geometry-and-trigonometry

[dire obsidian]

all good now!

Thanks for replying tho

[vagrant steeple]

i need help with applying demoivre's theorem here. i get that you have to apply the exponent to 3 and i'm lost from there

[onyx cloud]

isnt this do u have a calculator?

[vagrant steeple]

so just plug in and all good?

[onyx cloud]

just wondering, you're gonna end up doing a fourth binomial expansion so idk

[manic crown]

Can anyone explain this..

I don't get what they mean

any example?

[onyx cloud]

what don't you understand exactly?

all they are saying is that all real numbers can be a radian measure

[dark sparrow]

@manic crown imagine wrapping the real number line around the circle like a string

then the number 1 will land on the point for 1 radian, the number 2 will land on the point for 2 radians, etc

2π is a full turn so it'll land on the same point at 0

[manic crown]

Oh thats it ! Lol

Thanks

[dire obsidian]

lol ambiguous triangles are

ugh

are these two ambiguous?

[naive warren]

Use the law of sines to find the angle in between the sides, and see if there are two answers that create a triangle (the second answer would be 180-theta) If there are two values for the angle in between, they are ambiguous

Draw the triangles if it helps

[vast dune]

...........

Im I allowed to divide both sides by a trigonometric function while solving an identity?

exmaple:

[somber coyote]

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

[dark sparrow]

ok first off bad tex

$\sin, \cos$

[somber coyote]

Ann:

[dark sparrow]

not $sin, cos$

[somber coyote]

Ann:

[dark sparrow]

second, why not just factor out sin(x)

[vast dune]

good point

didnt see that lol 🤦‍♂️

[urban oxide]

how do i make z1 be in trigonometric form?

43 btw

[upper karma]

I need some help

@modern saffron

[surreal roost]

how do I do this lol

[upper karma]

@surreal roost do you know about law of sines?

[surreal roost]

probably but i dont recall that name

[upper karma]

see how 62 degrees is across from the 15?

hold on

nevermind

this is an interesting question I haven't come across this yet

[covert zinc]

@upper karma is the answer (x, y) (2,0) for Q?

[upper karma]

confused about ,, cot(7pi/3)

,, cot(7pi/3)

[somber coyote]

owen hooper:

[upper karma]

i gotto $2\pi + \frac{\pi}{3}

$2\pi + \frac{\pi}{3}

,, 2\pi + \frac{\pi}{3}

[somber coyote]

owen hooper:

[upper karma]

idk how to work this lol

but i got to this and figured 60 degrees

so

[somber coyote]

owen hooper:

\begin{gather*}
 \frac{1/2}/\frac{\sqrt{3}{2}
\end{gather*}
```Compile error! Output:

! File ended while scanning use of \gather*.
<inserted text>
\par
<*> 614626669755498531.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.

owen hooper:

[upper karma]

ffs

there we go

is that wrong?

[silent plank]

and then simply the fractions

[upper karma]

so the answer should be

,, \sqrt{3}/{3}

[somber coyote]

owen hooper:

[upper karma]

right?

[silent plank]

yes

$\frac {\sqrt{3}}{3}$

[somber coyote]

ramonov:

osmium:

[vestal oak]

Yes

[upper karma]

is that not the same thing

[silent plank]

the value is the same but rational denom is preferred

[upper karma]

multiple choice answer was the first one so thats what i went with

,, Given Sec(\theta) = \frac{9}{2} and \theta is in quadrant IV find Tan(\theta)

[somber coyote]

owen hooper:

[upper karma]

yikes

i think its readable but how would i go about this

[open sinew]

Given $\sec(\theta)=\frac{9}{2}$ and $\theta$ is in quadrant IV, find $\tan(\theta)$.

[somber coyote]

osmium:

[open sinew]

@upper karma

[silent plank]

find a relation between sec and tan
and is tan positive/negative in quad 4

[upper karma]

so i figured Cos t = 2/9 and since sin^2 = 1-cos^2 then sin^2 = 77/81

then $\sin(\theta)=\frac{\sqrt{77}}{9}$.

[somber coyote]

owen hooper:

[upper karma]

what do i do from here/

[silent plank]

not quite.

[upper karma]

oh

[silent plank]

is sin positive or negative in quad 4?

[upper karma]

negative

oops yeah that but negative

[silent plank]

and you also have cos(theta)

[upper karma]

which is -2/9 right

[silent plank]

no

you are given sec(theta) = 9/2
and you got cos(theta) = 2/9 from that

[upper karma]

right

[silent plank]

(also cos is positive in quad 4)

[upper karma]

oh right right

[silent plank]

what's the relation between sin, cos and tan?

[upper karma]

sin/cos

[silent plank]

which means your tan(theta) is:

[upper karma]

,, -\frac{\sqrt{77}}{9} / \frac{2}{9}

[somber coyote]

owen hooper:

[upper karma]

?

[silent plank]

which simplifies to:

[upper karma]

,, -\frac{sqrt{77}}{2}

[somber coyote]

owen hooper:

[silent plank]

\sqrt
yeh

[upper karma]

wait

oh yah

damn. i had the process right

my buddy's answer was just wrong

[silent plank]

you can also be more direct and apply: $\tan^2(\theta) + 1 = \sec^2(\theta)$

[somber coyote]

ramonov:

[upper karma]

Got another problem i cant figure out

Given $\csc(\theta) = \frac{-7}{3}$, $tan(\theta)$ is positive, and $0<\theta<360$; find $\theta$

[somber coyote]

owen hooper:

[upper karma]

got it down to

,,cos(\theta) = \frac{sqrt{40}}{7}, sin(\theta) = \frac{3}{-7}

,, cos(\theta) = \frac{sqrt{40}}{7}, sin(\theta) = \frac{3}{-7}

[somber coyote]

owen hooper:

[upper karma]

Think its right so far but now im stuck

[silent plank]

you don't need to find cos(theta)
look at what the question is asking for?

[upper karma]

$\theta$

[somber coyote]

owen hooper:

[silent plank]

csc (and sin is negative)
tan is positive.
which quadrant is theta in?

[upper karma]

3?

[silent plank]

yes

what would be your reference angle?

[upper karma]

no idea

[silent plank]

well you know
sin(theta) = -3/7

[upper karma]

i missed the whole reference angle thing

[silent plank]

the reference angle is the angle the terminal side makes with the x-axis

[upper karma]

still lost

[upper karma]

oh ok

understood

[silent plank]

i denoted the reference angle as alpha.

[upper karma]

so how do i find the reference angle

from the sinx

[silent plank]

it is the angle the terminal side makes with the x-axis

\sin(alpha) = |-3/7| = 3/7

[upper karma]

im sorry I still dont understand, where do we get the angle from the -3/7

like how does -3/7 relate to the angle

[silent plank]

you can get the angle from inverse trig

\sin(alpha) = 3/7
alpha = arcsin(3/7)

[upper karma]

Thanks so much man

Just aced that exam

[coral sapphire]

[shut mountain]

Starting from which line am I failing?

[twin prawn]

First <=>

[shut mountain]

Epic.

:D

[weak shoal]

Hmm, with your first line, it's problematic because you can't show that the equation is true by assuming that the equation is true. That equals sign needs to go away and you need to evaluate both the right-hand sides and left-hand sides separately.

[shut mountain]

Alright

Makes sense

[twin prawn]

You multiplied the RHS by, what, 8? Instead of 2

[vast dune]

why cant u assume its equal?

[weak shoal]

Cos you're trying to prove that they're equal.

[shut mountain]

What's RHS

[weak shoal]

Let p be a statement that you're trying to prove. Then, you can't use p => p. That's just a tautology. It's trivially true but p could still just be false.

[twin prawn]

Right-hand side

[weak shoal]

Ah, the right-hand side

[vast dune]

well if u assume its equal and they end up not being than u proved it?

[shut mountain]

Lol, that's true. Thanks, @twin prawn

[twin prawn]

np

your proof will become a lot shorter

[weak shoal]

@vast dune Okay, suppose they ask you to prove a proposition. You can't assume that the proposition is true and then show that the proposition is false. That's not even how contradictions work.

[twin prawn]

But you can prove that a certain statement is equivalent to a tautology

Which is what's happening here

[vast dune]

You assume the proposition is true, but u derive at an equation that isn't hence u proved that the proposition is false

[weak shoal]

Oh right right, my bad, my bad. Had a brain fart there. In that case, that would be a contradiction that you derived.

In this case, though, he's attempting to prove that the given proposition is true. You can't really assume the proposition in trying to prove that the proposition is true.

Here’s a shorter proof. In that situation, you don’t have to know what the right hand side is beforehand. You can just derive it 🙂

Eh, I should probably type that out in Latex for practice

[somber coyote]

Abhijeet Vats:

[twin prawn]

[weak shoal]

:(((((( sorry sorry, I'm dumb lol

[dark sparrow]

$\sinh(2x) = \frac{e^{2x} - e^{-2x}}{2} = \frac{(e^x - e^{-x})(e^x + e^{-x})}{2}$

[somber coyote]

Ann:

[dark sparrow]

watch & learn, i suppose

in latex, groups of symbols are delimited by {}

[weak shoal]

Yea man, that's what I'm trying to do. I just have to get used to it, I suppose.

Ah that's where I went wrong.

[dense pollen]

https://gyazo.com/43640fbb0466c86737ea927a3b385033

Can someone tell me if this is possible?

When I break the triangle part down I do 6.2 * (sin30) to find how much of the top part is used in my right triangle portion and I get 3.1

then 3.1-3.1 = 0 for the width of the remaining part

[dire rampart]

uhh

does that say the base of the triangle is 0?

[dense pollen]

O = opposite

I shouldve used x and y

Im pretty sure this isnt solvable, I drew it on autocad and makes no sense

[weak shoal]

,rotate

[weak shoal]

There's no point in solving problems like these in numbers first. Use variables to simplify your work as much as possible.

[dense pollen]

Did you try numbers yet?

[weak shoal]

Nope. Why?

[dense pollen]

I dont think the numbers work out

With r= 6.2 and the 60 degrees given it makes the base of the right angle = 3.1

and the whole top part is supposed to =3.1

[weak shoal]

Of course they don't. That's why we solve it using variables first so that we can change the numbers so that they make sense.

[dense pollen]

Im trying to find the area of the given shape

So its unsolvable..

[weak shoal]

With the given numbers, sure

[dense pollen]

I understand what you mean, but I just wanted to make sure I wasnt making a silly mistake lol

[weak shoal]

🙂

[surreal roost]

https://cdn.discordapp.com/attachments/326138757474680852/651856370232066070/Screenshot_1.png Yeah probably really easy lol.

[jolly storm]

Just curious. Is sin(θ-45)/cos(θ-45) = Tan(θ-45)?

[weary drift]

yes

[jolly storm]

ok thanks

I'm fairly new at trigonometry

we haven't properly studied it yet so I'm getting by with what I learn in applied math

[weary drift]

np. that's the definition: $\tan(x)\equiv\frac{\sin(x)}{\cos(x)}$

[somber coyote]

RokettoJanpu:

[jolly storm]

See I knew about that but I didn't know if it applied to compound angles

I greatly appreciate the help

Yeah I'm actually stupid

When I was getting the time with Sy=ut+1/2gt^2, I forgot to multiply the time and gravity by half

[long sapphire]

If we let x = theta - 45, it becomes more evident

[upper karma]

does (some number*pi)/3 always have a reference angle of 60

[weary drift]

nah

[upper karma]

why no

t

[weary drift]

0pi/3

[upper karma]

ok

not zero

[weary drift]

3pi/3

[upper karma]

not anything that simplifies

[weary drift]

i poked too many holes in what you said

[upper karma]

ok but what about if it doesnt simplify

[weary drift]

define exactly "doesn't simplify"

[upper karma]

like it cant be further simplified to somethijng that isnt a fraction because both of those do

[weary drift]

i'd rather you properly learn how to count angles in the unit circle

if it can't simplify to an integer multiple of pi, sure

[dire obsidian]

Can someone help@me set this up

The ship one

its a trig problem

ill have to draw a triangle

[void moat]

Use Pythagorean’s theum @dire obsidian

[dire obsidian]

im just having trouble determining the an-

[void moat]

Find hypthenuse

[dire obsidian]

lol its not pythagorean..

i need to know the angle of northwet

west

northwest

whats that in degrees

[void moat]

It’s trigonometry

Soh cah toa

[dire obsidian]

is it 135?

north west is 135

yes, i am aware but in order for me to use soh cah toa, i need to find an angle

[void moat]

Ok so use a compass

Northwest is 135

But use compass to make sure

[dire obsidian]

yeah 135

sorry for the late response

im gonna draw it out ina bit

i was working on it in class

and the bell rung

[sacred cradle]

no to find the area of the triangle use the formula A=s1s1sinx

[dire obsidian]

so ill come back w the answer soon

[sacred cradle]

s2*

[dire obsidian]

L O L say what now

[sacred cradle]

so bcsin135

[dire obsidian]

yeah no, my teacher says to use trig for it

the area thing is different

thats for the 2nd question

i was asking about the 1st question

oof

[sacred cradle]

oh nvm then

[dire obsidian]

its fine

ill come back w the answer

[sacred cradle]

whats ur question?

[dire obsidian]

it was the ship one

i just wanted to know the angle

and i had in mind that it was 135 but i wasnt sure

cause this is like a last minute thing my teacher brought up today cause the test is tomorrow

[sacred cradle]

oh yea im pretty sure its 135

[dire obsidian]

@void moat could u possibly work it out and get an answer so that when i report back w an answer, we can compare our results?

[void moat]

I don’t have pen and paper

[wooden compass]

My teacher mentioned to use cross multiplication or something for this?

I don’t really understand it?

[deft ingot]

(1/z) × z = 1

[wooden compass]

?

[sacred cradle]

yea cross multiply

a/b=c/d => ac=bd

[wooden compass]

Oh ok

So 20*5 = 3z?

[sacred cradle]

yes

[wooden compass]

Oh ok

So it would be 33.3333

How would I show the answer?

[deft ingot]

You got a wrong answer

[wooden compass]

20*5 = 100

Divide that by 3?

That is z?

[deft ingot]

No;

You will get 20×3 = 5×z

[wooden compass]

you mean 5z

[deft ingot]

Yes

[wooden compass]

i have a problem that is

(a -7)/7 = ?/?

[deft ingot]

Wait what?

Can you provide the source?

[deft ingot]

Do they give additional information for a?

[wooden compass]

Nope

Oh wait

[long sapphire]

what are you supposed to do

[wooden compass]

I am blind

The answer is simple

[long sapphire]

??

[dire obsidian]

oof

am i wrong or is my key wrong

is the exact value of cot-330 equal to root 3 or -root3

root 3 right

[long sapphire]

tangent is positive in quadrant 4

so \sqrt{3}

[dire obsidian]

ye

[split geode]

Can someone help me with this question?

[grim plank]

Not sure what its asking 🤔

Are you meant to expand them using angle formulas and relate them?

[wooden compass]

how would i solve this?

(4a+1)/7 = 2a/3

[grim plank]

How have you tried to solve it

Wrong channel as well

[wooden compass]

oof

4a+1 = 3 doesnt make sense

4a +1 = 2a doesnt make sense either

so i dont know what to do for this

also i asked in this channel because my class is IM2 and like most of what we do is geometry, sorry

[deft ingot]

#help-1

Ignore that ^

It is in use right now.

[wooden compass]

why do we have questions channel

[deft ingot]

To post questions.

[wooden compass]

like why cant we ask questions in here?

[deft ingot]

Because we have a questions channel.

[wooden compass]

ok

so do i move to a questions channel now?

[deft ingot]

Yes, prefer to do so.

[upper karma]

We can ask questions in here

It's what these channels are mainly used for

[deft ingot]

Yes; but the questions asked were not related to the topic.

[dire obsidian]

https://cdn.discordapp.com/attachments/326138757474680852/652302282095853579/image0.png

can someone confirm if this is approx 140 mi

[weary drift]

@upper karma don’t spam server invites

[weak shoal]

Show your attempt?

[fringe breach]

i’m not sure how to start

but i think the top angle for the triangle with the two 35s is 110

[upper karma]

Internal triangular degrees is 180z

180*

Typo

So

[fringe breach]

35+35 is 70so the other is 110

[upper karma]

180 - (35 +35)

180 - 70

110

[fringe breach]

so 110

[upper karma]

Yep

Correct

[fringe breach]

but i don’t know what to do after

[upper karma]

Let’s see...

Ima give this an attempt.

Probably going to fail but we’ll see where this goes.

The other two missing angles are 67.5 degrees.

Don’t really know why anyone needs to know that but some info anyways.

[fringe breach]

but can we just assume the two angles are congruent?

[upper karma]

Probably?

[fringe breach]

ok

[upper karma]

I’m as stuck on this one as you are

[fringe breach]

ah

thanks anyways then

[upper karma]

We also see that the three other missing sides are congruent as suggested by the II marks....

Have you tried plugging in values for X?

Trial and error?

ok so call the side with two marks y

mmk

[reef meadow]

just google squadron dog 106

[upper karma]

split both triangles in half

[reef meadow]

that will give you the answer

[upper karma]

There’s the idea I was formulating

so (3x-7) = ysin(22.5)

and (x+5)/2=ycos(35)

I see

(3x-7)/sin(22.5)=(x+5)/2*cos(35)

2*cos(35)(3x-7)=sin(22.5)(x+5)

and now for some algebra

solving equations.

this big boy

welp

He a thicc boy yea he is

we can do some approximating though

we need to know if this is less than 6 or not

Can that be simplified?

[long sapphire]

why can't it be simplified

[upper karma]

I’d say it can...

[long sapphire]

it's just anumber

[upper karma]

That’s what I thought

i factored out sin(22.5) on both numerator and denominator

and canceled

Mhm

idk if that helps

i feel like it does

It should

Plus after you get rid of the sin things you can divide the thing

i just set it as a variable

now i do polynomial division

Well it has a value but that makes sense.

Where’d the 7 come from?

so we know x is greater than 2

Nvm

It is greater than 2 but is it more than 6?

[reef meadow]

math is gay

[upper karma]

well the smaller a is, the greater x is

true.

so lets assume it is smaller than it actually is

so a=cos(35)/sin(22.5)

cos(35)=sin(55)

sin(55)/sin(22.5)

ok so this is clearly greater then 1

so lets plug in 1 for a

2+(2+7)/(6-1)

2+9/5

this is less then 6

therefore the answer is C

@fringe breach there’s your answe

R

Er whoever it was

if you had a calculator this would obviously be easier

@fringe breach

[fringe breach]

ah

[upper karma]

Sorry wrong ping

[fringe breach]

thanks

[upper karma]

He did it all, I contributed little.

A few angle measures perhaps

well these questions are good fun anyways

I agree

I should get better at them

time to speedrun 40 Kahn academy videos on trig

lol

I learn fast so this should be possible within let’s say 2 weeks.

[hallow flame]

is there anything I can do here?

[upper karma]

yes actually

tanx= sinx/cosx

so sinx^2/cosx=0.98

sin^2x=0.98cosx

1-cos^2x=0.98cosx

cos^2x+0.98cosx-1=0

use quadratic formula to find cosx

this is equal to cos(x)

X = -b +- <sqroot b^2 -4ac> /2a

oof

hold on

Donut forget the +-

there we go

idk how to use +- in desmos

but its implied ig

I see

Ok looks good to me

[hallow flame]

interesting

[upper karma]

you could approximate to 1 for 0.98 if thats your goal

and then you get the golden ratio minus 1

which is neat

1.61803398875...

Yes I memorized that

so either way

there you go

[weak shoal]

Dafuq? Why’d you memorize a number like that?

[upper karma]

people memorize pi

why not golden ratio

but i can't remember many digits of anything

3.141592

2.817

or actually 2.718

totally didn't look that up

3.1415926538

35?

Perhaps

That’s all the digits I remember

[summer plume]

I had a couple ideas how I can solve this, but when I try it, it doesn’t really look/ feel right. Anyone else know a way to solve this?

[silent plank]

what was you idea?

[summer plume]

Separating the whole numbers by putting them in parentheses. Like 3(cos theta + 1). Or maybe divide both sides by 2, and then I could square root the sin^2 theta

[silent plank]

that doesn't help you

try writing your whole equation in terms of cos(theta)

[weak shoal]

Remember the relation between sin(theta) and cos(theta)?

[summer plume]

Right, so I would change the 2sin^2(theta) to 2 - 2cos^2?

[weak shoal]

Correct

[silent plank]

cos^2**(theta)**

[weak shoal]

Then what do you do?

[silent plank]

you have a quadratic in cos(theta)

[summer plume]

I would want to bring over the 2cos^2(theta), but if I did that, how would 3cos(theta) be changed if at all?

[weak shoal]

It wouldn't get changed

[silent plank]

well what operation are you applying to remove the 2cos^2(theta) from the RHS?

[weak shoal]

The point is that you would form a quadratic equation in cos(theta)

Then, solve that quadratic equation for the values of cos(theta)

[silent plank]

^(will that affect the actual individual terms)

[summer plume]

This is what I have so far

[silent plank]

how do you usually solve quadratics?

everything to one side = 0 right?

if you aren't used to this, do a substitution and let
u = cos(theta)

[pliant vector]

So i have a question regarding co ordinate geometry

Basically section and mid point formula

A straight line passes through point P(2,1) and cuts the Y-axis and X-axis at A and B respectively if AP: PB =3:1 find
(i) The co ordinate of A and B
(ii) equation of the line AB

[weak shoal]

Show your working?

[pliant vector]

I did it till here cause i thought it was wro ng

Wrong*

[weak shoal]

Okay, the problem is that your working is just a hodgepodge of symbols that don't make any sense whatsoever.

1. It's a geometry problem that relies on the Cartesian Plane. So, draw the scenario in a Cartesian Plane, please.

[grim plank]

,rotate

[grim plank]

u labeled p wrong

sin rule might help

Label each side of the triangle and relate them with the sine rule to start off ...

sure.

i think u can simplify that 1st

one

sin 180 - x - y

Not sure - i dont remember any formulas or anything

Id just draw a graph of sin and see

After u simplify it, bear in mind you're after XY + YZ

the one with a vertical bar is called phi

the other is called theta

anyways, yes

and YZ?

wait a sec

what u have for XY isnt quite right

[grim plank]

there is no right angle in that triangle

necessarily

All you have is the sin rule

to help you out

Cant use sin theta = opposite / hypothenuse

[weak shoal]

Notice how my diagram is half the page?

Start by drawing large diagrams

In every single geometry problem, draw large diagrams. You will see things better.

Then, you can't assume anything to be true unless it is explicitly stated by the problem.

Everything else that you know has to be deduced from the given information as well as any theorems that may apply to the given situation.

In this case, you may use the Law of Sines to deduce the lengths of the two unknown sides. You may use the addition formula for the sine function and you may also use the value of the sine function for a specific x.

Proceed logically and have better handwriting. Your handwriting is terrible and needs to be improved upon. Your parantheses have to be in the correct places as well.

I know i'm being harsh but this is a necessity if you want to solve problems effectively.

[grim plank]

um

🙄

[weak shoal]

Use Brilliant.Org if you want more practice.

@grim plank Yessss shuri, anything ya wanna say??????

😄

[grim plank]

Yes, use a ruler

No, not really

[weak shoal]

Oh for my drawing?

Lol

[grim plank]

Draw a graph of sin x

Think what sin(180 - x) simplifies to

mhm

x but sure

[weak shoal]

^help la

@upper karma If there's one thing you should definitely take away, it's that drawing big diagrams allows you to see what's going wrong or right with your solution.

🙂

[weak shoal]

Draw a bigger diagram on a blank piece of paper

Don't worry about it, it's normal to get stuck

There you go

Just draw one line so that it's perpendicular to PT

[grim plank]

r for radius

right angled triangle

[weak shoal]

Do you understand the logic I used?

[vast dune]

Can someone confirm this is not correct?

shouldnnt it be instead?

[weak shoal]

Woah, what is that handwriting?

[vast dune]

not mine 🤷‍♂️

[weak shoal]

Nah nah nah, that ain't acceptable. Write it out nicely for my poor granddaddy eyes

[vast dune]

whah thats my teachers handwriting lol

[upper karma]

I can't even decipher that my god

[weak shoal]

Your teacher should consider getting his/her handwriting fixed.

[upper karma]

Is your prof a doctor?

[weak shoal]

or replaced

[vast dune]

xddd

[weak shoal]

whichever of the two is more feasible

[upper karma]

I have disgraphia and my handwriting is better

[weak shoal]

I have myopia and I can't see.

😄

Come come, write it out nicely on a piece of paper so that I can caress it lovingly examine it in greater detail.

[vast dune]

btw if u thought that was bad

I had to decipher this a bit ago

all just wigly lines lel

[weak shoal]

Ah I see now

Okay, so what needs to be done, my lord @vast dune

[vast dune]

Was wondering if that step was correct

the signs should be what I wrote in brown I thought?

[weak shoal]

huh

you have -(sin(x))^3-(cos(x))^3 and you wrote that as a^3-b^3?

[vast dune]

nono thats just the formula I used

[weak shoal]

I think you were right in what you did but it may have been a rather convoluted way of doing it.

You'll have to show me the intermediate steps.

[vast dune]

How did u go from + to * there?

oh ok

I think u just forgot it there

[silent plank]

applying the sum of 2 cubes is fine
but there shouldve been an intermediate step of

1 - (a^3 + b^3)/(a+b)

[upper karma]

i need help with suimilarity

[weak shoal]

@vast dune yea i just forgot it lol

To be fair, I did that at 5am lkl

lol

?? Where did angle x come in?

Urm, write down your reasoning on a piece of paper

Because in that second part

It seems like you’re taking the cosine of the sine of an angle

Which is not an angle

You’re taking the cosine of a length

Typically, the notation used is arcsin(), not asin(). The latter gives off the impression that you’re multiplying sin() by some constant

Yeap

But you didn’t need to use cosines and sines

Pythagoras would’ve been enough

Well

Draw a line from the center of the circle to one of the endpoints of the chord

Do you see a right triangle?

🙂

[subtle spear]

Hi guys, how would I rewrite this in terms of sine and cosine with no quotients?

I managed to figure out that it equals cot(x)

[weary drift]

whatcha got against quotients?

[dark sparrow]

$\cos(x)(\sin(x))^{-1}$

[somber coyote]

Ann:

[subtle spear]

Idk @weary drift haha, textbook says no quotients

@dark sparrow thanks!

[wispy pawn]

Can someone help me out?

[signal hornet]

C is supposed to be 106deg. I can't seems to get the right answer

[thorn talon]

What's your working?

[signal hornet]

I know I got the side b right

Where did I go wrong?

[silent plank]

there are 2 possible angles C in (0,180) where
sin C = 4sin45° / sqrt( 20 - 8sqrt(2)
calculating it like that using the sin rule creates some ambiguity

[thorn talon]

180 - C

[silent plank]

you can choose the correct value by observing the angles.
or avoid this by applying cosine rule again

[signal hornet]

I used the sin rule the second time

Only used the cosine rule for side b

[silent plank]

but using the sin rule the second time ignores the info of the side length 2, and there would be ambiguity

sin(73.68°) approx 4sin45° / sqrt( 20 - 8sqrt(2))
but
sin(106.32°) is also approximately the same value

[signal hornet]

Yeah I got the answer now

Thanks

[marble grove]

can someone explain projecting basis vectors onto other vectors to me

this shits been hurtin my head

and apparently you acheive this with the cross product

[thorn talon]

what about it?

[marble grove]

right, well say i have a vector ab and i want to get a new vector thats orthogonal to i hat

how the hell does that work

I think i would do something like what i described before?

[umbral snow]

@marble grove
If you're working in R³, you can get a vector orthogonal to both a and b by finding a×b

If you're in R², (-b,a) is orthogonal to (a,b)

[marble grove]

when you say r^3 what are you referring to?

[umbral snow]

I mean real vectors with three entries

[marble grove]

whats considered not a real vector? points?

[umbral snow]

Like (3,4,8)

[marble grove]

oh

[umbral snow]

A vector that isn't real is something that includes i, for example

[marble grove]

i see yea

why does multiplying them give an orthogonal vector?

[long sapphire]

it's not component-wise multiplication, it's the cross product

cross product of a and b, by definition, will result in a vector that is orthogonal to both a and b

[marble grove]

thats helpful to kno

[fathom zenith]

i've been using khan academy and professor leonard for math in general, but i need some serious trig review from the ground up. can anyone recommend something that isn't either of those?\

[devout shell]

probably some book about trig or a precacl book

[weak shoal]

I'd recommend the book The Method OF Coordinates by Gelfand, as well as Trigonometry by Gelfand. They are brilliant 🙂

Then, if you're up for something a little more challenging, try Coordinate Geometry by Luther Pfahler Eisenhart.

[fathom zenith]

thank you!

[robust socket]

,, 2\cos(x)+2\sin(x)=\sqrt{3}+1

[somber coyote]

Umma.Gumma:

[robust socket]

I'm wondering which method would be the best one to solve this

that radical is extremely annoying and after a few attempts, I always get stuck because of it

[weak shoal]

Well

How about squaring both sides and using identities to simplify?

[robust socket]

ok let me check

,, (2\cos(x)+2\sin(x))^2=(\sqrt{3}+1)^2

is this what you would do?

[weak shoal]

Mmmh you missed a 2 on your cos(x)

[robust socket]

It gets too messy

[vast dune]

cant this be solved with t formulas and substitution?

[robust socket]

tried that as well

it ends up with a fucked up quadratic eq

[dark sparrow]

lmao just rewrite $\cos(x)+\sin(x)$ as $\sqrt{2}\cos\paren{x + \frac\pi4}$

[somber coyote]

Ann:

[vast dune]

[robust socket]

care to explain how you did get there?

[dark sparrow]

er

should be sin, not cos. my bad.

https://brilliant.org/wiki/trigonometric-r-method/

[vast dune]

shouldnt it be $\cos(x)+\sin(x)$ as $\sqrt{8}\cos\paren{x + \frac\pi4}$ than?

[somber coyote]

Abdos:

[vast dune]

since $R = \sqrt{2^2+2^2}$

[somber coyote]

Abdos:

[weak shoal]

Haiz, just square both sides and use sin^2(x) + cos^2(x) = 1

You'll also get 2sin(x)cos(x) = sin(2x) lol

Then, solve sin(2x) = whatever for x

[dark sparrow]

squaring both sides is not to be done nonchalantly like that

it introduces extraneous solutions

[robust socket]

if you end up with sin^2(x) + cos^2(x) = 1 you are doing it absolutely wrong

[weak shoal]

Ya ya of course, just account for those extra solutions

No, no no you'll get something of that sort

Like, something that looks like that

Wait, actually, staring at it a bit now.

Isn't it a bit obvious that x = pi/6 is a solution lol. So is x = pi/3

[robust socket]

I mean looking at it yes, but I don't like 'obvious' in math

[weak shoal]

:/ lel you can do it analytically but identifying obvious solutions is always a good start.

You've identified two particular solutions. Use those to construct more solutions

[robust socket]

those are the actual two solutions of this crappy equation

but I'm interested in doing it analytically tbh

[weak shoal]

^The above is an analytical way of doing it lmao

Guessing solutions to equations and then adding more solutions based on those guesses is done quite a lot.

Okay, so you know that pi/3 and pi/6 are solutions. That means that if you take pi/3 + 2kpi and pi/6 + 2kpi for natural values of k, those will be solutions as well.

Now, extend it backwards to the negative integers as well.

Check if it extends in the same way. If it doesn't, then you can solve the equation using the method I suggested or Ann's method. Get more solutions out of it.

[robust socket]

ok thanks

[fierce fulcrum]

hello there,
While learning about perceptron learning algorithm, I stumbled upon this

# This function determines the cross product between a line and a given point.
# Returns 1 if above the line and -1 if below the line.
def targetFunction(x1,y1,x2,y2,x3,y3):
    u = (x2-x1)*(y3-y1) - (y2-y1)*(x3-x1)
    if u >= 0:
        return 1
    elif u < 0:
        return -1

So it determines if the point is above or below a line. My question now is where does (x2-x1)*(y3-y1) - (y2-y1)*(x3-x1)
come from

I found something about Heron's formula online that is probably related to this

[slender nacelle]

What do you mean by cross product between a line and a point?

@fierce fulcrum

[fierce fulcrum]

@slender nacelle I dont know. the code is from https://github.com/kirbs-/edX-Learning-From-Data-Solutions/blob/master/Homework_1/Python/homework_1_by_kirbs.py

[slender nacelle]

Oh I see. It's a way to see if point is above or below the line

Do you understand what they mean by that?

@fierce fulcrum

[fierce fulcrum]

yes I get the idea

but dont understand where it comes from

and was wondering if anyone could give some explanation

[slender nacelle]

So do you know 2 point formula for straight line?

[fierce fulcrum]

yes

[slender nacelle]

$\frac{y-y_1}{y_2-y_1}=\frac{x-x_1}{x_2-x_1}$

[somber coyote]

oscillatingEquilibrium:

[slender nacelle]

Do you see that given expression in the code is reformulation of that?

At point $(x_3,y_3)$

[somber coyote]

oscillatingEquilibrium:

[fierce fulcrum]

ohh

wait let me think

[slender nacelle]

Sure

[fierce fulcrum]

but where comes the x3 and y3 in? @slender nacelle

the y and x?

[slender nacelle]

So, points 1 and 2 are points ehich make a line

[fierce fulcrum]

yes

[slender nacelle]

If point 3 was on the same line, it will satisfy the expression given in code and give 0

[fierce fulcrum]

makes sense

[slender nacelle]

But if putting that point gives something other than 0, it means it is not on the line

If it gives positive value, it is on one side of the line

If it gives negative value, it is on different side of that line

[fierce fulcrum]

yep

[slender nacelle]

Now, let's say if you put 2 different points in that expression

Both are of same sign, it means both are on same side of the line

What does upper part of a line means? Lime containing point$ (0, \infty)$

[somber coyote]

oscillatingEquilibrium:

[slender nacelle]

If you put $(0,\infty)$ in that expression, the result is $\infty$, which is positive.

[somber coyote]

oscillatingEquilibrium:

[slender nacelle]

So if that expression is positive, it means the point is on upper half

If it's negative, it's on lower half

And if it's zero, it's on the line

@fierce fulcrum

[fierce fulcrum]

what do you mean by putting 2 different points in the expression

one after another?

[slender nacelle]

Yes

And both gives same sign (positive, positive) or (negative, negative). They both are on same half

[fierce fulcrum]

okay wait. I got that. However I didnt know the formula u presented up there for the 2 points. I learned from this: https://www.mathsisfun.com/algebra/line-equation-point-slope.html

[slender nacelle]

Oh that's great! You learned something new today

[fierce fulcrum]

so where is the relation between yours and that link?

[slender nacelle]

Of course. You know $(y-y_1)=m(x-x_1)$

[somber coyote]

oscillatingEquilibrium:

[slender nacelle]

And what is m?

[fierce fulcrum]

slope

[slender nacelle]

$m=\frac{y_2-y_1}{x_2-x_1}$

[somber coyote]

oscillatingEquilibrium:

[slender nacelle]

If you substitute this, you'll get my equation

[fierce fulcrum]

oh

so in conclusion

$\frac{y-y_1}{y_2-y_1}=\frac{x-x_1}{x_2-x_1}$

[somber coyote]

YoureStein:

[fierce fulcrum]

is what I can use for checking if a point is either above or below a certain line

[slender nacelle]

Yes

For any line equation ax+by+c=0

If putting a point in ax+by+c

If you get 0, the point is on that line

If you get positive value, it is on a side of the line (don't know which)

And it is on opposite side if you get a negative value

[fierce fulcrum]

wow this conversation cleared a lot of things up for me in my math world

thanks

[slender nacelle]

You are welcome!

[narrow storm]

cos^-1(cos(3)) is equal to 3 right?

[surreal bolt]

is 3 in radians?

[narrow storm]

no

[surreal bolt]

seriously?

[narrow storm]

?

[surreal bolt]

if 3 is in degrees yes, if it is in radians, no.

[narrow storm]

i need to solve for t in the equation:
x/2 = cos(2t)

how do i remove the cos

[surreal bolt]

are there any restrictions on t?

[fierce fulcrum]

arc cos @narrow storm

[narrow storm]

mathway says its not 4

@fierce fulcrum

[dark sparrow]

if 3 is in degrees yes, if it is in radians, no.
actually, no, arccos(cos(3)) is 3. because 3 is in [0, π].

@narrow storm that's weird

how did it even get that decimal

should be 2.28 ish

[somber coyote]

The following error occured while calculating:
Error: (intermediate value)(intermediate value)(intermediate value) is not a function

[upper karma]

can anyone help me how to start off this question?

Use equations on the top to help you out

[silent plank]

where did you get stuck?

[jolly storm]

So basically I'm supposed to get a value for theta (a) but I just don't know how

I have the answer here but idk how to get to it

[somber coyote]

nickyteddy:

$2tan(a)=\frac{cos(a)}{sin(a)}$

[jolly storm]

I know that cot(a) = cos(a)/sin(a) but idk where I would go from there

[silent plank]

multiply both sides by tan(a)

(which remove the trig functions on the rhs)

are you told which quadrant your angle is in?

[jolly storm]

not at all

[silent plank]

what do you get after following that step above?

[jolly storm]

Assuming I did everything right I got:

$tan^2(a)=0$

[somber coyote]

nickyteddy:

[silent plank]

how are you getting 0?

[jolly storm]

What is the rhs btw?

[silent plank]

right hand side

[jolly storm]

ah

[silent plank]

tan and cot are recipricals.

a * 1/a = ?

[jolly storm]

Ah I don't know that much about trigonometry

[silent plank]

this action is algebra

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

[jolly storm]

ah thanks

I'll take that down in my book then

[silent plank]

i mean you implied that you already knew that

[jolly storm]

oh no this is the first time I'm coming across this

ok I simplified it. Thanks for the help!

[silent plank]

unless you were told specific information,
tan(theta) can also be -1/sqrt(2)

[wind heart]

How would I do 12? I kind of forgot

[silent plank]

the thing that looks like a "1" should be a lowercase (L)

are those your calculations?

[rich wolf]

i don't know why that "1" is even there

you don't need it

[silent plank]

maybe they want you to find 2 variables.
but there's no ° so probably shouldn't be there

[rich wolf]

yeah that's what i was thinking

it doesn't look like it's supposed to represent a length either

[silent plank]

and to find x, use the angle sum of a triangle

[hybrid jasper]

idk why but I'm not seeing how a = 1/2
I know that sin(-210) = a/2a = 1/2 but how does that make a = 1/2

[dark sparrow]

2a=1

[hybrid jasper]

yeah but if sine is opp/hyp isn't 2a = 2 @dark sparrow

this is so easy but honestly I don't know why I can't get this lmao

[surreal bolt]

sin here is a/(2a) or (1/2)/1

[dark sparrow]

no? just because a/b=c/d doesn't mean a=c or b=d

[upper karma]

opp is a, and hyp is 2a right, so a/2a is 1/2 and sin is a

sin is the "height" of the triangle and cos is the "base" since the hypotenuse is 1

[hybrid jasper]

wait am I supposed to assume this is on the unit circle

[upper karma]

Well you dont have to but if the hypotenuse is 1 then it is on the unit circle

[surreal bolt]

it happens to be on the unit circle yes because r = 2a = 1.

(but as Decartes says you can use any circle if you want to consider cases where the hypotenuse or radius is Not 1)

[upper karma]

however this solution works no matter the scale, just that a wont be equal to the sine

[hybrid jasper]

ah when you said (1/2)/1 it made sense

yeah this shouldn't have taken me this long lol

[upper karma]

trigonometry is confusing to everyone I think

[heady jetty]

Thats so very true

[hybrid jasper]

Euler was probably chilling

[heady jetty]

I keep note cards with all the trig identitys on my person 24/7

[hybrid jasper]

lmao

[upper karma]

Big brain tactic is having to relearn them all the time

[heady jetty]

Im being 100% serious

You can never tell when someones gonna mug you and you need the triple angle identity to get away

[dark sparrow]

it's bad form to write t's like that in the presence of plus signs with which said t's are likely to be confused.

[heady jetty]

Thats true

[upper karma]

They are presumably not for anyone to read though

[heady jetty]

Also true

Trigonometry was my first class that really required memorizing a bunch of formulas, and i got lazy

And memorizing a Laplace transformation table isnt any easier

[upper karma]

I dont usually memorize a lot

[eternal obsidian]

AED and ADC are straight lines, find angle c

Anyone know how to do this?

[weak shoal]

show your working

Cos look very closely at your diagram. Can you find anything that relates a, b & c?

Anything at all.

[eternal obsidian]

I know how to do this if the big triangle is iscoceles but we dont know

Is the question defective

we know ED and BC are parallel lines so we can use some rules there

and a + b + c = 180

i still dont know how to do tis lmao

[weak shoal]

a + b +c = 180

Notice that (180-2a) + (180-2b) = 150?

Then, can you find a+b?

[eternal obsidian]

huh howd u get (180-2a) + (180-2b) = 150

oh wait jk yeah

and a + b = 180-c

[weak shoal]

Lel

So can you find c?

[eternal obsidian]

lmao no

HAHAHAHAH

wait nvm im dumb

ok i got it

thanks

[weak shoal]

You're not dumb lol

[tribal rose]

O_O

[weak shoal]

Just needed a nudge in the right direction.

[tribal rose]

(180-2a) + (180-2b) = 150

how did you get 150?

[weak shoal]

Well

Look at that triangle in the left corner very closely

[tribal rose]

uh huh

[weak shoal]

Can you see that there's one angle that's 180-2a and the other angle is 180 - 2b?

Notice that (180-2b)+(180-2a)+30 = 180?

[tribal rose]

yeah I see what youre looking at now thank u

[weak shoal]

🙂

[quiet bloom]

dont memorise

just implement an interactive unit circle into your head

then you can forget everything but it'll always stay there

No need to remember the signs when you just look at them inside and say 'oh that's -ve, oh that's positive'

[upper karma]

How do I find this solution?

so circumference is 2r.pi

area is pi.r²

given is 50.24 = circumference

which means

50.24 = 2r.pi

Right?

How do I continue from here?

<@&286206848099549185>

?

Can I do 50.24 / 2pi to get r?

[summer spire]

what's the fomula for the area of a circle?

[upper karma]

This is what I end with

Pi.r² = area

[summer spire]

do you have the radius?

[upper karma]

25.12/pi

Answer should be area around 200.8581218

Right?

200.86 then?

units ²

[summer spire]

roughly, depending on how you round

[upper karma]

how should I round it,

?

oh it's correct

awesome thanks for veryfying

[weak shoal]

Draw a diagram lmao

[wind heart]

Did I do these right

[drowsy timber]

doing math in pen

i think for your #5 its supposed to be angle A approximately equals C @wind heart

thats the transitive property where if a = b and b = c then a = c

[wind heart]

Oh shittttt I didn’t catch that

If you know how to do this, can you help me with it? I’m not having a problem with the triangle ones for some reason but these polygon ones are weird

[drowsy timber]

which one?

[wind heart]

6 and 7

(It connects with the polygons)

[drowsy timber]

angle A = 100 not 80

you can tell its greater than 90 degrees by looking at it

so 2x + 4 = 100, x = 96/2

and its the same for 7, 3y - 3 = 12 and solve for y

[wind heart]

Hm?

[drowsy timber]

Angle A aprox equals Angle P and side DC aprox equals RS

[wind heart]

Oh

I see

So we have to go by what it’s congruent to from the angle letters

[drowsy timber]

yeah

[wind heart]

Thanks for the help. I appreciate it

[drowsy timber]

np

[wind heart]

I’m making sure I understand the rest of the math going on since I got my first C on a Geometry test

Anyways

Cya

[drowsy timber]

gl

[dusty badge]

can you add a point to a vector?

[umbral snow]

Arguably all points are really vectors

Depends what you're doing lol

[neat crane]

yo whos good at triangle proofs

i need mad help

damn ok

[drowsy timber]

@neat crane post it

[spice ridge]

point slope form lines are y-y1=m(x-x1)+b

right?

nvm

without B

[signal hornet]

hi

shouldn't the line be going to opposite way?

because it's -2 not 2