geometry-and-trigonometry

[dark sparrow]

ok right so we are just asked for angle AYR

in that case, what progress have you made thus far?

[rare fossil]

One message removed from a suspended account.

[rare fossil]

One message removed from a suspended account.

One message removed from a suspended account.

[dark sparrow]

so you already had some answers and wanted somebody to check them

next time post them along with the question

anyway which ones are you unsure about and why

[rare fossil]

One message removed from a suspended account.

[rare fossil]

One message removed from a suspended account.

[brisk ore]

@exotic yarrow hey you know how we were discussing obtuse angles in trig yesterday, i ran into another question where i didnt realise it. i think im having a hard time visualising why we subtract our angle found from 180 if the angle we got is already acute so it cannot be obtuse. heres the question

oh wait is it because the answer (149.32) satisfies the inequality 90 < x < 180 which it needs to satisfy for angle SPQ to be obtuse

[dark sparrow]

@brisk ore can you show your work up to and including the spot you're asking about

[brisk ore]

I got stuck at after P = 30.68

[dark sparrow]

ok hold on

this transition is troublesome

you get sin(P) = 25/49 but it is now that you need to account for P being obtuse, and take 180° - arcsin(25/49) instead of arcsin(25/49) itself.

[brisk ore]

Ohh

Yeah that makes more sense thanks

Btw

The first half was wrong because I was only taught one formula for area;
Area = 1/2 ab sin C

Are there other derivations of this formula I need to know

[dark sparrow]

i mean if you remove the 1/2 factor then you'll get the area of the entire parallelogram and not half of it

and understand that 1/2 ab sin(P) would give you the area of SPQ rather thnen PQRS

[brisk ore]

Yeah 👍 thanks

[brisk ore]

when i solve the quadratic and i get theta is 34.48 and 180 - it to get the answer, i usually take that to be the final answer and its finished but for some reason the mark scheme wants me to include 34.48 and 34.48+360 in my final answer. Does anyone know why that is the case

[nocturne remnant]

because 34.48+360 is also a solution

...?

what are you confused about

[native jetty]

because it specifies 0<theta<450

you need all solutions in that domain

[brisk ore]

im confused because in this question we only take 520.5

[brisk ore]

oh yeah

nvm i get it now

thanks for the help @nocturne remnant @native jetty

do you guys know how to do this

[faint comet]

Can somone please explain to me why he would change 2pie/3 into pie/3.

[dark sparrow]

pi, not pie. and we didn't "change 2π/3 into π/3"; that's not a thing.

what they did was write cos(2π/3) and sin(2π/3) as -cos(π/3) and sin(π/3) respectively

however without context it is hard to tell why

@faint comet gonna need to see the original problem in full, please

[faint comet]

Like the argumentative of the angle ?

@dark sparrow yes I understand how we change it and they equal the same but I don’t get why, the question is to write it in polar for

Form

[dark sparrow]

if the question is to write z1/z2 in polar form then the last step is pointless

the line written in red directly converts into $\frac{1}{2} e^{2\pi i/3}$

[somber coyote]

Ann

[faint comet]

No In caretsion form sorry

[dark sparrow]

...

ok then maybe that step is meant to be intermediate to calculate cos and sin of 2pi/3.

assuming you don't know these directly.

[dark sparrow]

also you're definitely not using the word "argumentative" correctly here.

[faint comet]

Like the argument angle or whatever where you minus it from 360

[dark sparrow]

i think you are confusing yourself.

[faint comet]

How is -sqaure 3/1 the tan of -pi/3

[dark sparrow]

did you really mean "square" (^2) or did you mean "square root" (√)?

@faint comet

[upper karma]

Hey guys

If i divide both sides of trig equation by trig function, i need restrict function domain, so that it doesn't equal 0 for sure, right?

[obsidian harness]

Yes, you need to deal with the case where the denominator is 0 separately

[upper karma]

For example i want to divide by sin x, but what if the solution for an equation is pi beforehand, but when i divide by a trig function, i basically don't allow the solution to be pi by default. Am i right?

[dark sparrow]

well if you take note separately that x=π is a solution then you can divide by sin(x)

[obsidian harness]

Ah, so you're transforming an equation where x = pi is allowed into one where x = pi is not allowed

[dark sparrow]

you'd have to look at all values x = kπ though so as not to miss anything

[obsidian harness]

So you're losing the solution x = pi, and so you need to check it

[obsidian harness]

Also this, remember there are infinitely many solutions to sin(x) = 0

[upper karma]

But i dont know if the solution is pi beforehand

[dark sparrow]

(all of which i just listed)

[obsidian harness]

Yes, so you definitely need to check it

Cause you don't know

[dark sparrow]

can you show us the equation you are looking at?

[upper karma]

No, i was asking generally

But, i think i got one

[dark sparrow]

before dividing both sides of an equation by g(x), you have to account for all values of x where g(x) = 0, and see whether or not they are solutions

for this it does not matter whether g is trigonometric or not

[upper karma]

Here is the equation: 2sin x * cos x = 3cos x

[dark sparrow]

dividing by sin(x) would be unhelpful here

[upper karma]

The solution is pi/2, but by dividing both sides by cos x, i doesnt allow x to be the solution right?

And i will get sin x = 3/2

So it's wrong

[dark sparrow]

dividing both sides by cos(x) and failing to account for x = π/2 (and in fact x = π/2 + kπ for any k ∈ Z)

will lose that family of solutions

[upper karma]

How to divide by function but not lose the solution?

[dark sparrow]

again, account for them separately beforehand

you can just say that if x is a point s.t. cos(x) = 0 then both sides of your equation will be equal to 0

[upper karma]

So what would it give me?

I understand that they will be equal to 0

[dark sparrow]

so you get a family of solutions. and you don't know yet whether that's all your solutions, or only a proper subset. but you have them, and now you can divide by cos(x) to look at what happens with all the other values of x.

[upper karma]

But at what point do i get those first solutions? How do i get them if i only know how to solve the equation via dividing by function.

[dark sparrow]

...

either you're overthinking it or there is a language barrier

actually, if you don't mind me asking... what is your native language?

[upper karma]

@dark sparrow so, how to solve this equation properly?

What is the method?

[dark sparrow]

are you implying that there's only 1 correct method and all other methods are wrong?

[upper karma]

No

[dark sparrow]

there's two methods that i can tell you about

[upper karma]

If you don't mind, can you tell me them?

[dark sparrow]

method 1 is to write in plain text the following: "We want to divide both sides by cos(x), but this risks losing solutions when cos(x) = 0. Thus we will check those explicitly, then proceed with division by cos(x) when looking at all values of x except those."

then solve the equation cos(x) = 0

and then verify that all solutions of it are also solutions of the equation

and then, once you have done that, divide both sides of 2sin(x)cos(x) = 3cos(x) by cos(x), and solve the equation 2sin(x) = 3.

method 2 is to not bother with division by cos(x) at all, and instead to subtract 3cos(x) from both sides, then factorize:

2sin(x) cos(x) - 3 cos(x) = 0
<=> cos(x) (2 sin(x) - 3) = 0
<=> [cos(x) = 0 OR 2sin(x) - 3 = 0]
and then solve each of the two smaller equations separately, and union their solution sets at the end.

[upper karma]

By "union their solution sets" you mean writing a general solution?

[dark sparrow]

"general solution" and "solution set" are synonyms as far as i am concerned.

and union means union.

like the set-theoretic operation called union

∪ <- this

[upper karma]

Ok, got it

Thanks

[upper karma]

Generally, do you consider trig equations to be hard? And is it necessary for calculus?

[dark sparrow]

why reply-ping to an irrelevant message? if you wanted to get my attention, you could have just @-mentioned me.

[upper karma]

@Ann like that?

[dark sparrow]

@ann_dec

[upper karma]

Ok

[dark sparrow]

but to answer your two questions, in that order:

1. trig equations are not all hard, any more than algebraic ones could be. there are easy ones and there are hard ones
2. familiarity with trig is definitely necessary for calculus, but i am not sure to what extent. but the more the better.

[upper karma]

Ok, thanks

[faint comet]

Root sorry

[dark sparrow]

ok, so what is the best description of your confusion?

(A) "I don't know how to find tan(-π/3) at all."
(B) "I think tan(-π/3) is NOT -sqrt(3)."
(C) "Something else -- I will explain myself now."

[faint comet]

B

[dark sparrow]

ok, why do you think tan(-π/3) ≠ -sqrt(3) ?

[faint comet]

Becuse sin/cos of -pi/3 how does that = that ?

[dark sparrow]

what is sin(-pi/3)?

@faint comet ?

[leaden valve]

could i get some help

[dark sparrow]

!da2a

[lime crown]

No need to ask “Can I ask…?” or “Does anyone know about…?”—it’s faster for everyone if you just ask your question! See https://dontasktoask.com/

[upper karma]

tan x + tan 2x- 3tan3x transform in product

[forest blade]

Would someone be able to explain to me why this uses inverses?

The video I was given to watch just went through the steps of finding the answer but didn't actually explain why that process was being done

[muted remnant]

cheeseburger 😔

[trail tendon]

do you get that tan(theta) = h/5

[forest blade]

Yes, because it's opposite over adjacent

[trail tendon]

yes

do you know what the inverse of tangent is

?

[forest blade]

I know it's expressed as arctan or tan^-1

[trail tendon]

yes

[forest blade]

and that's it's expressed within a range

or well it falls within a range

[trail tendon]

yes, but that doesn't matter here really

so

arctan(tan(theta)) = theta

inverse functions mean if you plug one function of x into its inverse function, you just get x

does that kinda make sense or nah?

[forest blade]

well I kinda assumed that the arctan cancels itself out on the left side when they were doing the video

[trail tendon]

yes

here's a better way to think about it

[forest blade]

I was just confused on why they even did an inverse function in the first place

[trail tendon]

they're finding theta

[forest blade]

was it because theta is outside with the h inside of the parentheses?

so they are reversed in this instance

[trail tendon]

take f(x) = 2x + 3. lets say you want to get x alone.
let f(x) = y. so y = 2x + 3
if you swap y and x you get the inverse function., you get x = 2y + 3. now you want to get y alone again. so,
2y = x - 3
y = (x-3)/2
thus, the inverse of f(x) = 2x + 3 is f(x) = (x-3)/2

does that kinda make sense?

the inverse of f(x) is also denoted as f^(-1)(x)

which looks weird on discord lol

anyway

arctangent is defined to be the inverse of tangent

[forest blade]

I get that. Maybe it's the wording of the problem that's confusing me. Because if I saw this on a test, I wouldn't even know what they want me to do.

Like what's the tell that you're supposed to be finding the inverse here?

The fact that it's theta(h)

?

[trail tendon]

they want you to express it in terms of theta

[trail tendon]

yeah

"express the angle of elevation theta as..."

so that means

theta = ....

they want you to express it as a function?

theta(h) = ....

[forest blade]

oh I think I see now

so since the angle of elevation is tangent by default

it kinda wants you to get rid of that tangent by doing an inverse to cancel it out

so it's only theta

[trail tendon]

basically yeah

[forest blade]

okay that helps

thanks!

[upper karma]

Hey guys

I got a question

If one of the solutions for a trig equation is, for example, -pi/6 +2pin, should it be this, or should it be 11pi/6 + 2pin?

Are they both correct?

[dark sparrow]

they are both correct

[alpine jungle]

Guys I really need help

I got 4 packets and I dont really understand them

Packet A

Packet B

Packet C

Packet D

Pleaseee help!!!!

[mystic quartz]

My mind actually broke in half over this question. I know how to do it, I just really CAN NOT keep track of anything. I'd really appreciate if someone double checked my answers 😭

[trail tendon]

sure, but it might be easier for you if you drew it

[ripe bison]

Hm

[mystic quartz]

this is horrendously wrong but i think i got the hang of it now

[trail tendon]

wdym horrendously wrong

how much of it is wrong

i thought most of it was right, except im not sure about those last two

[mystic quartz]

last 2 are wrong, rest is right

[crimson thunder]

could someone double check this

[mystic quartz]

i dont like things being wrong

[trail tendon]

that's what i thought 🤣

[trail tendon]

that's generally a good thing to not like 👍

[mystic quartz]

one of these is apparently wrong, but my teacher has a history of getting typos or mistakes on their answer keys so idk if it's right or wrong

[trail tendon]

are yall classmates or smthn 🤣

[crimson thunder]

yes

[mystic quartz]

yea we struggling tho fr 😭

[crimson thunder]

fr

[trail tendon]

let's see

yeah nah i dont see anything wrong with it

i think its correct

[crimson thunder]

yea but when clancy submitted it, it said something was wrong on this question

[trail tendon]

i'd bet your teacher's answer key has a mistake

or else all three of us got it wrong 💀

i got the same answers you did

[crimson thunder]

💀

welp i'll wait till clancy gets a response from the teacher

thanks for the help

[trail tendon]

yea ok bet 👍

[crimson thunder]

this happens sometimes

[trail tendon]

yeah 🤣

[crimson thunder]

oh well, thanks again for the help

[vocal bison]

help ping plz

[upper karma]

Hmmmm

Help pls?

I don't want to mess this one up

I got one more time

😭

[trail tendon]

i despise proofs

proofs are the absolute literal worst

they make no sense

like you "prove" something, and then the actual proof is like "no, actually..." and then they say exactly what you just said but in the most complicated way possible

like bruh 💀

like how am i supposed to know all the details to include???

bruh

[trail tendon]

we know the triangles that make up the quadrilateral are similar because the share two angles and one side

ASA similarity

and the side length is the same

thus, the proportionality is 1 to 1

[upper karma]

Hmmmm

[trail tendon]

thus, the triangles have the same side lengths

and uhh

QT is proportional to RS in the similarity context

i don't know how to say that in a way that formally makes sense

but it makes sense in my head

thus, QT = RS

thats the general gist of the proof maybe

unless they did something weird

[trail tendon]

oh is that right side your answer?

[upper karma]

Bro

you literally saved my life

[trail tendon]

LOL WUT

[upper karma]

YES BRO

[trail tendon]

XD

[upper karma]

😭 😭 😭 😭 😭 😭 😭 😭 😭 😭 😭

[trail tendon]

im glad i helped...

[upper karma]

Can I kiss you

[trail tendon]

no idea what i did tho 💀

[trail tendon]

gay

💀

[upper karma]

After you said ASA

and I realized

[trail tendon]

ohh

you were tryna use SSS

but its ASA

[upper karma]

Yea

[trail tendon]

i see

[upper karma]

First

I tried using SAS

[trail tendon]

out of curiosity, what is CPCTC

[upper karma]

corresponding parts of congruent triangles are congruent

From google

[alpine jungle]

Pls helpppp I found out how to do D and B

[upper karma]

Hey guys

I got no idea how to solve this trig equation:

tan (x+1)/tan(x-1) = 2 + sqrt(3)

Can anyone explain how to solve it?

[dark sparrow]

$\frac{\tan(x+1)}{\tan(x-1)} = 2 + \sqrt{3}$

[somber coyote]

Ann

[dark sparrow]

this?

like the angles inside the tans are x + 1 radian and x - 1 radian?

@upper karma

[upper karma]

Yes

[dark sparrow]

are you 120% certain?

[upper karma]

Yes

[dark sparrow]

can you show a picture of the problem

cause i have serious suspicions

[upper karma]

Yes

[dark sparrow]

just as i thought...

this says $\frac{\tan(x) + 1}{\tan(x) - 1}$, NOT $\frac{\tan(x+1)}{\tan(x-1)}$.

[somber coyote]

Ann

[upper karma]

Ok, got it

But

Can you explain how to solve

[dark sparrow]

there's two ways

either you first isolate tan(x) to reduce the equation into the form tan(x) = a, or you find a way to apply the tan(x+y) identity to (tan(x) + 1)/(tan(x) - 1)

the latter will take a little bit of tinkering

[spice hornet]

Hi

[dark sparrow]

hello, do you have a math question to ask?

[spice hornet]

Hi! Yes

[dark sparrow]

http://nohello.net/

[spice hornet]

Uhm... Formulas for shapes... Like there were areas... perimeter... volumes

[spice hornet]

Sry

[dark sparrow]

so what's your question...?

[spice hornet]

1... what other things are asked for shapes when it comes to formulas?

the thing is I need a refreshner to all sorts of shapes

Cylinder, cones,

It's such a broad question, but I don't know how to ask.

[dark sparrow]

go to http://khanacademy.org/ honestly

[spice hornet]

It's been a long time

[dark sparrow]

look through their geometry stuff

[upper karma]

arcsin(x) - arccos(x) = pi/6

Can you solve it please

I think it is sqrt(3)/2

But my method was quite intuitive, i dont know how to solve it properly

[bronze wind]

Are these all correct?

[rugged shuttle]

How to prove 54

[dark sparrow]

your trig ratios seem correct but some of those values are suspicious

especially #6 -- how did you get a negative answer? are you sure your calculator is in degree mode?

[dark sparrow]

have you tried anything on your own?

[rugged shuttle]

I was trying to see if cotx-cscx or cosx + 1 would cancel out

It didn’t work and I don’t know how to go from there

Also I just can’t get these other 2 for the life of me

[dark sparrow]

ok that's 3 problems, we have to handle them one at a time

[rugged shuttle]

Yes #54 first

[rugged shuttle]

(If u want)

[dark sparrow]

right

for 54, and other problems like it, it often makes sense to write both sides in terms of only sin(x) and cos(x)

i suggest you do that

on the left, obviously

actually the "write in terms of sin and cos" thing goes for all 3

[rugged shuttle]

Ok I got this

[round vault]

Hello, can someone explain to me how this is calculated?

h x p = hp + qh / ab?

[rugged shuttle]

Yo i got 54 thx @dark sparrow

Do u have time to help me with the other?

[dark sparrow]

the others, you mean?

there's two of them.

again, write them in terms of sin and cos.

see where that takes you.

[rugged shuttle]

.-. I just got 85 tysm !!! But I have no clue how to do 74 which has x and y

[dark sparrow]

same shit

write all the tangents and cotangents in terms of sin and cos anyway

[empty yew]

Where is 74? I would like to see the question too!

I don't know much about trigonometry but I try my best.

[rugged shuttle]

Ok 👍

[rugged shuttle]

Here

[empty yew]

I got it

Did you got it? @rugged shuttle

[rugged shuttle]

Not yet this is my first time learning this stuff

:/

[empty yew]

Did you done this?

[empty yew]

@rugged shuttle

[rugged shuttle]

Ah I did it

[empty yew]

Then just cross multiple

[rugged shuttle]

Here is my work @empty yew

[empty yew]

And

[empty yew]

You can't square the cos having different x and y

It just remain as $\sin x \sin y$

[somber coyote]

KingDanger

[empty yew]

$\cos x \cos y$

[somber coyote]

KingDanger

[rugged shuttle]

Oh

Let me retry then

[empty yew]

It's simple then.

[rugged shuttle]

Oh wait it’s the same thing

[empty yew]

🤔

[dark sparrow]

$\cos(x) \cos(y) \neq \cos^2(xy)$

[somber coyote]

Ann

[dark sparrow]

reminder that cos(x) is not the product of x with a number called "cos"

[rugged shuttle]

Because instead of writing sin x (sin y) as sin^2 xy, u can just write it as sin x sin y for every time u multiply and it turns out the same

[rugged shuttle]

Yes I was confused for a sec

[dark sparrow]

instead of writing sin x (sin y) as sin^2 xy
sin(x) sin(y) is not equal to sin^2(xy).

[rugged shuttle]

👍

[empty yew]

But how is 85?

[dark sparrow]

$\frac{1-\sin(x)}{1 + \sin(x)} = (\sec(x) - \tan(x))^2$

[somber coyote]

Ann

[dark sparrow]

this one?

[empty yew]

Yes

[rugged shuttle]

@empty yew this is what I got for 85

[dark sparrow]

i feel this is a complicated route to go down

also your lighting makes it kind of hard to read your work

your pen is too faint

i recommend once again writing everything in terms of sin and cos

in this case the RHS becomes $\paren{\frac{1 - \sin(x)}{\cos(x)}}^2$.

[somber coyote]

Ann

[rugged shuttle]

Sorry

[rugged shuttle]

Wow I’m dumb

;(

[empty yew]

I'm pretty much confused with your work

Did you converted $1 + \sin(x) = \cos^2(x)$ ?

[dark sparrow]

\sin and \cos

[somber coyote]

KingDanger

[empty yew]

@dark sparrow please explain this.

[dark sparrow]

explain what?

[empty yew]

This.

[dark sparrow]

that's a copy of the questoin

[rugged shuttle]

I think that’s where I used cross multiplication

[dark sparrow]

question*

[empty yew]

I don't know how to solve the question!

[dark sparrow]

then you could have asked me "please explain how to do this question"

[dark sparrow]

bad

[empty yew]

Yes

That's what I wanted

Sorry for bad wording!!

[dark sparrow]

$\text{RHS} = \paren{\frac{1 - \sin(x)}{\cos(x)}}^2 = \frac{(1-\sin(x))^2}{\cos^2(x)} = \frac{(1-\sin(x))^2}{1-\sin^2(x)}$

[somber coyote]

Ann

[dark sparrow]

two more steps remain, which i hope you and cyberr are both able to see.

[empty yew]

Is $\frac{1-\sin(x)}{1+\sin(x)} = \left(\frac{1 - \sin(x)}{\cos(x)}\right)^2}$?

[somber coyote]

KingDanger
Compile Error! Click the reaction for more information.
(You may edit your message to recompile.)

[dark sparrow]

yes

but i don't understand why you're asking this

[rugged shuttle]

I don’t get it do u cross multiply with lhs or simplify rhs further

[empty yew]

Sorry I misunderstood something very dumb but now I'm clear

[dark sparrow]

i do not cross multiply anything

in fact i recommend against ever doing such a thing when proving identities.

[rugged shuttle]

OH WAIT

IM STUPID AS HELL

Yea I get it now

[rugged shuttle]

Is there a special reason why not

[dark sparrow]

because then your work makes it read as if your very first step is assuming that your goal is true

which is extremely not kosher, to say the least

and in general

if you wish to write down the identity that you aim to prove (ie your goal)

at least put a question mark directly over the equals sign

[empty yew]

Kosher?

[rugged shuttle]

Alright fair enough I missed my teachers lecture so this is very helpful

[empty yew]

What's the meaning?

[empty yew]

Ohh thank you!

[rugged shuttle]

There’s one more problem…

@empty yew wanna try and solve this before we ask Ann

Or I ask Ann?

[clear anvil]

what's the question?

[clear anvil]

i know that

what are the instructions

what do you want to do lmao

[rugged shuttle]

Prove :(

@clear anvil

[clear anvil]

okay

[clear anvil]

why don't you start with difference of squares on the left

[rugged shuttle]

Ok

[clear anvil]

replace tan(u) with sin(u)/cos(u) because tan(u) = sin(u)/cos(u)

once u do that then simplify the whole thing

[rugged shuttle]

Bruh

[clear anvil]

what?

[clear anvil]

simplify this

and this

[upper karma]

tan x + tan 2x- 3tan3x transform in product

[clear anvil]

yes

you're done

factor out the sin^2 if that wasn't already evident

then use the pythagorean identity

[dark sparrow]

btw

you should not write <u> the way you've done there

you're setting yourself up for confusing it with v

which is bad!

[rugged shuttle]

? How

[rugged shuttle]

Shoot ur right

[clear anvil]

[rugged shuttle]

🥺

[clear anvil]

,, (\sin (u) \cos(u) )^2 = \sin^2 (u) \cos^2 (u)

[somber coyote]

nyxie9151

[rugged shuttle]

@clear anvil

[clear anvil]

why did you even do that

just keep as this

but okay sure

[clear anvil]

if u want to work with this

wait no what do you want to work with first?

[rugged shuttle]

I guess this…

[clear anvil]

Like i mentioned factor out the sin^2

[clear anvil]

cmon i already told you what to do here

it's like 1 last step

[clear anvil]

^

[rugged shuttle]

Ok lemme think

[clear anvil]

there's nothing to think

especially when i told you

[rugged shuttle]

Wait

[rugged shuttle]

I get this

[clear anvil]

that's not how multiplication works

,, \frac{\sin^2 (u)}{\cos^2 (u)} \cdot (1 - \cos^2 (u)) \neq \frac{\sin^2 (u)}{\cos^2 (u)} \cdot \frac{1 - \cos^2 (u)}{\cos^2 (u)}

[somber coyote]

nyxie9151

[rugged shuttle]

Wait why

[clear anvil]

,, \frac{5}{4} = \frac 14 \cdot 5

[somber coyote]

nyxie9151

[clear anvil]

not $\frac 54 = \frac 14 \cdot \frac54$

[somber coyote]

nyxie9151

[rugged shuttle]

https://tenor.com/view/galaxy-brain-meme-gif-25947987

He too good

@clear anvil thx :)

[clear anvil]

sure np

[empty yew]

Did you solve?

Sry I'm gone then.

[alpine jungle]

Why won't anyone help me? 💔😔

[wanton edge]

what is the problem?

[lime crown]

Helpers are just people volunteering their time to help you. Be polite and patient.

[alpine jungle]

I've had it sent since yesterday

[alpine jungle]

It's these packets

@wanton edge

I don't understand them at all😔

[upper karma]

Hello guys

How to find general solution for these two solutions

Sometimes i don't understand how to get a general solution

[clear anvil]

that is the general solution

for k in Z

[upper karma]

The general solution is 2pik/3

This is what photomath gave me

But i dont know how it added these both solutions and got the general one

[clear anvil]

no

[wanton edge]

Sorry but what is the context here

[clear anvil]

yeah the context helps too

actually yeah just post the actual trig question u got this from

don't want to operate with half the context

[wanton edge]

That means cos(x) = 0 or 4cos(x) + 2 = 0

[upper karma]

I know

[dim steppe]

good

[proud nest]

thats the same answer, right?

[wanton edge]

yes

[dark sparrow]

yes, but perhaps there was a format requirement to leave your answer as a fraction

[proud nest]

nope it said i can have integers or fractions

[dark sparrow]

do you still have the exact text of the format req

just to triplecheck

[clear anvil]

right so there's 4 "pairs" of general solutions

cos(x) = 0 -> gives you 2 pairs of general solutions

4cos(x) + 2 = 0 -> gives you the other 2 pairs of general solutions

[upper karma]

cos(x) = 0 gives me pi/2 + 2pik and 3pi/2 + 2pik. But i can combine these two solutions, convert them into pi/2 + pik. Where k is an integer

Maybe i don't what these "combined" solutions are called

So, instead of two solutions i can get one

[wanton edge]

union

[upper karma]

Yes, thanks

[wanton edge]

np

[upper karma]

So how do i get the union solution for this

[wanton edge]

I think there is nothing you can do. Since
$$\frac{4\pi}{3} - \frac{2\pi}{3} \neq \left(\frac{2\pi}{3}+2\pi\right) - \frac{4\pi}{3}$$

[somber coyote]

Arya

[upper karma]

Photomath said this

The union solution was 2pik/3

[wanton edge]

That includes 0 which is not the solution to the equation

[upper karma]

But where does it include 0

At which point

u mean k?

[wanton edge]

When k = 0, 2πk/3 = 0

[upper karma]

Ok

But is there any method of unifying several solutions?

[wanton edge]

i don't know

[wanton edge]

I am very skeptical about that since Photomath itself does not say that the general solution is (2πk)/3

[upper karma]

Maybe I confused it with something else

Ok, thanks

[wanton edge]

np

[upper karma]

By the way

If during solving an equation i get smth like this: ax^2 + bx + c = 0

Should i solve it as a quadratic equation?

Or factorize this non-perfect trinomial?

Which method is faster

[clear anvil]

depends

also btw sorry i didn't see your message for the previous question

u realize that you don't have to condense your solutions right?

it's equally valid to leave all 4 solutions as is

so you don't have to trouble yourself with "omg how to consolidate this"

[upper karma]

Ok, thanks

But depends on what?

[upper karma]

I didnt know it

[clear anvil]

on the actual equation

you can't just generalize

quadratic formula always works but with the downside that it's slower

[upper karma]

Especially using discriminant

[clear anvil]

factoring, occasionally works (maybe it works most of the time in your case), but when it does work then you save alot of time

[upper karma]

I mean D = b^2 - 4ac

[upper karma]

Ok

[clear anvil]

what about it?

Do you have a specific question in mind?

[upper karma]

No

[clear anvil]

Then try to factorize ig

Idk why you mentioned discriminant

[sudden shell]

Is me always the answer it’s not in the choices

[dark sparrow]

... are you in grade 6?

[dark sparrow]

,rccw

[dark sparrow]

how old are you?

... <@&268886789983436800> underage, but an edge case?

we can't allow users under 13 years of age on discord

[civic seal]

When

Ok, see you in a week then, sorry chief

actually I'm just gonna ban you for longer, terrible pronouns

[dark sparrow]

oh yikes

[civic seal]

rip bozo

[dark sparrow]

rip bozo indeed 🫡

[sudden shell]

What did Bro do?😭

[low oar]

rib bozo 🫡

[low oar]

be a child

[dark sparrow]

"only two genders no lgbtq" in pronouns field

[sudden shell]

Oh

[sudden shell]

?

[dark sparrow]

that's the command for rotating an image counterclockwise 90 degrees

Rotate CounterClockWise

cause yours was misorientedd

anyway the answer is among the choices -- it just is not quite immediate

[sudden shell]

But wouldn’t it be tan 40 because Ab is the opposite and bc is the adjacent and was supposed to find which function will get us the value of AB

[dark sparrow]

both can be used

consider that tan(40°) = AB/BC, but tan(50°) = BC/AB is just as good

[rugged shuttle]

Yea I did do u want to see my work

[forest blade]

Is writing quadrants as Q1, Q2, Q3, and Q4 a bad habit or does it not matter?

I just find it faster than doing the roman numerals but don't know if later on in math this is gonna mess me up lol

[wanton edge]

,rotate -90

[trail tendon]

depends on context

if you draw a cartesian plane (the normal plane with four quadrants) then you usually don't need to write the quadrant. if you are referring to the quadrants, i would say its better to say "quadrant four" instead of Q4. you don't have to say "quadrant IV". then again im missing the context

anyway if you can't think of the roman numerals or something its probably ok, just other people might not know what you mean by Q4

i wouldnt say its particularly a bad habit though

[empty yew]

No need🙂

I do know i just wanted to check that you completed or not 😀

[forest blade]

it's for trig and trig equations where you need to know which quadrant has +- of sine or cosine

[trail tendon]

oh

visual is the best imo

[forest blade]

I've just been writing them as Q1 to speed up note taking but have been wondering if that forms a bad habit for later if Q is used for something in math later on

[trail tendon]

if you're doing homework your teacher might not know what Q1 means

[trail tendon]

oh i see, i wouldn't worry about it

as long you aren't doing homework and the person grading your homework doesn't like that you don't use the roman numeral notation

for some reason

[rugged shuttle]

👍

[alpine jungle]

Do y'all know how to do #7?

[trail tendon]

do you know the ratio between the sides of a 30 60 90 triangle

[alpine jungle]

Wym?

1√3?

[trail tendon]

yea kinda

its uhh

the sides of a 30-60-90 triangle have a ratio of 1 : (sqrt3) : 2

where 1 is the shortest and 2 is the longest (hypotenuse)

right?

[alpine jungle]

Yea

[trail tendon]

so you have a 30 60 90 triangle there right

can you find the hypotenuse?

[alpine jungle]

Yes I have one, I'm not sure how to, I transferred to this school not long ago and she just shoved it in my face to explanation no nothing

[trail tendon]

ok 1 sec

[alpine jungle]

Okay!

[trail tendon]

this is the ratio between the sides of a 30-60-90 triangle

btw do you know what a hypotenuse is?

[alpine jungle]

No not really, I know that there's a leg

I think I know the equation tho

C= √a²+b²

[trail tendon]

oh yeah ok um

[trail tendon]

we can't use that here because

we only know one side

we'd need two sides to do that

[alpine jungle]

Ohh okay

[trail tendon]

yeah

but

with triangles that have the angles of 30 60 90

we have just a ratio between the sides

[alpine jungle]

Okay

[trail tendon]

so the hypotenuse is just the biggest length of the triangle

theres three lengths

the hypotenuse is the biggest

does that make sense so far

XD

[dark sparrow]

sqrt(3), not (sqrt3)

[trail tendon]

its the same...

i was just trying to seperate the colons but

XD

i have a feeling im not making sense

[lunar cave]

Someone please give me the answer

[trail tendon]

no

can help u tho

[lunar cave]

Ok that’s fine

🥲

[trail tendon]

does this help

🤣

[lunar cave]

No

[trail tendon]

alr

hmm

[dark sparrow]

if you try to put parentheses like you did but with a more complicated expression, then (sqrtx+1) becomes ambiguous

[trail tendon]

yea

but theres only x

[dark sparrow]

you could have meant $\sqrt{x+1}$ but it could be misread as $\sqrt{x} + 1$

[trail tendon]

in this case

[somber coyote]

Ann

[trail tendon]

right

[dark sparrow]

i know, but good habits.

[trail tendon]

yeah thanks 👍 🤣

[trail tendon]

what don't you understand about the problem?

[lunar cave]

I’m slow

[trail tendon]

um

[trail tendon]

do you understand what i did

[lunar cave]

Yes

[trail tendon]

ok cool

[lunar cave]

(1,4)(2,2)

Right

[trail tendon]

nah its uhh

(1,2) means the x value is 1 and the y value is 2

does that make sense or nah

[lunar cave]

Yes

So um

Do I add those numbers together or

[trail tendon]

not exactly

uhh

how far apart are the points

like look at the graph

💀

i'm struggling to explain

uhhhhh

[alpine jungle]

sorry my phone died lol

[trail tendon]

no prob 🤣

[alpine jungle]

it takes a century for it to get to 1%😔

[trail tendon]

💀

[alpine jungle]

depressing😔

[lunar cave]

You should js give the answer so u don’t have to stress urself tryna explain 💯

[alpine jungle]

Yea kinda lol

[trail tendon]

😭

[alpine jungle]

But how r u gunna learn it with just the answer? What if u got a quiz on it or smth ( ik I'm not gunna understand this lowkey I need an answer too)

[trail tendon]

with a 30 60 90 triangle
if the shortest length is 1, the longest length is 2
if the shortest length is 2, the longest length is 4
if the shortest length is 3, the longest length is 6

in other words, the longest length is always 2 times the shortest length.

similarly, the medium sized length is always sqrt(3) times the shortest length

[lunar cave]

wait you kinda right imma be confused when I have a quiz 😭 (I skip math class whenever I have a quiz)

[dark sparrow]

!noans

[lime crown]

The purpose of this server is to help you learn, not to hand out answers. Do not ask someone to give you the answer directly.

[trail tendon]

😭

[alpine jungle]

Okay

[trail tendon]

so with your triangle, you have one length, right?

[alpine jungle]

Rightt

[trail tendon]

is that the shortest, medium, or longest length?

if you're not sure lmk XD

[alpine jungle]

..

I'm not sure💔

[trail tendon]

alr 1 sec

[alpine jungle]

Ur teaching an infant

[trail tendon]

🤣

reposting problem

ok

so in your triangle you have a 90 degree angle and a 60 degree angle, whats the last angle?

[alpine jungle]

30°

[trail tendon]

yea

so is the length opposite the 30 degree angle going to be long or short

would the length between A and B be longer or shorter than the length between C and D

[alpine jungle]

Wait wait wait would the answer be 20?

[trail tendon]

no

i just realized they probably wanted you to do trig

LOL

💀

[dark sparrow]

it might help to know that in any triangle (not just 30-60-90, and not just right ones), longer sides are opposite to bigger angles.

[lunar cave]

Sooo can someone help me please

[alpine jungle]

Wanna see the directions it got on there?

[trail tendon]

sure yeah

[alpine jungle]

Heres the whole paper

There's directions on the top and bottom

[trail tendon]

ok it says trigonometry

so im assuming

we should probably do it the trig way

lol

and it looks like you've been doing trig

so yeah

[alpine jungle]

I copied everything u see on tht paper.. 💀

[trail tendon]

💀

dang ok

uhh

do you know sin(30 degrees) = ?

[alpine jungle]

I know the equation for sin

[trail tendon]

what is it?

[alpine jungle]

Sin∅=a/h

[trail tendon]

good cool

a = adjacent and h = hypotenuse, right?

[alpine jungle]

Yea

[trail tendon]

do you know the equation for cosine?

or cos

[alpine jungle]

Uhh I think

[trail tendon]

wait i'm dumb

[trail tendon]

that's actually cos

[alpine jungle]

Oh rlly

[trail tendon]

cos(theta) = a/h

yeah

sin(theta) = o/h

o = opposite

[alpine jungle]

Oh

[trail tendon]

but we actually need the cosine one lol

[alpine jungle]

Oh lol

[trail tendon]

so yeah
cos(theta) = a/h

[alpine jungle]

Dang it's 1 am

[trail tendon]

do you know the unit circle?

[trail tendon]

bruh

maybe you should sleep and try again tmr

🤣

[alpine jungle]

It's due tmmr😫

Should I js leave it blank?

[trail tendon]

bruh

[alpine jungle]

And be like whoopsie I didn't see tht one

[trail tendon]

🤣

uhh

i'm not sure

are you getting taught by a teacher?

do you get the
sin^2(a) + 9/25 = 1
part?

[alpine jungle]

Yea, a new one

[wanton edge]

1 - 9/25 = 16/25

[trail tendon]

you might want to go up to them and tell them you don't understand it so that they may be able to help you understand it 🤷‍♂️

they just subtracted 9/25 by both sides

[alpine jungle]

Man I switched out of my old class cuz it was a new teacher and he didn't know wht he was doing and the put me into another class with another new teacher

[trail tendon]

💀

[trail tendon]

BR 💀

[alpine jungle]

Good idea 😓

[alpine jungle]

It's mental 😫💔

[trail tendon]

the teachers are there to help you, even if they suck 😭

💀

it happens lol

[alpine jungle]

This one time we were taking a test and the smart kid finished early and the teacher graded it and gave it back to him while we were taking the test and everyone flooded the smart kids desk getting the answers from him

Well he wasent rlly a smart kid

[trail tendon]

LOL

[alpine jungle]

He had the answers on his phone

[trail tendon]

LOLOLOL

[alpine jungle]

Lol

[trail tendon]

the plot twist i wasn't expecting 💀💀

[alpine jungle]

Lmaoo

[trail tendon]

nah but uhh

if the teacher can't teach

try to understand it

and/or

you're screwed

is a nice way to look at it

[alpine jungle]

I can get help from another teacher

[trail tendon]

🤣 💀 half joking

[trail tendon]

oh

[alpine jungle]

LOL

[trail tendon]

thats cool

do that

[alpine jungle]

Ive got one more teacher if he can't help me then I'm rlly screwed

Hes on the younger side so maybe he can relate to me more

[trail tendon]

yeah

[trail tendon]

just dont stop looking for help

unless it really is hopeless

[wise creek]

do you need help w this

[alpine jungle]

I would text my old tutor but she's just as helpless as me in math

[trail tendon]

💀💀

[alpine jungle]

Sure but it's impossible to teach me lol

[trail tendon]

nah its just not getting taught to u in the right way

if many ppl try maybe it will click 🤷‍♂️

[wise creek]

are u just learning trig

[alpine jungle]

Yea

[wise creek]

so just sin cos tan

no secant stuff

[alpine jungle]

Yea

[wise creek]

did they tell you soh cah toa

that is easiest way to remember

[alpine jungle]

No they didn't

[wise creek]

Do you think you can remember that phrase

Say it in ur head a few times if u need

[alpine jungle]

Yea i can

[wise creek]

So that will make sure u don't forget which one u need to use

$\sin(\theta)=\frac{opposite}{hypotenuse}\\cos(\theta)=\frac{adjacent}{hypotenuse}\\tan(\theta)=\frac{opposite}{adjacent}$

[alpine jungle]

Alrighty!

[somber coyote]

SLURPZZZ

[wise creek]

Try to do that one

Well first, what function do u need based on the sides that they give u and want u to find

[brisk ore]

i cant find a mechanics channel so ill post it here. when we resolve the weight into the triangle and find the opposite value by doing hypsin(theta) what direction do we take the opposite value to be going in?

[brisk ore]

my teacher has a thick accent and is bad at teaching she cant do most of the hard textbook questions but we just have to manage

[sharp lava]

Made her do trig with surds but they weren’t bothered to teach soh cah toa that’s crazy 😭😭

[onyx kraken]

me when sohcahtoa

[elder pivot]

given some length of string and you use that string and pull it taught to make different shapes
would the area change?
like a circle and square with the same permitter is the area the same?

[dark sparrow]

no, the area is not the same

even if you limit yourself to something simple like rectangles

4 m of string can be made into the shape of a square with side 1 m, for an area of 1 m^2

but the same length of string could also be formed into a 1.99m by 0.01m rectangle, for an area of only 0.0199 m^2

[gilded sand]

I’m not sure if I understand your question but when you resolve the components of the weight force, the perpendicular component would be opposing the normal force of the surface, and the parallel component would be alone the direction of the incline

Like logically if you were standing on a steep incline you’re naturally going to feel like you’re sliding down, that’s what the component of gravity does

[brisk ore]

yeah that makes more sense

thanks

[stuck pelican]

I can't solve this.

[tardy orchid]

ye same here

lol

[stuck pelican]

Average jee questions 😭💀

[fickle cosmos]

am i doing this right**?**

[ionic viper]

Yes

Your doing great

Your great

[upper karma]

Any of these chapters gonna help with geometry and trigonometry?

[rugged shuttle]

How would u simplify the fraction

[silent plank]

complementary relationship between sine and cosine

[fickle cosmos]

I'm not sure if I did this right or not**...**

can someone help me**?**

[trail tendon]

let me se...

that looks about right yea

[vocal bison]

Help pls ping 🙏

[trail tendon]

can you draw a picture?

or are you having trouble visualizing it

[vocal bison]

huh oh theresd a picture im just getting the wrong answer

[fickle cosmos]

Well**...** don**'t you think its a bit off to you?**

[vocal bison]

nvm i got it

[trail tendon]

i think it looks correct

it COULD be slightly off but idk

[pure root]

geometry algerbra 1 and 2

[spice hornet]

How do you know when to use soh cah toa, or inverse soh cah toa?

[trail tendon]

it depends on the problem

[spice hornet]

I wish to learn this... even if it takes step by step, please!

[trail tendon]

if you want the opposite over hypotenuse, use sin. if you want the hypotenuse over opposite, use csc.

or wait

did you mean inverse sine, or cosecant

[trail tendon]

step 1: we need a problem so i know what you need/how to help you lol

[spice hornet]

I'm trying to take math seriously

Lolll

[trail tendon]

that's good XD

[fickle cosmos]

3**/1.4 feels a bit off to me...**

[spice hornet]

Because I need to be prepared how to know basic math

[trail tendon]

you might want to try to write it in terms of pi and estimate it

pi is about 3.14

[spice hornet]

I started trigonometry. I learned the a^2 + b^2 = c^2... that was easy

Then if there was the angle... you look for the opposite

adjecent, hypothenuse... adjust those to the "formulas"

[fickle cosmos]

Learning Trig graph is so confusing and hard at the same time**...**

😭

[spice hornet]

but idk which one to use as sin, cos, tan, or inverse

[upper karma]

The constant inside cos() doesn’t look rational

Try pi

Then scale the graph accordingly

[trail tendon]

ok so uh

[upper karma]

Errm you don’t need to reset the whole thing

[fickle cosmos]

oh**.**

[upper karma]

Go back to this and replace 3/1.4 x with pi x

[trail tendon]

theres:
sin = opposite/hypotenuse (soh)
cos = adjacent/hypotenuse (cah)
tan = opposite/adjacent (toa)
csc = hypotenuse/opposite
sec = hypotenuse/adjacent
cot = opposite/adjacent

[upper karma]

Why -3

You already figured out -2 and -2 in the graph above

[trail tendon]

because the middle line is 3 below the x axis...?

[upper karma]

@trail tendon they already figured two constants out here and there’s no need to change it

[trail tendon]

maybe i don't understand what the prompt is

[fickle cosmos]

So**...** -2 stays the same**..** ?

[upper karma]

Yeah, both -2

[trail tendon]

ohh are you trying to match the dotted line?

[upper karma]

Yeah

[trail tendon]

ohh

im dumb 💀

[upper karma]

you can see that the complete line graph is just the dotted line graph but compressed by 3/2

(By observing their periods)

With this, you can change the pi into 2pi/3

[fickle cosmos]

I've found my amplitude and midline**..** ( i think )

[upper karma]

The amplitude here is 3

[upper karma]

Remove the -1

[rugged shuttle]

Sin cos and tan give u ratios based on an angle and inverse sin cos tan gives u an angle based off a ratio

[queen mango]

help, with this problem, the solution comes out as a function of a,b and c
<@&286206848099549185>

[frail robin]

sin(2alpha)=a/b or am I getting something confused

[queen mango]

I think cos(2alpha)=a/b, can you help me more please

[frail robin]

yeah it is

I accidentally wrote sin instead of cos

rip

wait so you’re trying to get b

[queen mango]

don´t worry

[frail robin]

isn’t that solved

b=a/cos(2alpha)

or do you need another form

oh yeah right

no alpha

[queen mango]

no, i trying to get this

in function of a,b and c

[spice hornet]

So you're saying you're trying to know the complete ratio... both sides... in order to do that, work with the given angle

[frail robin]

what’s tg

tan?

[queen mango]

yes

[rugged shuttle]

For inverse sin cos tan yes

[cedar cradle]

im so sorry

i ready

[wispy warren]

if you want read my help forum

[wispy warren]

pls help me

i'm dying lol

[cedar cradle]

i dont know bro im sorry

i HATE i ready

[wispy warren]

pls try

[faint comet]

How do I write the general Solution like should I try to combine the answers or no

[spice hornet]

wait I realized there's sin cos tan, there's csc, sec, and cot and an inverse for all these :/

[obsidian harness]

:/

[rugged shuttle]

Lol yeah sec csc and cot are reciprocals of cos sin tan (respectively) and also give a ratio given an angle

I was scared for my life when I realized this

[dark sparrow]

honestly barely anybody gives a shit about arcsec and arccsc

arccot is on thin ice already