geometry-and-trigonometry

[vast dune]

the text translated: Using the drawing we are going to calculate |PQ|^2 with 2 methods

[signal hornet]

I need help figuring out the next step

[upper karma]

anyone know of an arctangent approximation?

or a website with a table of values 0-90

[vast dune]

@signal hornet

2nd step

[signal hornet]

ahh

an algebra error

[signal hornet]

what now?

[upper karma]

@signal hornet step two is wrong

[signal hornet]

how so?

[upper karma]

should be: sin^2 - 2cos * sin + cos^2

[gritty sail]

^

then you know that sin(2theta)=2costhetasintheta

[signal hornet]

I did not know that @gritty sail we were not given that identity

Is there any to get the answer from here?

[gritty sail]

welp

[signal hornet]

I lied

[gritty sail]

use what i said

[signal hornet]

I did know that identiy

[gritty sail]

nice

[signal hornet]

haven't used it until now though

but I did write it down

so this is what I got, but my calc says that is not the answer, it's telling me 7pi/4

[flint osprey]

it's good until the very end

where you write 2theta = 3pi/2

it should read 2*theta = 3*pi/2 + 2*pi*k

the choice k = 0 you implicitly made corresponds to sin(theta) - cos(theta) = + sqrt(2)

[signal hornet]

this is just for 0 to 2pi

[upper karma]

sin²(x) = 2 - cos(x)

I tried replacing sin²(x) with 1 - cos²(x) but it leads to a quadratic which can't be solved

It led to this: cos²(x) - cos(x) + 1

btw for 0 to 2 pi

[flint osprey]

@signal hornet even if you only care about theta between 0 and 2*pi then you still need to consider k = 1 in the formula I wrote

[signal hornet]

so both 3pi/4 and 7pi/4

[upper karma]

so...

[signal hornet]

@flint osprey is my calc wrong?

[flint osprey]

nope everything is correct

you have shown that if theta solves your equation, then it must be 3pi/4 or 7pi/4

and now in the screenshot you posted, you have also check that 7pi/4 solves your equation and 3pi/4 does not

so the solutions are 7pi/4 and no others (between 0 and 2pi)

[tropic tide]

Wait

I realised

What is the point of trigonometry here if you have trigonometry topic in precalc

Or is this like the trigonometry before precalc

Or something

[upper karma]

not sure but there's a problem up there which needs to be solved

[tropic tide]

¯_(ツ)_/¯

[gritty sail]

@upper karma that means no real solutiosn

[upper karma]

yeah

but it think im on to something brb

ok nvm

but are youi sure there is no other way

[gritty sail]

if u think about it grpahically

-cosx+2

will go from [1,3]

while sin is [-1,1]

so the only possible is at 1

but if u graph it u will see it wont ever intersect at 1

oh wiat it is squared

mb

uh sin^2x is from sin x

since sinx is from [-1,1], sin^2x is from [0,1]

but yeah no solution

[upper karma]

k

[gritty sail]

use sin(45-30)

idk hwat ur question is asking

[split otter]

sin 15° = sin (45° - 30°) = sin 45° cos 30° - cos 45° sin 30°, which are special angles whose sines and cosines have values that can be determined by a right isosceles triangle and a 30-60-90 triangle.

[lusty sluice]

how do you prove

Sin^2x + cos^4x = cos^2 + sin^4x

[tawdry pivot]

$\sin^2(x)-\cos^2(x)=(\sin^2(x)-\cos^2(x))(\sin^2(x)+\cos^2(x))$

[somber coyote]

EpicGuy4227:

[split otter]

Isn't that sin⁴ x - cos⁴ x?

[tawdry pivot]

yes

@lusty sluice

[weary drift]

splitting p & q into components helps

use the magnitudes and bearings to find them

[shut mountain]

I'm trying to solve tan x = 1 here.

I got this, and

and this as the general form

Which isn't correct. Where did I go wrong? :D

[dark sparrow]

(1/4)2π? where did that come from

[shut mountain]

tan = 1 every 1/4th of a whole round 2π

[dark sparrow]

uh... no.

tan(π/4 + π/2) is -1, not 1.

[shut mountain]

ah i see thanks

so simple

yet couldnt see it

[frank spear]

can someone prove to me that these two functions are the same

i know they are because i plotted it but

a proof would be nice

[vast dune]

why is it at x/2 only +kpi and not k2pi?

looking at the unit circle I cant make sense of it

[twin prawn]

Omdat tan periodiek is met periode pi, niet 2pi

[normal harness]

Can someone tell me if this stereographic projection formula from 4D down to 3D is correct?

Assuming x, y, z, w are a 4D coordinate input and L_x, L_y, L_z, and L_w are the respective coordinates of the light source

Most stereographic projections involve a light source with a fixed point along the new axis, such as the Z or W axis, but I want to create a stereographic projection that allows a free range light source at any real coordinate pair

[open sinew]

using the double angle identity for tan: tan(2x) = (2 tan x)/(1 - tan^2 x)
let 2^-(k+1) = a <=> 2^-k = 2a, then we have:
(2 + tan 2ax tan ax)/(tan 2ax) = cot ax
= (2/tan 2ax) + tan ax = cot ax
= 2(1 - tan^2 ax)/(2 tan ax) + tan ax = cot ax (using the above identity here)
= cot ax - tan ax + tan ax = cot ax
@atharv.#2987

wut hes on the server but his name isnt highlighted

is my discord glitching

also proof of double angle identity for tan using the ones for cos and sin:
cos(a+b) = cos a cos b - sin a sin b
sin(a+b) = sin a cos b + cos a sin b
tan(2x) = sin(x+x)/cos(x+x) = (2 sin x cos x)/(cos^2 x - sin^2 x)
= (2 sin x / cos x)/((cos^2 x - sin^2 x)/cos^2 x)
= (2 tan x)/(1 - tan^2 x)

[open sinew]

@frank spear

huh worked that time

[frank spear]

Cheers

[shut mountain]

Are the upper and lower row the same? Just checking if I can even read these things :D

[shut mountain]

How about this? Is it correct?

[twin prawn]

In most common notation conventions, yes

[shut mountain]

ok thanks

[upper karma]

I know P and I know a
I need to know b

Can someone help me refactor this? I have no idea where to begin

plug in

kek

I need a B = ... formula

but Idk how to move B out

[shut mountain]

so what's a

[silk crown]

Could I get some help with 39.?

[dark sparrow]

is/are

[silk crown]

yeah it's supposed to have multiple answers from ABCD

[dark sparrow]

this doesn't sound like a multiple-applicable-answers question

[silk crown]

So what I think I need to do is find the number of points of intersection of the hyperbola and the director circle of the circle

[dark sparrow]

the

what

[silk crown]

the circle from which perpendicular tangents can be drawn to the circle

as in, if I pick a point on the director circle and draw the two tangents to the circle x²+y²=a², the two tangents will be at an angle 90° to each other

[dark sparrow]

the circle from which perpendicular tangents can be drawn to the circle
,,, so the circle with a radius sqrt(2) times as big?

yeah

sounds correct

[silk crown]

yeahh

so the director circle here would be x²+y²=2a²

Now how do I find the points of intersection of this and the hyperbola

[dark sparrow]

$\begin{cases} \frac{x^2}{a^2} - \frac{y^2}{b^2} = 3 \ \frac{x^2}{a^2} + \frac{y^2}{a^2} = 2 \end{cases}$

[somber coyote]

Ann:

[twin prawn]

The circle and the hyperbola can never intersect

[silk crown]

$y^2=-\frac{a^2 b^2}{a^2 + b^2}$? So would it be 2 points?

[somber coyote]

CoolShot:

[dark sparrow]

no it'd be y^2 = (something negative)

[silk crown]

oh

so no real points

got it thanks

Could I also get some help with this one?(Q1.)

So I got the equation of the normal, then I substituted (9,0) into the line, which gave me a²/b²=18, so I plugged that into the formula for the eccentricity but that doesn't give me the right answer

It gives me √(17/18)

[robust socket]

,, tan(x) = -\sqrt{3}

[somber coyote]

Umma.Gumma:

[robust socket]

how do I know which of the two functions(i.e. sin or cos) is negative?

In this case for instance both 2nd and 4th quadrant angles would be adequate

[jaunty sail]

Cos runs along the x axis and sin along the y, so just think about x and y would be positive and negative on a coordinate plane @robust socket

My explanation might not be completely accurate but it's how i understand it

[robust socket]

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

both numerator and denominator can be negative here

[jaunty sail]

That would be 1st and 3rd quadrant

its tan(x)=sqrt-3

so either the numerator or denominator need to be negative

[trail minnow]

\tan

pls

[robust socket]

corrected

[jaunty sail]

ya so 2nd or 4th would work like you said

[somber coyote]

Umma.Gumma:

[trail minnow]

also surround with $ for latex

lol someone is getting help rn

you need to wait a bit

[robust socket]

so

either num or den would be negative, that is clear

which one though?

[trail minnow]

given something like tan(x) = sqrt(3)

[jaunty sail]

there are two angles that you could answer with

[robust socket]

yes, thing is, I always get one solution only in the books I'm checking

[trail minnow]

there are many more than 2 angles

[robust socket]

which makes me wonder whether either of the answer is correct

[trail minnow]

there are an infinite number of tem

*them

[robust socket]

yes all angles +2k\pi

but I'm interested in the 'first' one

[jaunty sail]

its in the 2nd quadrant

[robust socket]

how can you tell that?

2nd and not 4th

[trail minnow]

i assume for tan you mean between -pi/2 and pi/2

in which case, tan(x) = sqrt(3) implies that 0 < x < pi/2

so just think about the values of sin and cos between 0 and pi/2

[robust socket]

there are no assumptions

[trail minnow]

what

you asked for the principal angle

for tan

[robust socket]

no, I stated that the equation I wrote has 2 legitimate solutions, which are different depending whether which of sin or cos are negative

[trail minnow]

which equation exactly?

tan(x) = sqrt(3)?

[robust socket]

no, tan(x) = -sqrt(3)

[trail minnow]

ok

that has an infinite number of solutions

[robust socket]

yes, that is clear

[trail minnow]

but it only has 1 solution if you ask for the principal anlge

[robust socket]

,, \alpha + k\pi

[trail minnow]

*angle

[somber coyote]

Umma.Gumma:

[trail minnow]

you understand that the range of the principal angle for tan is -pi/2 < x < pi/2 right

cause theres only 1 solution in that interval

[robust socket]

yes

just one doubt

[trail minnow]

so im a bit confused as to what you mean when you say there are "two solutions"

because there are not 2 solutions over that interval

[robust socket]

yes, about this

,, \pi-\frac{\pi}{3}+k\pi

[somber coyote]

Umma.Gumma:

[robust socket]

this is the first one I thought about

[trail minnow]

can you explain to me what exactly you are trying to do here

solve tan(x) = -sqrt(3)?

[robust socket]

please give me 1 min I'll write on paper and take a picture

nevermind, solved

thanks a lot anyway

[trail minnow]

lol

[grizzled crater]

is this a place i can ask for help?

[upper karma]

ye

[grizzled crater]

Erm

trig identities

can i post the picture in here?

[upper karma]

ye

[grizzled crater]

iT'S Just that last one

thanks in advance

[trail minnow]

pls

pls just take a screenshot

[umbral snow]

Usual first step, convert tan to sin/cos

[grizzled crater]

NO sorry it is just that last one thanks

[umbral snow]

Oh mb

That's not an identity lol. Have you distributed?

[grizzled crater]

Yeah.

I always end up buggering myself up

[umbral snow]

Actually, try squaring everything. That's my suggestion

[grizzled crater]

Tbh this is only a gcse question

I am not exactly sure it needs much depth

I am probably skipping over something really obvious

I have tried the hint

but i do not see where that gets me

[umbral snow]

5sin²(x) + 5sin(x)cos(x) = 3sin²(x) + 3cos²(x)
2sin²(x) + 5sin(x)cos(x) - 3cos²(x) = 0

[grizzled crater]

ahhh

[umbral snow]

You could make it all sin² or cos²

Then factoring is your method after that

Hmm

[grizzled crater]

thanks so much

I cant believe i didnt notice that

🙂

[umbral snow]

It's still not done oop. And I'm not sure how to best finish it

[signal hornet]

is this room free now?

$\sin(\theta) = \frac{3}{5}$

[somber coyote]

Jason_Bjorn:

[dire rampart]

@grizzled crater turn it into a quadratic in terms of cos

Actually nvm

[signal hornet]

find $\sin(2\theta)$

[somber coyote]

Jason_Bjorn:

[signal hornet]

I got 12/5

but I have no idea if I am right or wrong

so I though i'd check in my calc

I typed in sin(2*arcssin(3/5))

[dire rampart]

Uh

[hard gale]

12/5 wat

[dire rampart]

Are u given any info about theta

[signal hornet]

theta is between 0 and pi/2

just 12/5

[dire rampart]

That's more than 1 tho

[signal hornet]

I know

[dire rampart]

What did u do

[signal hornet]

one sec

I got the right answer

nvm

I messed up some basic arithematic

[grizzled crater]

ive done it now

solutions are 26.565 degrees and 206.565 degrees

[signal hornet]

$cos(\alpha) = - \frac{\sqrt(15)}{4}$

[somber coyote]

Jason_Bjorn:

[signal hornet]

I know alpha is in quadrant 3

I need to find tan(alpha/2)

how do I know which quadrany tan(alpha/2)` is in?

I am not sure how to find out if it;s gonna be plus or minus

<@&286206848099549185>

[river forge]

draw a triangle

[signal hornet]

I did

this is just for alpha

I know tangent is positive in quadrant 3

no idea where a/2 is though

[river forge]

oh this problem is slightly harder than i first read

use half angle formula

[signal hornet]

I did that

just not sure if the answer is positive or negative

that is my question

[river forge]

oh

well let's think of bounds of half the angle of the third quadrant anges

angles

namely the largest is 3pi/2 and the smallest is pi

half of 3pi/2 is 3pi/4, and half of pi is pi/2

so your angle must lie in between these two angles

[signal hornet]

ok, I follow that

so negative?

because (pi/2, 3pi/4) is quadrant 2

[signal hornet]

I am stuck

what is the next step. I have gone as far as I can on both sides

[median crown]

what are you trying to do?

[signal hornet]

prove the identity

[regal crescent]

of what?

[signal hornet]

the two statements on the top

[median crown]

just do the sum of angles for tangent

[signal hornet]

can you explain?

you mean like 2theta plus theta?

[median crown]

yes

and then use double angle for tangent

inside of that sum of angles

that'll probably make it easier

[signal hornet]

interesting approach

I will try that

[signal hornet]

I don't see how this is easier

[median crown]

do what it says

add fractions

lol wdym you don't see how this is easier

you aren't even done simplifying

nobody goes halfway through a problem and then says, oh well, i'm not even done, but ill just not finish it

@signal hornet you got it?

[signal hornet]

yep

thanks

[median crown]

[signal hornet]

didn't even have to toch the right side

good intuition on your part

[robust socket]

can someone please check whether this is correct?

[dark sparrow]

yes

[robust socket]

thanks :)

[coral mural]

https://www.youtube.com/playlist?list=PLRUjgfP7UXwErm9DDNXA0UDxAlSl_LUfS

[main totem]

If BD=AD and DE || AC. Is this enough to say DE is the midline of the triangle?

[coral mural]

Yes mid point theorem

i believe

[main totem]

Ok thanks

[coral mural]

this occurs when de is parallel to ac

only subject to this condition

[robust socket]

,, \sin(x)\cos(x)=0

woudl appreciate some hints to solve this one

[somber coyote]

Umma.Gumma:

[robust socket]

I tried with the fundamental relation of trig but no success so far

[tardy junco]

Sin2x = 2sinxcosx

@robust socket

[robust socket]

yeah but it's not 2sinxcosx

[idle bloom]

Yea but it's really close and in fact you could get it to look like 2sin(x)cos(x)

[robust socket]

no idea

[dark sparrow]

hint: $1 = \frac12 \cdot 2$

[somber coyote]

Ann:

[idle bloom]

$ x = 0 $

[somber coyote]

Spamakin🎷:

[idle bloom]

$ 2x = 0 $

[somber coyote]

Spamakin🎷:

[robust socket]

,, \frac{2\sin(x)\cos(x)}{2}=0

[somber coyote]

Umma.Gumma:

[robust socket]

,, \sin(2x)=0

[somber coyote]

Umma.Gumma:

[robust socket]

,, x=\frac{k\pi}{2}

[somber coyote]

Umma.Gumma:

[robust socket]

nailed it, thanks a lot

[dark sparrow]

i mean tbh

you could've done without that identity too

you could've solved sin(x) = 0 and cos(x) = 0, and taken the union of their solution sets

[robust socket]

well yes, definitely

[robust socket]

I've been struggling a bit with a few of these

I always get close to the end, but at that point I'm not sure how to proceed

must be trivial but I'm not realising how to get it done

[split otter]

How about factoring out the sin x instead of applying a double-angle formula?

In the fourth line there.

[robust socket]

let me try

[split otter]

Actually, now that I think about it, you could probably straight away factored out tan x in the first step.

[robust socket]

solved thanks, however

how would you solved it from the last line I wrote?

factoring out sinx seems more reasonable but I'm curious at this point

[split otter]

I don't think you can? I don't think I know how to, anyways.

Because it's quite hard to solve a trigonometric equation with 2 different angles and no way to factor them individually.

There's no real easy way to relate sin 2x to sin x, unless we write it as 2 sin x cos x and factor out sin x.

[robust socket]

ok, got it, thank you!

[median crown]

you're welcome

[tardy junco]

I'm not getting the answer for this..

[dark sparrow]

this.... is false

[tardy junco]

Wait a minute wrong question

[dark sparrow]

doesn't even work for x=0

[tardy junco]

Nvm got it

I was trying to prove it for any angle so I wasn't getting it

[robust socket]

,, \cos^2(x)+\frac{2}{\tan^2(x)}=\frac{5}{2}

[somber coyote]

Umma.Gumma:

[robust socket]

I've tried to solve this not less than a dozen times..

the constant term is a bit annoying as it does not help in factoring out either sin or cos

but maybe there's something I'm missing

[dark sparrow]

why not substitute u := cos^2(x)

then you'll have $u + \frac{2u}{1-u} = \frac52$

[somber coyote]

Ann:

[robust socket]

let me check

[robust socket]

I think I solved it but I've got a doubt

I end up with

,, \cos^2(x)=\frac{1}{2}

[somber coyote]

Umma.Gumma:

[robust socket]

which is

,, \cos(x)=+/-\frac{\sqrt{2}}{2}

[somber coyote]

Umma.Gumma:

[robust socket]

...which is $ x= \pm \frac{\pi}{4}+k\pi $

question is: why k\pi instead of 2k\pi?

[dark sparrow]

$\pm$

[somber coyote]

Ann:

[dark sparrow]

and it's because there are actually four solutions in [0, 2π]

π/4, 3π/4, 5π/4 and 7π/4

[robust socket]

ok makes a lot of sense

thanks a lot

[elder violet]

hello?

[elder violet]

oh ok

how to use them?

[silk crown]

How to use what

[elder violet]

ah man im so lost

[silk crown]

Me too, dont know what your doubt is lmao

Are you asking how trigonometric ratios are useful?

[elder violet]

no

my exam is tomorow and im freaking out

[silk crown]

ok but I can't help unless I know what the problem is lol

"how to use them" is basically asking "teach me trig"

[elder violet]

yeah sorry bout that

i should've a question somewhere

oh wait

my teacher gathered my notebook

sorry to take your time, i think i'll just gonna fail badly this time

[silk crown]

nah bro if you have a specific question just ask we'll help

@upper karma virtually every teacher I've seen has taught that tho...

[split otter]

Yeah. If you have an example problem, go ahead and post it here and we'll try to help.

[upper karma]

cuz your teacher is a good one

[split otter]

Yeah. My teacher taught us that trick too.

[silk crown]

Tbh you also should've been able to see this obvious pattern lol

[split otter]

8th grade me was extremely dumb and would not have picked it up.

Or was it 9th? I don't remember.

[upper karma]

it’s just something i found on IG

[elder violet]

i was actually good at math till SIN, COS,etc came along

[silk crown]

I mean

What grade are you in

[elder violet]

10th

in Indonesia

[silk crown]

Just practice and improve your trig, I guess

[split otter]

Hello, Asian friend! You'll get used to trigonometry eventually. Takes some getting used to.

[elder violet]

well i basically stuck now, since i don't have my notebook with me. i dunno wat to do.

i could practice when the exam started lmao

thats basically illegal

[split otter]

Khan Academy is pretty good, if you need some video explanations to refresh your memory.

[elder violet]

im gonna check it out, thx for refrence

[silk crown]

Khan is good for learning but if you want to practice you should try looking for some resources

[elder violet]

like questuon practices?

[silk crown]

Yeah

[split otter]

Oh yeah. Khan Academy's practice exercises are quite basic.

[elder violet]

cool

[silk crown]

Just try looking for practice worksheets/textbooks for the curriculum you're studying in on trigonometry

[elder violet]

thx again lads

[silk crown]

np mate

[split otter]

All the best for the test!

[chrome quarry]

is there anyway to solve a matrix like this with the TI-84?

assuming all the co-ordinates are given

well i meant a determinant*

[silk crown]

Is there a way to ge the equation of the cycloid?

(in blue)

[summer spire]

@chrome quarry have you consulted the manual?

[chrome quarry]

@summer spire i haven't found any way to insert the x, y and z

to obtain an equation looking like ux + vy + wz + t = 0

[summer spire]

,rotate

[chrome quarry]

@summer spire ended up writing a program manually

[silk crown]

<@&286206848099549185>

Is there a way to ge the equation of the cycloid?

[umbral snow]

No. There's no way to get it

http://jwilson.coe.uga.edu/EMAT6680Fa07/Gilbert/Assignment 10/Gayle&Greg-10.htm

[silk crown]

thnkxs

[verbal forum]

can anyone help me find a and tell me the method?

[coral mural]

It is easy

[silk crown]

Try using the following properties: the opposite corners of a cyclic quadrilateral are supplementary, and if a triangle has 2 sides equal and one side 60°, its an equilateral one @verbal forum

[coral mural]

Check out this : https://www.youtube.com/playlist?list=PLRUjgfP7UXwErm9DDNXA0UDxAlSl_LUfS

[silk crown]

No

@coral mural nope

[coral mural]

Your choice. I am not compelling 😂

[silk crown]

aight

That video does not relate to the problem in a direct way

[coral mural]

I know that.

[silk crown]

...

[coral mural]

but I am planning to make on these topics.

[silk crown]

This problem can be solved without any complex trig

[coral mural]

I know

it was damn easy

[silk crown]

If its your channel, stop advertising it when it is irrelevant to the discussion mate

[coral mural]

sure

[silk crown]

cool

[grizzled crater]

Sorry to bother again, but can anyone clarify when I can and when I can't cancel out functions when proving trig identities?

[idle bloom]

Can you send examples of what you mean because that's very vague

[upper karma]

does algebra 2 count here?

need help with a few problems

[dark sparrow]

#prealg-and-algebra

[upper karma]

ah ok

[dire obsidian]

For #39, i dont get how cos is -root2/2

[dark sparrow]

what do you think cos(135°) should be instead

[dire obsidian]

I got root2/2

[dark sparrow]

what quadrant is 135° in?

[dire obsidian]

2nd

[dark sparrow]

and what is the sign of cos in the second quadrant?

[dire obsidian]

o.

but

[dark sparrow]

well then.

[dire obsidian]

like

whats weird is

my teacher makes us draw triangles

so for this

I believe its a 45-45-90 triangle

and we use the quadrantal point things

[dark sparrow]

the "reference angle", as it were, is 45°

[dire obsidian]

like 0,1

yes

the reference angle is 45

[dark sparrow]

but 135° itself is in quadrant 2

so its cos is negative

gotta account for that

[dire obsidian]

oh

I see

ok thank you

opened my eyes to a better way of loking at it

looking*

[split otter]

Yes. The value of a trigonometric function evaluated at an angle is the same as that evaluated at the reference angle, but the sign may differ. This is based on whichever quadrant the angle is.

[wooden compass]

I am in Integrated Math 2 which seems to be mostly geometry

Can someone link me a book, not a workbook, but a book that teaches me the concepts that I am learning

[median crown]

one second

let me connect my brain to your teacher's computer

to see what you are learnng

[upper karma]

Which step is the same when constructing an inscribed square and an inscribed regular hexagon?

Construct a circle first.
Construct a line first.
Set the compass to the radius.
Set the compass to greater than half the diameter.
first option am i right?

[wooden compass]

@median crown it seems to be high school geo so I would like a book or something on it

[median crown]

just google it

high schoo geo math books

pdf

or something

i'm not really a math book conoisseur

[upper karma]

dont mind helping me out with this one? im not quite sure if this is the right channel to post in

[dark sparrow]

@upper karma yes this is the right channel

how did you get the answer you wrote up there?

[upper karma]

@dark sparrow 25*cos(52)

[dark sparrow]

and where did that come from?

no

i'm not asking where the problem came from

i'm asking where you got it from that F = D*cos(52°)

[upper karma]

bruh

am not sure, i just used the formula to find missing length for trig

[dark sparrow]

it should have been a red flag to you that you ended up with the hypotenuse (F) somehow being SHORTER than one of the legs (D)

am not sure, i just used the formula to find missing length for trig
what formula did you use?

[upper karma]

i honestly dont know where i got the formula from all they gave me was this diagram

i just used what came into my mind

[dark sparrow]

ok great cos(θ) = adj/hyp

what is the hypotenuse here

[upper karma]

40.61 was the correct ans for the hyp

[dark sparrow]

no

not what i'm asking you

what is the hypotenuse

[upper karma]

the longest point

[dark sparrow]

which one of D, E and F on the picture

"longest point"? points don't have lengths

[upper karma]

F

[dark sparrow]

alright, great, F is the hypotenuse.
and what's the leg adjacent to θ?

[upper karma]

E?

[dark sparrow]

do you know what ADJACENT means

[upper karma]

D

[dark sparrow]

ugh.

fine.

[upper karma]

bruh

[dark sparrow]

the adjacent side here is D.

cos(θ) = D/F.

solve for F.

[upper karma]

thx

[vestal oak]

So passive aggressive, Jesus

[green roost]

i know, like come on it might be simple to you now ann, because yeah its basics but explain it, i always got confused with what the adjacent was when i started trig

[past ibex]

Need some help with this one

https://gyazo.com/3e22cfdbcb165d16e7507fba4bbd6401

ive got it to cosx/1+sinx = 1-sinx/cosx

but not sure where to go from here

[silent plank]

parentheses

(1 - sin(x))/cos(x)

you're just 1 step away

do you know what sec and tan are?

[past ibex]

yes

for the most part

1/cos and sin/cos

[silent plank]

yes, and currently what do you have?

[past ibex]

1-sinx/cos

[silent plank]

parentehses

[past ibex]

(1-sin(x))/(cos(x))

[silent plank]

$\frac{1 - \sin(x)}{\cos(x)} = \frac{1}{\cos(x)} - \frac{\sin(x)}{\cos(x)}$

[somber coyote]

ramonov:

[past ibex]

right

thats what i had before and then simplified

[silent plank]

this is supposed to be an identity proof right?

[past ibex]

correct

[silent plank]

OH, you started from the rhs

[past ibex]

rhs?

[silent plank]

right

[past ibex]

oh yeah

i did

[silent plank]

its easier to start from the left for this

[past ibex]

(cos(x))/1 + (cos(x)/sin(x))

is that in the right direction

[silent plank]

although you can use the similar approach of using conjugates

[past ibex]

Not sure how i would do that

[silent plank]

no, fractions don't work like that

the idea is to multiply the numerator and denominator by:
the conjugate of 1 + sin(x) and then apply the identity
sin^2(x) + cos^2(x) = 1

alternatively, if you are starting with the right, multiply by the conjugate of 1 - sin(x)

[past ibex]

so (cos(x)) * (1-sin(x))?

[silent plank]

that would only need be the numerator. you would need to multiply the denominator by that too

[past ibex]

right

((cos(x))*(1-sin(x)))/(1-sin(x)^2)

[silent plank]

try not to write the ^2 like that in plain text

[past ibex]

What would i do from here?

[silent plank]

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

[past ibex]

where do i have that?

[silent plank]

1-sin^2(x) = ?

[past ibex]

oh

that = cos^2(x)

and then i can drop the cosx from numerator and leave 1-sinx?

/cosx

[silent plank]

yes

[past ibex]

tysm

Not sure what it means by trig identity family used though

gotta list it beside the proofs

[ruby lotus]

hello?

[gritty sail]

how does this make sense
i addded up the areas
and got
(1+sqrt(2)+sqrt(6))/2
which is not (3+sqrt(3))/2
and i also did it with shoelace and got (1+sqrt(2)+sqrt(6))/2
can someone else confirm

nvm im pretty sure its wrong

[vast dune]

How did sin(A+B) turn into 2sin(..)cos(..) here?

[upper karma]

@vast dune I can't even read that, so I'm having to guess

double angle identitity?

$\sin(2x) = 2\sin(x) \cos(x)$

[somber coyote]

Sun:

[vast dune]

but its $\sin(A+B)$

[somber coyote]

Abdos:

[vast dune]

doesnt make sense to me

[upper karma]

$A + B = C, \sin(A+B) = \sin(C)$?

[somber coyote]

Sun:

[vast dune]

uh

$\sin(A+B) = 2sin(\frac{A+B}{2})cos(\frac{A-B}{2})$

[somber coyote]

Abdos:

[upper karma]

I don't believe that's true

[vast dune]

actually its $\sin(A+B) = 2sin(\frac{A+B}{2})cos(\frac{A+B}{2})$

[somber coyote]

Abdos:

[upper karma]

That does seem correct, consider D = (1/2)C

sin(2D) = ?

[vast dune]

sin(2(1/2)C)?

[upper karma]

sin(2D) = sin(C) = sin(A+B)

in the form of sin(2D) you can immidiately apply the identity

$\sin(2x) = 2\sin(x) \cos(x)$

[somber coyote]

Sun:

[vast dune]

ok but

A and B are 2 diffrent angles

[upper karma]

they are added before the sin is applied to them

there's no distinction between sin(90) and sin(45+45)

[vast dune]

oh I see it now

[upper karma]

weird question

given two sides can you find an angle? terrible at geometry

like i dont know how to find this

like would you ahve to assume it's a right triangle?

[silent plank]

unless you make assumption(s), you can't find theta

[spice ridge]

how do i do this

what does x_1 mean?

[weary drift]

there are supposedly two values of x that satisfy the eqn, x_1 and x_2. you're given x_1, so find k and x_2

[spice ridge]

ohhhh

[ivory vortex]

$sin^6x+cos^6x+3cos^2x-3cos^4x=1$

how can you prove this trig identitiy

alternatively, if someone has wolfram pro and could enter this in,,..,

[somber coyote]

Voyager:

[silent plank]

sum of 2 cubes
then reduce everything down to ^2

@ivory vortex

[ivory vortex]

how did you come to that thought

[silent plank]

when you have high powers you want to reduce them so that you can apply stuff like
sin^2(x) + cos^2(x) = 1

[ivory vortex]

oo

thanks

[ivory vortex]

$\frac{tan3x}{tanx}=\frac{3-tan^2x}{1-3tan^2x}$

[somber coyote]

Voyager:

[ivory vortex]

what is the opening big brain move to prove this

im thinking making tan3x to tan(x+2x) but idk if thats legal

[somber coyote]

RokettoJanpu:

[dire obsidian]

can someone explain how to do these

ive done them and sometimes i get the right answers and sometimes i dont

[slender nacelle]

One way to do it is, draw a unit circle, identify regions where sin, cos, tan are positive and negative. And deal with the problem according to what you want.

And it's also helpful to remember that sin is on y coordinate and cos is on x coordinate

[dire obsidian]

@slender nacelle teacher hasnt gone over that sin is on y axis and cos is on x. so could u explain how that helps

[slender nacelle]

Do you know where sin is positive and where cos is?

[dire obsidian]

yes

its tht astc thing

we use

[slender nacelle]

Do you know why?

[dire obsidian]

all students take calculus

no

i just know its astc lol

and which are positive and negative in each quadrant

[slender nacelle]

It's because sin corresponds to y axis. So in any wuadrant, y coordinate is positive, sin is positive

Which is in 1st and 2nd quadrant

That's why sin is positive there

[dire obsidian]

oo

[slender nacelle]

Look at this image. Do you understand why sin is y coordinate and cos is x

[dire obsidian]

Yes

[slender nacelle]

So it's just a more fundamental way to understand signs of trig ratios than astc

[dire obsidian]

Sin is y, y is positive

[slender nacelle]

Yes

[dire obsidian]

Yeah

So sin is y/r

Cos is x/r

[slender nacelle]

Yup, r= 1 for unit circle

[dire obsidian]

Yeah

[slender nacelle]

So any of these questions, you can relate it with unit circle

[dire obsidian]

so i was told to use this

[slender nacelle]

Yes, it is the same thing

[dire obsidian]

and to know that 1,0 = 0;360 degrees

0,1 is 90 degrees

-1,0 is 180

[slender nacelle]

Yes

[dire obsidian]

0,-1 is 270

ok

i just dont understna dhow to identify the angle(s) between 0 and 360

i refer back to the ratio

but for some of them

it confuses the crap out of me

[slender nacelle]

Let's try an example

Which one do you want to go about?

[dire obsidian]

#57

[slender nacelle]

Sure

[dire obsidian]

tan is y/x

[slender nacelle]

Yes, so where tan theta= 1, y=x

[dire obsidian]

er

[slender nacelle]

One such angle is
45 degrees

[dire obsidian]

but how

i dont understand that lol

i cant visualize it

[slender nacelle]

Do you know your trigonometric tables?

[dire obsidian]

uh

like the

[slender nacelle]

What is sin (45) and cos(45)?

[dire obsidian]

like this?

[slender nacelle]

Yes these

[dire obsidian]

cause thats the only one ive done

we only did them with quadrantal angles

[slender nacelle]

But for 0,30,45,60,90

[dire obsidian]

nop

wouldnt those be

special triangles

[slender nacelle]

You should remember them in order to do these questions

[dire obsidian]

so sin45 would be like 1/root2 or root2/2

[slender nacelle]

Yes

So sin and cos both are equal at 45

[dire obsidian]

yeah

[slender nacelle]

Because it's an isosceles triangle

So tan theta =1

[dire obsidian]

is at 45

[slender nacelle]

Yes

[dire obsidian]

o i see

[slender nacelle]

But that's only between 0 to 90

[dire obsidian]

cause y/x in the 45-45-90 triangle

is

1

[slender nacelle]

Yes

[dire obsidian]

so

45 is one of the angles

[slender nacelle]

Yup

[dire obsidian]

so then is it 180-45

and u get-

[slender nacelle]

Is it?

[dire obsidian]

or

180+45

180+45

[slender nacelle]

Yes

[dire obsidian]

to get 225?

[slender nacelle]

180+45

[dire obsidian]

ok i kinda get it

[slender nacelle]

Why not 180-45

Or 360-45?

[dire obsidian]

tan is positive in quad 3 and 1

[slender nacelle]

Great

[dire obsidian]

quad 3 and 1 are the 180 thing

[slender nacelle]

Yes

[dire obsidian]

whereas quad 4 is 360

so if it were cos and positive, it would be +360 right?

meh?

not really

wait

-360?

cause if u add, theres a chance it goes over 360

which is not what u want

[slender nacelle]

If you subtract, it'll be negative

[dire obsidian]

o

[slender nacelle]

So it's not -360. It's 360-

[dire obsidian]

oh yeah

got it

uh can we try

59

[slender nacelle]

Sure

[dire obsidian]

oops

sin is negative in quad 3 and 4

[slender nacelle]

Yes

[dire obsidian]

is this a 30-60-90?

not 45-45-90

[slender nacelle]

Forget the triangles

[dire obsidian]

lol aight

so sine is

y/r

y has to be -1

[slender nacelle]

Yup

[dire obsidian]

y is -1 at (0,-1)

[slender nacelle]

Correct

[dire obsidian]

which is 270

[slender nacelle]

Yes!

[dire obsidian]

ok

[slender nacelle]

That's your answer!

[dire obsidian]

lol

uh one more?

[slender nacelle]

Alright

[median crown]

270?

270 radians is a lot dude

[dire obsidian]

i can check my key

[slender nacelle]

It is correct, he's just messing with you!

[dire obsidian]

o

oof

[median crown]

lol

[slender nacelle]

You should write degrees

[dire obsidian]

o can we do 54

[slender nacelle]

Sure

[dire obsidian]

o yeah i do write degrees, just lazy to type it out on my laptop

[silent plank]

alt+0176

[slender nacelle]

That's alright

[dire obsidian]

lol i have a mac

idk if thats working

yeah no

i dont have the numbers on the right

º

option 0

[slender nacelle]

It's okay, just write degrees 😂

[dire obsidian]

LOL aight

so how would you approach a problem if its = to 0?

[slender nacelle]

What is cos?

[dire obsidian]

does it just sign-

x/r

so x is 0

[slender nacelle]

Yes

[dire obsidian]

cos i-

would u consider 0 positive

[slender nacelle]

Sp which point(s)?

[dire obsidian]

ik thats a dumb

question

but

LOL

(0,-1) and (0,1)

[slender nacelle]

0 isn't positive or negative (ie point will be on boundries of quadrant, not inside a quadrant)

Yes, so which angles?

[dire obsidian]

thats 180 and 90?

or 0?

[slender nacelle]

Is it 180?

[dire obsidian]

180 and 0

wait

holda

LOL

[slender nacelle]

(0,1) and (0,-1)

[dire obsidian]

90 and 270

yeah mb

[slender nacelle]

Yup

[dire obsidian]

i was staring at the other points

[slender nacelle]

😅

[dire obsidian]

ok i get it kinda

just gotta practice some more

thank you!

much appreciated

[slender nacelle]

Alright! You're welcome

[shut mountain]

Some visualizations:

Can you help me understand what is achieved by plugging in x=0?

In the first picture

[idle bloom]

As far as I can tell nothing is achieved by plugging in 0

Maybe you meant you plug in not x = 0

But y=0

Because that's what you need to do to find where's positive

[shut mountain]

Alright. I guess I'll need to take a look at some formula sheet.

Thanks.

Trying to decrypt some notes I took without a thought :D

[upper karma]

When making a inscribed hexagon I know that we have to set it to the radius but do we need to set the compass for the radius or just draw it first

my guess is that we need to set it first right?

[clever pike]

Can anyone help me with number three

[split otter]

,rotate 90

[split otter]

Really sorry, have to sleep now. Hope someone will be able to help.

[surreal bolt]

okay I'm in. Still there?

@clever pike hiya

[sweet charm]

I’m wondering if somebody will be able to help me, I have exams next week, and I going over the practice final and I’m really lost, I haven’t been doing to good on the class in general, and could really use some help

It’s just high school geometry, so it shouldn’t be hard, but I haven’t had great understanding of it so far.

<@&286206848099549185>

[clever pike]

@surreal bolt yea

[surreal bolt]

Are you allowed to use law of cosines?

@clever pike

[clever pike]

@surreal bolt yes anything

[surreal bolt]

This solution is pretty convoluted ...

AH can be found using law of cosines: we have MA and MH and angle AMH is 120.

Further, call AT "x" and TH "y". Draw in MT. Both right triangles have hypotenuse MT.

Equation 1: 100 + x^2 = 1600 + y^2

Also we have AH and ATH = 60 degrees.

Equations 2: (law of cosines for ATH using x and y)

Two equations with 2 unknowns. ... It's theoretically possible to get both lengths?

[clever pike]

So I would solve for y? @surreal bolt

[upper karma]

Anyone want to help me check my math?

I’m not educated and I’m attempting to learn

[upper karma]

@upper karma ok

[swift grotto]

@gritty siren t'es toujours disponible?

[upper karma]

Anybody know/recommend a high school geometry workbook? I am looking for one to study for finals.

I want to calculate what I should and shouldn’t be able to see over a body of water. Distance would be 52 miles.

I thought it was calculating the Sagitta

But there’s this line of sight issue due to curvature of the earth

But there’s a “hidden” height that I want to calculate that I’m having trouble with

@upper karma

[vast dune]

could someone explain what step exactly happend here

[placid pagoda]

Question: "Find the values of x in the interval 0°<x<360° for which 2sinx+2cosx=0". If someone could help me with this I would appreciate it a lot

[vast dune]

u can rewrite that as cos x = -sin x

now when is the cos of an angle equal to negative sin of that angle?

[placid pagoda]

idk

[vast dune]

cos(90+x) = -sin x

[placid pagoda]

ahhh okay

so from that how would i find the values? because idk how i would place that on CAST

[vast dune]

CAST?

[open sinew]

i prefer ASTC

[uncut nova]

Ok, I'm stupid. I have to proof that in triangle where AC < BC, BAC = a (alpha), ABC=b (beta) and straight CD divide AB on two same parts and creates with AB acute angle t (theta), tan(t) = 2sin(a)sin(b)/sin(b-a).

Im a bit confused, is sin(b-a) positive or negative in triangle when b<a? Aren't angles a and b otherwise directed? When I try to proof that exercise I recieve negative sign on right side of equation.

[open sinew]

@placid pagoda 2 sin x + 2 cos x = 0 => sin x = - cos x
=> (sin x)/(cos x) = -1 => tan x = -1 => x = arctan(-1)

you know that (sin x)/(cos x) = tan x right?

i was a bit mystified staring at it for a bit at first tho tbf

[placid pagoda]

ahhh okay

yeah i know that

does arctan mean tan^-1?

[open sinew]

ye, i thought it might be less confusing

i prefer tan^-1 actually but then it gets confused with cos^2 x meaning (cos x)^2 not cos(cos x)

[placid pagoda]

yeee

[vast dune]

U arent alllowed to divide by cos x

[open sinew]

or more importantly, with cos^-1 x meaning inverse cos not sec x

[vast dune]

it could be 0

[placid pagoda]

its bigger than 0

[open sinew]

yeah, so you have to check if the 0s of cos works

[placid pagoda]

i think

0°<x<360°

[open sinew]

2sin(pi/2) + 2cos(pi/2) isnt 0, neither is 2sin(3pi/2) + 2cos(3pi/2)

corrected

[vast dune]

Had to solve similar question some time ago, but didnt have the 0°<x<360° in the question so couldnt do that

U probs can than I guess

[open sinew]

i mean, you can still do it

keep in mind you could just do pi/2 + n tau and 3pi/2 + n tau

[placid pagoda]

x = arctan(-1)
for this, once i get x, how do i know what of the four CAST sections to put it in?

[open sinew]

(where tau = 2 pi)

what does cast have to do with the answer

[placid pagoda]

oh idk tbh

[open sinew]

the only thing you should be worrying about is if there are other answers in your range other than the principle one it gives

[vast dune]

isnt it 3pi/2 - n?

[placid pagoda]

i have to find the values of x though

idk how else to do that

[open sinew]

$(\text{[trig.\ func.]}(x) = \text{[trig.\ func.]}(x+2\pi n))\forall(n\in\mathbb{Z})$

[somber coyote]

osmium:

[open sinew]

@placid pagoda you have various identities for trig. func.'s, e.g. cos(x) = cos(-x), sin(x) = - sin(-x), tan(x) = tan(x+pi), use them to generate more answers that satisfy the equation youre given to try to find more that might be in the range 0 degrees to 360 degrees (or equivalently 0 to 2pi radians)

tbh i prefer to always work in radians when using trig stuff and then convert at the end if needed

2cos x + 2sin x = 0 has a solution at tan^-1 (-1) right? so tan^-1 (-1) + pi is also a solution, as is tan^-1 (-1) + 2pi or tan^-1 (-1) - pi

[placid pagoda]

okay ill try that tyy

[spice ridge]

how does i work

i was absent in my trig class today due to illness

and they decide to learn about numbers that don't exist

[open sinew]

complex numbers in a trig class? isnt that a different topic to trig

i mean they do connect but i mean, they should teach u abt complex numbers before using them with trig stuff no?

[split otter]

Maybe they're teaching it first to prepare for the polar form of complex numbers?

[dire obsidian]

Is this channel occupied w a question rn>

?*

[onyx cloud]

@dire obsidian kinda late but shoot