so i wouldnt substitute the n for anything right?

since it would just mean any number

to cover all solutions, etc

[torn swift]

n is any integer

so any integer will do

[brisk frigate]

Yeah im getting it a lot more now

tysm for the help, trig is prob the hardest thing i've come across in school tbh

[harsh cipher]

Hi guys

may I interrupt?

why do I not move 5 infront of "z" but move (z3) to the numerator?

[native timber]

exponents before multiplication

the ^-5 is applied to z and only z

while in (3x)^-2, it's applied to both since it's grouped by a parenthesis

[harsh cipher]

and the 3 is applied to both? -_-

aha let me look at exponent laws again ^^

[native timber]

5z^-5 is 5(z^-5) due to exponentiation binding tighter than multiplication

[brisk frigate]

Aparently there are 4 answers but I could only get 2 and I think one of them is outside the range the question is asking for, is there something I did wrong that I am not noticing?

I just used the addition formula to turn the equation into "cos(4x)" and then solved it

arccos(0)= 1pi/2, ---> (1pi/2)/4 = 1pi/8 <--- First answer, then I added pi to it and got 3pi/8 as my second answer

[trim fable]

I need help im stuck on a question

so far I did

csc^2(1+cosx)
=1+cot^x (Idk what to replace 1+cosx with)

I have to prove that using trig identities

[rigid beacon]

My hint would be using conjugates @trim fable

[trim fable]

oh

ok

where tho

@rigid beacon

[rigid beacon]

look on the left and right sides of the equation

my hint relates to what you said "Idk what to replace 1+cosx with"

[trim fable]

ye

ok

so like

csc^2(1+cosx) 1+cosx)
=---------------- * -------------
1 1+cosx)

csc^2(1+2cosx+cos^2x)
= -------------------------
1+cosx)

= csc^2(cosx+cos^2x)

ye im kinda stuck there

lol

is it even right so far

[uncut mulch]

its the wrong approach

[trim fable]

aww

[uncut mulch]

the idea is the simplify the denominator on the left

[trim fable]

oh

so i start with the Left side

not right side?

[uncut mulch]

usually you start with the more complicated side

[trim fable]

yeah

ik

which is the

one im

working with

[uncut mulch]

fractions are more complicated if the denominator consists of a sum

[trim fable]

ye

[uncut mulch]

the idea of using the conjugate is to turn
the denominator into 1-cos^2(x) which can be represented as a single term

[trim fable]

ye

so multiply it by

1-cos^2(x)?

[uncut mulch]

no

[trim fable]

oh

[uncut mulch]

as mentioned by spamakin, it would involve conjugates

[trim fable]

yes

but the conjugates i used was

(1+cosx)

[uncut mulch]

what would you multiply to
(1 - cos(x)) to get (1- cos^2(x))?

you multiplied the conjugate of the lhs to the rhs which doesn't really help you

[trim fable]

1-cosx?

[uncut mulch]

no

[trim fable]

wait

uh

by

cosx

[uncut mulch]

what is the conjugate of
(1- cos(x))

[trim fable]

(1-cos(x))

[uncut mulch]

no

[trim fable]

(1+cos(x))

[uncut mulch]

yes

and you should multiply both the numerator and denominator of the lhs by that

you should not have multiplied that on the rhs

[plush matrix]

guys quick question about the application problem

[trim fable]

oh

[uncut mulch]

$\frac{1}{1-\cos(x)} \cdot \frac{1+\cos(x)}{1+\cos(x)}$

[obsidian monolith]

ramonov:

[trim fable]

oh

so ur

working with both sides?

[uncut mulch]

no

[trim fable]

rs?

[uncut mulch]

1/(1-cos(x)) is your left side

[trim fable]

or ls

yeah

[uncut mulch]

by multiplying by
(1 + cos(x))/(1 + cos(x))
you are essentially multiplying by 1

[trim fable]

yeah

thats what i meant tho

[uncut mulch]

and after some more manipulation/ simplification
you will reach the rhs

[trim fable]

by 1+cosx

i just didn't write it twice lol

oh ok

so i work with the left sidee

[uncut mulch]

yes

[trim fable]

oh ok

i was working with the rs 😮

my bad

[uncut mulch]

you could work from the right, but the steps would be different and its harder to see what you're supposed to do

[plush matrix]

nvm I got it

[trim fable]

@uncut mulch

ok so

i got

1+cox
=-------------------
1-2cosx+cos^2x

[uncut mulch]

no

[trim fable]

huh what

[uncut mulch]

what is
(a - b)(a + b)?

[trim fable]

square the first

OH

WAIT

difference of

squares

oops

1-cos^2x?

[uncut mulch]

yes

that's pretty much the reason why you use the conjugate

[trim fable]

yeee

1-cos^2x is equal to

sin^2x

so replace?

then get csc

[uncut mulch]

do you mean replace 1/(sin^2(x)) with csc^2(x)?

[trim fable]

thank u

ye

i got it

🙂

will u help me with another

i attempted but

got stuck

[uncut mulch]

sure

[trim fable]

so what i did was

worked with the ls

=csc^2x+1

1+cscx

then can i cancel?

to cscx

then

1/sinx

i just need the -sinx at the top

so did i mess up somewhere

[uncut mulch]

yes

[trim fable]

oh where

lol

[uncut mulch]

from the start

also inappropriate cancellation

what's the identity relating
cot and csc

[trim fable]

cot=1/tanx

csc=1/sinx

[uncut mulch]

not what i'm looking for.

[trim fable]

oh

oh ik

1+cotx=csc

[uncut mulch]

missing the ^2

[trim fable]

ye but

u didn't ask for cot^2x hehe

ok tho

😛

[uncut mulch]

no excuse

[trim fable]

aww

sorry

ok

[uncut mulch]

1+cotx=csc would be false

[trim fable]

ok

1+cot^2x=csc^2x

there 🙂

[uncut mulch]

no

[trim fable]

aww

but its a

[uncut mulch]

you can't just make the equal sign dissapear

[trim fable]

oh ok

my bad

oops i meant = not -

[uncut mulch]

so what would cot^2(x) be equal to?

[trim fable]

it will equal

csc^2x-1

[uncut mulch]

yes

[trim fable]

oh why did i write +?????

wowwwww

what a dumb mistakeeeee

I BROKE A LAW OF MATH

[proud jetty]

CAn ya all tell me why this doesn’t work? If I’m simplifying

[trim fable]

heyyy

trig identities

im doing that too

hehe

[proud jetty]

lol nice

[trim fable]

lol

[proud jetty]

I got unit test and idk jack shit

[trim fable]

oof

[uncut mulch]

are you able to continue from
(csc^2(x)-1)/(csc(x) + 1)?

[trim fable]

uh conjugates?

or no

[uncut mulch]

factorise the numerator. same idea as before (difference of 2 squares)

[proud jetty]

if ya people tryna help @me

[trim fable]

oh ok

[proud jetty]

star u got unit test tomorrow?

[trim fable]

no

lol

[proud jetty]

Lmao fk me than wait can ya help me with my shit?

[trim fable]

uh i just learned it

[uncut mulch]

what's the original question asking for?

[proud jetty]

Factor and simplify

[uncut mulch]

also cos^2(x) - 1 isn't sin^2(x)

[trim fable]

ok

trueeee

[uncut mulch]

1 sec

looks like the answer can have multiple forms

[proud jetty]

So I’m right?

[uncut mulch]

also cos^2(x) - 1 isn't sin^2(x)

[trim fable]

question

[proud jetty]

Isn’t it if u manipulate the identity

[trim fable]

csc^2x-1 (1-cscx)?

over

1+cscx(1-cscx)?

[proud jetty]

2-85 Ramon I use right

Star ? U doing math pre calculus 12?

[trim fable]

ye

[uncut mulch]

you don't really need to multiply by the conjugate for that

[trim fable]

well advanced functions

[uncut mulch]

csc^2(x) - 1
is a difference of 2 squares which you can factorise

[trim fable]

?

just the

denominator?

[uncut mulch]

(also if you were to do that, parentheses)

[trim fable]

oh sorry

hehe

im lazyyy

[uncut mulch]

wdym,
factorise "csc^2(x) - 1"

[trim fable]

umm ima skip that for now

is itbetter to work with the

[uncut mulch]

difference of 2 squares

[trim fable]

left side?

[uncut mulch]

you'd would've been 2-3 steps away from finishing it

[trim fable]

oh well

[uncut mulch]

i mean multiplying by the conjugate takes the long way

[trim fable]

im doing this first lol

[uncut mulch]

but it would also work

[trim fable]

ok

so what

about that

ls?

rs?

[uncut mulch]

@proud jetty manipulate the identity again,
s^2 + c^2 = 1
c^2 - 1 = ?

[proud jetty]

yea

but s^2 would be neg

[uncut mulch]

yes

[trim fable]

im getting so close

to the answer

[proud jetty]

yea im slow but i'm getting there

[trim fable]

but its sinx/1-sinx

not cosx/1-sinx

ahh

[uncut mulch]

what do you get after fixing your signs?

[trim fable]

?

[uncut mulch]

at mlg

[trim fable]

oh ok

[uncut mulch]

you would obtain the other form from factorising the denominator as a difference of two squares

though the one you got is simpler than the solution obtained from that method

[proud jetty]

its sx/cx+1

[trim fable]

secx+tanx
=sinx

cos^2x
=sinx

1-sinx

so what am i doing wrong

[uncut mulch]

parentheses

,w sin(x)/(cos(x)+1) = (1-cos(x))/sin(x)

[obsidian monolith]

Display results online and refine query

Requested by ramonov#0921

[trim fable]

?

this is the question

[uncut mulch]

that was mlg's

[trim fable]

😛

LOL

oh oopsss

[uncut mulch]

explain how you got the first line

[trim fable]

who

[uncut mulch]

you

[trim fable]

oh

ok so

sec(x)=1/cos(x)

so i brought cos(x)

to the bottom then

tan=sin/cos

so then wrote that

giving me sin(x)/cos(x)cos(x)

sin(x)/cos^2(x)

[uncut mulch]

no

[proud jetty]

ram if their is tanx-3-4
can I add them?

[trim fable]

oh

[uncut mulch]

its +

[proud jetty]

like the -3-4

[uncut mulch]

not multiply

[trim fable]

yee

so -7

right

is that what ur asking?

[proud jetty]

yea

[trim fable]

oh ask @uncut mulch to make sure tho

i dont wanna tell

u the wrong thing

[proud jetty]

lol whens ur unit test

[uncut mulch]

what's tan being applied to? only the x?
tan(x) - 3 -4?

[proud jetty]

yea

I factored one tan out

[uncut mulch]

can you post the original problem

[trim fable]

next week

[proud jetty]

ik I did it wrong

[uncut mulch]

yeh so its
sec(x) + tan(x)

[trim fable]

me?

right

[uncut mulch]

you shouldn't factor like that

[trim fable]

who

[uncut mulch]

mlg

[proud jetty]

me i guess

[trim fable]

lol

[uncut mulch]

the numerator is a quadratic in tan(x)

[trim fable]

brb

[proud jetty]

ramonvo could u highlight the problems

like using paint brush or sumthing

and just type wut i can't do cuz i aren't getting this at all

[proud jetty]

bet lemma try to make it legal

[uncut mulch]

if you only factor two of the terms, there's still addition involved and you can't really divide

you would need to factor the whole thing

similarly, factor the denominator too

(cancelling tan like that was illegal, anything after that is irrelevant)

[proud jetty]

shit

I did it

that would work on exam right?

I am not doing anything just changing how it looks

@uncut mulch do u think if i work like this it will cause me problems with identities?

[uncut mulch]

you back substituted the wrong things

[proud jetty]

oh shit besides that

lmafo

[lethal oracle]

ya its sin x in the denominator

[proud jetty]

yea lol i see dat but will cause me problems with identities?

if i continue to work like dat?

[uncut mulch]

as long as you're clearly stating your substitution it should be fine

[proud jetty]

bet thx

[uncut mulch]

but do try and get to a point where you can do it without sub

[lethal oracle]

r=6cosx+6sinx

[proud jetty]

yea i'll try

[lethal oracle]

how would i change this to rectangular form

i had this on an exam on wednesday and i basically left it blank cause i couldn't remember how

[viscid thistle]

Is arccot the same as tan?

[willow bear]

no

[shadow glacier]

could anyone help me with this problem https://gyazo.com/08593826e4900e6e4a9024d75d2f1759

[willow bear]

15-minute rule.

[shadow glacier]

sorry i didn't read the rules, just read them and edited msg

[willow bear]

i didn't read the rules

ok whatever

what have you tried so far

and where are you stuck

[obsidian monolith]

Digicat195:

[serene heath]

,w graph y=(2(x+1)(x-3))/((x-1)(x+3))

[obsidian monolith]

Display results online and refine query

Requested by the one n only#3070

[serene heath]

should use #bots tho

[trail badge]

Yeah thats it

Thanks

What?

Why?

@Jeff#6412 I wrote this equation above, i don't have sure if it is right, but i think so

Ohh he was gone ;-;

[restive bridge]

how does dividing by -36 cause a hyperbola? it should be a parabola

no?

when you divide by -36, the -9 because a positive 4.

OOOOHH

the first one becomes negative, second one because positive

rearrange it fixes the issue

[quick pond]

Yo question, for this, can someone explain the proof behind say...

10^log(x)=x

This rule

[merry sphinx]

@quick pond

Do you want the intuitive proof, or the algebraic one?

[quick pond]

Uhh, the simplest one to understand, I know the other rules but this one is still eluding me

Most likely algebraic then

[merry sphinx]

Okay, so think about what a log is asking for

[quick pond]

It wants us to simplify into K

[merry sphinx]

For example, $\log_2 64$

[obsidian monolith]

pocofrosty12:

[merry sphinx]

What does it simplify to

[quick pond]

2^6

So 6

[merry sphinx]

So it is asking "for what power does 2 need to be raised to equal 6"

Does that make sense?

[quick pond]

Yeah it does

[merry sphinx]

So for $\log_a b$, it is asking for what power of a to equal b

[obsidian monolith]

pocofrosty12:

[quick pond]

Yes

[merry sphinx]

a^ "what power of a to equal b" = b

[quick pond]

Mhmm that I get

[merry sphinx]

This is the intuitive proof

So do you understand the property now?

[stuck lark]

$a^{\log_a(b)}=b$

[obsidian monolith]

RokettoJanpu:

[quick pond]

I think I do

Ohhhh now I get it

[merry sphinx]

Look at canceling exponents section on here for algebraic proof

https://en.wikipedia.org/wiki/List_of_logarithmic_identities#Trivial_identities

[quick pond]

So log_a(b) means x

And a^x=b

Because x is essentially the power that raises a up to b

I get it now

[merry sphinx]

Algebra proof is easy too, but I think the intuitive explanation develops a better understanding

[rich ocean]

thats the first time i fully understood that intuitively

[hollow bluff]

Would this be correct?

my first step was making them all (x+2)(x+2-i)(x+2+i)

then i multiplied

[uncut mulch]

factor theorem wasn't applied properly

[hollow bluff]

how do i apply it correctly

Oh wait

Its supposed to be the opposite sign

(x-2)(x-2-i)(x-2+i)

is that better

[willow bear]

each parenthesis is supposed to be (x - root)

as long as you recognize x-2-i doesn't correspond to the root 2-i you're good

[hollow bluff]

What do you mean

I feel like in know what your talking about but im not quite grasping it

[uncut mulch]

even those are the factors, it is unclear whether you are applying the factor theorem properly for the complex roots

[hollow bluff]

alright

so even if i get the right factors I still might get it wrong

So have a made a mistake?

[uncut mulch]

depends a bit on how you arrived at those factors

[hollow bluff]

ah

Well im going to revise my work right now

ill let you know if i need help

thank you!

[uncut mulch]

it would be better represented as
(x-2)( x - (2 + i))(x - (2- i))

[odd helm]

Why is there a shift to the right 172

This is what I drew idk why we need to shift wouldn’t this work ok?

[viscid thistle]

Because you have to start at day 0, time starts at day 0 goes to day 365.

I can't see the full question but its likley because of that. We use phase shifts so that the sinusodal shape will match up

[hollow bluff]

is this correct?

[fading token]

Seems like it

Although

[hollow bluff]

although?

I think this is correct

like im doubting myself

[fading token]

3 is a real zero, so R can't have 8 non-real zeros

[hollow bluff]

uh huh

so 6?

[fading token]

That seems more likely

And I think the max 4 real zeros was correct

[hollow bluff]

Did i do anything wrong?

[abstract dome]

@hollow bluff yes

Your -5 *R1 step

You said (-1)*(-5) = -5

[hollow bluff]

oh man im an idiot

so that mean it would be infinite solution?

[abstract dome]

yes

@hollow bluff

[hollow bluff]

darn

[hallow flicker]

Can someone calculate this without using the derivative rule?

[pale kettle]

yes

what have you tried

[hallow flicker]

Not much

I just used algebra

But all of them came to zero

[pale kettle]

to do what

[hallow flicker]

To simplify it

[pale kettle]

how do know that 0 isnt the right answer

[hallow flicker]

To solve it

Because of the derivative rule

That’s suppose to be the answer

But it uses the derivative rule

[uncut mulch]

are you familiar with conjugates?

[hallow flicker]

No

[uncut mulch]

do you have any idea what they are?

or heard the term before

[hallow flicker]

I’m searching them up right now

Can you do the math on paper

Then show me?

[uncut mulch]

read up what you can on conjugates
the idea is to rationalise the numerator

then divide by n,
then take the limit

try and do that yourself and come back if you're still stuck

[hallow flicker]

Ok thanks dude I figured it out

Ur the best

[lethal oracle]

ramonov is indeed the best

[stuck ivy]

Need help with filling out this table

[abstract dome]

f'' is the concavity - essentially, the graph being shaped like a U

f'' is the slope of the slope

so if it's an upward facing U, f'' is positive

and if it's a downward facing U, f'' is negative

[stuck ivy]

I'm just wondering how concavity affects a

like

f''(x) makes it 0

then the derivative of that would make it positive or negative

[abstract dome]

for that point (at a vertex), look just before and just after the point

just before, the slope is positive, but decreasingly so (the slope is becoming a smaller negative number)

and just after, the slope is negative and increasingly so (it's becoming a steeper/larger negative number)

[stuck ivy]

so it would be positive?

[abstract dome]

if it's an upward facing U, f'' is positive
and if it's a downward facing U, f'' is negative

[viscid thistle]

[stuck ivy]

So it would be negative

because the slope slows down

for f'(x) and speeds up after?

[abstract dome]

yes

it becomes increasingly negative after

and decreasingly positive before

[stuck ivy]

for part c is this correct? @abstract dome

[hoary valley]

What's the difference between

And

[hallow flicker]

what is this guy asking?

[hoary valley]

My teacher says they are the same.

[hallow flicker]

how

[hoary valley]

Idk, I'm asking what's the difference.

[hallow flicker]

you have two different variables at the bottom

the bottom is a derivative

the top has two different variables

[abstract dome]

yes @stuck ivy

[hallow flicker]

what

ok i am going to make the assumtion that a and x are the same

[stuck ivy]

sweet

[hallow flicker]

i am assuming that because you saw the lim of h aproch zero i think your mistaking it for zero

thats not really the case

until you simplify the expression always think of h as a value

because in reality it almost never really is 0

its only when you simplify it that the h is so insignificant that it pretty much becomes zero

the top alway sums up to zero while the bottom doesn't

assuming that x and a are the same i guess

[patent beacon]

@hoary valley
They are the same. Let x = a + h to go from the first to the second

Note the first should have a limit as x → a

[hoary valley]

@patent beacon Right. so why we have two formulas ?

[patent beacon]

I tend to stick with the second as "the definition" but they're the same formula