#geometry-and-trigonometry

1 messages · Page 277 of 1

upper karma
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watch this

brisk ginkgo
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Nope

dusky surge
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Ah, it's an triangle inequality problem

upper karma
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@brisk ginkgo are you familiar with the law of sines?

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try to determine the "?" angles

dusky surge
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This isn't right...

brisk ginkgo
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I haven’t eaten in days

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Do u know how much work I have

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I don’t sleep

upper karma
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watch the video i sent

acoustic jungle
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i HaVeN'T eAtEn iN dAyS

brisk ginkgo
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@acoustic jungle I am bombarded with work and work

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Like stop

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@upper karma thank you!

upper karma
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basically the video says: a + b > c; a+ c > b; and b + c > c in order for a triangle to exist

acoustic jungle
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typo

upper karma
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no

acoustic jungle
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yes.

upper karma
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no

acoustic jungle
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b + c > c

upper karma
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thats not a typo

acoustic jungle
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then he made a typo

upper karma
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no

dark sparrow
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this looks like a test

night summit
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that it is

dark sparrow
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this is against the rules of the server

night summit
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really?

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oh dam

dark sparrow
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yes

night summit
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sry

upper karma
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damn ann

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laying down the rules

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@upper karma are you still interested in knowing how to solve it?

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ugh.... again with the offline

eternal crag
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just to clarify @upper karma , the video has a small mistake, Fishraider is right

upper karma
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yikes

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was it supposed to be b + c > a ?

eternal crag
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yeah

upper karma
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first one is easy, since sin(90) = 1

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so 31/1 = 31

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but you don't know it's a right triangle tho

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you can determine angle A and C using the inverse sin

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and if all angles add up to 180 (since it's a triangle), you're good

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it means it's a right triangle

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nothin in the question incidated it was a right traingle

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he deleted the question 😐

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ik

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the question was if it's a right triangle or not

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u sure? i thought it was if it forms a triangle or not

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@brisk ginkgo which is it?

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pff, emojis in the name

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that's so gangsta

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um and his sides were 2,2 and 10, not 15,17 and 31

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that's the problem he posted

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then he removed it

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and probably replaced it with another one with those values

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I don't know, man...

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Geometry is awesome

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sigh... kids

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Kids?

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i think we are talking about two different questions

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I love the law of sines

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i love you

left nebula
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Help

hard gale
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i mean for a right triangle there's just pythagoras lol

left nebula
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Can someone help me with some problems

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5 problems

hard gale
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check that 15²+17² = 31²

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if it is boom you got a right triangle

upper karma
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yah just post juju smith schuster

left nebula
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Que

upper karma
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UwU

left nebula
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5 questions

upper karma
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between the lines

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and he deleted the question, because he's a child

left nebula
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Can someone help me with my question

upper karma
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Ya I don’t know it

left nebula
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How

upper karma
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it's a test

left nebula
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Is this discord not helpful

upper karma
left nebula
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It’s not actually a test

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Please help

upper karma
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angle C looks awfully similar to angle F...

left nebula
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Is it just half of it

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Nvm

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Oh it’s 140 right

upper karma
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Idk

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isn't there any rules you can use for triangles like these...

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V

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A is right?

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@upper karma there is, yes

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they're called circle theorems

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@upper karma is this from a test?

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why

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because we're not allowed to answer test questions

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oh cuz we can't help you if its a test

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you can get banned if they catch you

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anyways, do you know what the 1 in the equation means?

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You add 1 to sin

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right

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not really

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it determines how much the graph shifts up/down

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so just to be clear, it's not a test, right?

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its also the equation of the axis

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no d

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b isnt the answer

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if its not a or b

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ill give you a hint: b would be the answer if it was 6sinx + 1

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lmao

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oh wait

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you were close tho; theres something in graph b thats also included in the correct graph

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it is A

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yes

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lol

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you need to learn what an amplitude and a midline is

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then you can solve any one of these

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^

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this one is A aswell?

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because isnt a vertical shift up + 2

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ooo thicc trigonometry functions...

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yes

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squid you're thicc

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👌

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Graph a regular function and translate it until you get your target function.

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because I solved it 🙂

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How do i do this one

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Ok let’s help you—

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First is this a test or just practice?

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@upper karma

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practice quiz

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Ok

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Well

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Do you know what a vertical shift is?

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aka "midline"

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yes

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-2 is the midline

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no

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...

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i thouight it was the first

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number

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it's not

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oh mb

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its -1

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Take the function f(x) = sin(x) + 1. What is the vertical shift (aka midline in trigonometry).

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+1

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Good.

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Now what’s the phase shift?

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cos1/2

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no

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-2 cos 1/2(x+pi) - 1
can be written as:
-2 cos (x/2 + pi/2) - 1

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o

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so is the phase shift -2 cos

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no

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F(x) = sin(bx + c)

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Set bx + c = 0

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Solve for x.

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That will be your phase shift, according to my spotty knowledge...

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no

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"x" is not the "phase shift" (aka the horizontal offset)

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is the answer b

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Brother

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Look here

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Wait

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Perhaps

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if this really is a practice test, then you're just shooting yourself in the foot by answering it without knowing what the hell it is

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Hmmmmmmmmmm

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i wanna know how 2 do it

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just rewatched the thing, says “set bx + c = 0 and solve for x” which the guy does...

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Granted the variable isn’t the phase shift but it’s a “systematic way of doing it” that gives you the phase shift.

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@upper karma am I wrong that setting bx + c = 0 and solve for x will give you your phase shift?

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b is not the phase shift

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no I know that

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Solving for x will give you your phase shift if you set the bx + c = 0 and solve for x.

(Sin(bx + c))

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yes?

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solving for x gives you the value for the point of the graph at that particular time

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did you watch the video or am I understanding it wrong?

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<@&286206848099549185> I’m having a brain fade with phase shift

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Could I request some help?

edgy quail
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Okay

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What don’t you get about phase shift

upper karma
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see @1:57

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Um

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How do you find it?

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I saw in a video that if you set (bx+c) = 0

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You get it

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By solving for x

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But apparently not?

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I’m just having trouble understanding with all of this conflicting information I’m confronting.

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Also sorry for going offline briefly, was updating notebook.

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(2x - pi/2) turns into 2(x - pi/4) in both videos

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ok what does that have to do with anything?

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That appears to me as being basic distributive property at work.

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it is, yes

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Ok.

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So... why is it important?

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Relative to the current situation

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because you need to turn it into that in order to find the "phase shift"

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hmm...

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ok.

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y = A sin [k(theta - b)] + c

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mhm

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Parent function.

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"k" is the periodicity factor

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b is the phase shift

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but b is only accurate once you have the periodicity factor

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if it's negative, it shifts to the right

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if it's positive, it shifts to the left

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I suppose... that is correct. The other method also returns the same value though, and thus they are both correct. I will agree that your method is more elegant and thus I will use it.

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Thank you for your help.

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use whichever one makes more sense to you

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the video has a link to an entire playlist of this stuff

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he explains it very well

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They both make sense, yours is faster though.

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I can agree with that.

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Thanks again.

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sure

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55 videos of pure trig

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he makes a few mistakes here and there, but they're easy to spot

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and always read the comments

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Copy that.

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I was gonna walk you through it and everything, but apparently you ded

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gonna play some vidya and go to bed now

wind heart
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yo yo yo

upper karma
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that

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Is the most

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Beautiful solution

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For perimeter

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I have literally ever seen

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👏

acoustic jungle
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The question was solved 4 hours ago

upper karma
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Yeah the bread is right

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big brain

stone tulip
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how do i find

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how do I find the angle I marked

upper karma
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show the question

stone tulip
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the question doesnt matter

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I need to find the angle

upper karma
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theres a rule for angles like those

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the z rule

viscid ginkgo
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(secθ+1)/tanθ = tanθ/(secθ-1)

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please help me with this proof

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I tried many solutions, such as

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(secθ+1)/tanθ = (1/cos+1)/tanθ = (1+cosθ/cosθ)*(cosθ/sinθ) = (1+cosθ/sinθ) = 1 + tanθ = sec2 θ

upper karma
acoustic jungle
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have you tried conjugate.

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@viscid ginkgo

viscid ginkgo
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I'm pretty sure that I did yes.

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Let me do that right now

brisk ginkgo
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I CAN USE SIN

upper karma
brisk ginkgo
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@upper karma right?

upper karma
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No

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Cos

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Cos = adj / hyp

brisk ginkgo
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@upper karma THANK YOU

upper karma
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Soh Cah Toa

viscid ginkgo
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yeah, the conjugate gets me (sec^2 t -1) / (tan t sec t - tan t)

brisk ginkgo
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The hypothesis

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Nose

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Is what

viscid ginkgo
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so tan t/sec t - tan t

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it doesn't work

acoustic jungle
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k let me try

viscid ginkgo
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I tried conjugating the right side

upper karma
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🤔

acoustic jungle
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worked for me

brisk ginkgo
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So I use COS?

viscid ginkgo
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(tanθ(secθ+1))/(secθ-1(secθ+1)) = (tanθsecθ+tanθ/sec^2θ-1)

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=secθ+1

acoustic jungle
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I'll take a photo.

upper karma
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Bruh use LaTex nezukoAngery

brisk ginkgo
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COS OR NO

acoustic jungle
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ok

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LHS=RHS

upper karma
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It was fun!

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There is a sin lonely without theta

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Wtf happened on the 3rd line?

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Did you multiply the (1-cos0) to the other side?

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Or multiply (1-cos) to the numerator and denominator.

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LHS doesn't equal to RHS if RHS doesn't even exist.

acoustic jungle
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I multiplied by 1-cos

upper karma
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To the other side?

acoustic jungle
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Sorry I don't know what you are saying

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what are you saying is wrong?

upper karma
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$$\frac{1+\cos\theta}{\sin\theta} = \frac{\sin\theta}{1-\cos\theta}$$
$ \ $
$$\frac{1+\cos\theta}{\sin\theta} \cdot (1-\cos\theta) = \sin\theta$$
OR $\$
$$\frac{1+\cos\theta}{\sin\theta} \cdot \frac{(1-\cos\theta)}{(1-\cos\theta)} = \frac{\sin\theta}{1-\cos\theta}$$

somber coyoteBOT
upper karma
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Nvm.

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Just right the RHS for the 3rd and 4th line.

acoustic jungle
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what is wrong

viscid ginkgo
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@acoustic jungle that's not right because here is the original problem: (secθ+1)/tanθ = tanθ/(secθ - 1)

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I'm trying to make the left side equal the right side, or the right side equal the left side.

acoustic jungle
viscid ginkgo
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I'm not able to do this.

upper karma
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Wait who's the one that needs help.

acoustic jungle
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you are saying the third line is wrong?

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what is wrong.

upper karma
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3rd line is nonexistent.

viscid ginkgo
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yes but sin/1-cosθ is not = to one of the sides

upper karma
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You've stated before that LHS = RHS.

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Well, where's the RHS?

viscid ginkgo
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you u sed the left side of the equation,

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so it should equal the right side of the equation.

acoustic jungle
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the question is is (secθ+1)/tanθ equal to tanθ/(secθ - 1)

upper karma
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$\frac{(\secθ+1)}{\tanθ} = \frac{\tanθ}{(\secθ - 1)}$

viscid ginkgo
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yes

acoustic jungle
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exactly

somber coyoteBOT
acoustic jungle
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could you please circle the apparent mistake I made

viscid ginkgo
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the result is not equal to tanθ/[secθ - 1]

acoustic jungle
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proving these equations are equal can be done because both are equal to sin/(1-cos)

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Oh did you think I cross multiplied?

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I didn't do that

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I multiplied by (1-cos)/(1-cos)

viscid ginkgo
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you mean distribute?

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yeah of coursed

upper karma
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Okay.

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First off.

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You guys are making this more complicated than it seems.

viscid ginkgo
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I know right?

acoustic jungle
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you are saying it's wrong

viscid ginkgo
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I've been doing this problem for like 6 hours

acoustic jungle
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what is wrong

upper karma
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Remember that sec^2-1 = tan^2

viscid ginkgo
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because

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the left side equals the right side

upper karma
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Here.

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$$\frac{(\secθ+1)}{\tanθ} = \frac{\tanθ}{(\secθ - 1)}$$
$\$
$$\frac{(\secθ+1)}{\tanθ} \cdot \frac{\sec\theta-1}{\sec\theta-1} = \frac{\tanθ}{(\secθ - 1)}$$
$\$
$$\frac{\tan\theta}{\sec\theta-1} = \frac{\tanθ}{(\secθ - 1)}$$

somber coyoteBOT
acoustic jungle
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you are saying something is wrong with my solution

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what is it.

upper karma
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Your solution does not have a RHS.

viscid ginkgo
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the result lol

acoustic jungle
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RH is right hand side

viscid ginkgo
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it doesn't = the right side of my problem

upper karma
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You know what, I'm not gonna argue about notation.

acoustic jungle
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but they both equal to sin/(1-cos)

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therefore they are both equal

upper karma
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Fishraider, for the love of God, rewrite the RHS at the bottom.

acoustic jungle
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you can reverse the process to get back to rhs if you want

upper karma
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@viscid ginkgo @acoustic jungle Both of you stop kthx.

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Jkjk.

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It's been solved a long time ago.

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Stop stressing over it.

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@viscid ginkgo He converted sec and tan, to their cos and sin counterparts.

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Chill.

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@acoustic jungle Right the RHS again at the bottom. If RHS really = LHS, then make sure there's a RHS actually there.

acoustic jungle
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k

viscid ginkgo
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ok how does secθ + 1(secθ - 1) = tan?

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(secθ+1)(secθ-1)/tanθ(secθ-1)

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=(sec^2θ - 1) / tan(secθ-1) = secθ-1/tanθ

upper karma
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For the god sake use latex

viscid ginkgo
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me?

upper karma
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It may take you even less time

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Yesss

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And it'll be clearer for everyone

viscid ginkgo
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well

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I need to learn now

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\theta

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\alpha

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\cos (2\theta)

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?

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how to use

upper karma
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Look, $\theta$ $\cos{\theta}$

somber coyoteBOT
upper karma
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"{}" are really important in LaTex

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Sometimes are needed

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For theta for example not

viscid ginkgo
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ok

upper karma
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Same for all trig (csc, sec, etc.)

viscid ginkgo
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and a fraction

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?

upper karma
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$\frac{numerator}{denominator}$

somber coyoteBOT
upper karma
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Its not obligatory to use LaTex but it saves a lot of times to us, the helpers

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The normal format its just messy

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Anyway

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--------(free channel)------------

viscid ginkgo
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ty

rich wolf
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$\frac{\mathrm{numerator}}{\mathrm{denominator}}$

somber coyoteBOT
rich wolf
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@upper karma

pallid shadow
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Hey guys I have a question

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How do you prove angles at the centre and circumference proof?

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I have searched on the internet

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But all the proofs dont make sense when the origin of the circle is outside of the beams

versed river
pallid shadow
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like in this case

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angles at the centre and circumference proof

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yes, this one

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all the proofs cut things into 2 isoceles triangles

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but in this case you cant

versed river
pallid shadow
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because the origin is outside

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Thakns man

versed river
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oh this one uses isoceles triangle

pallid shadow
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well it doesnt work for the example I posted 😦

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As you can see, it is a huge challenge.

versed river
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case c

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in the khan academy article?

pallid shadow
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alright

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I will check

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yes

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it is legit

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thaks

versed river
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np mate

pallid shadow
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I am so sorry Sneaky, I just spent a couple of minutes trying to search for this proof

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All the proofs only do one case XD

versed river
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Don't worry about it

upper karma
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How far does a 15 cm long minute hand travel around a clock in 10 minutes?
the clock is circular
the minute hand stretches from center to edge

quiet mason
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what you tried?

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try to calculate the angle 10 minutes subtends at the centre

upper karma
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@upper karma find the circumference on the clock

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No I know how to solve it

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I solved it

quiet mason
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?

upper karma
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I am waiting for someone else to solve it

quiet mason
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oh

upper karma
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To see how y’all do it

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show us what u have

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5pi cm

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if I plugged in values correctly...

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Yea seems right.

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how did u solve it

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I went the long way, and instead of using formulas, I just made proportions and such...

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So

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Basically

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I first found the circumference of the clock.

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Then

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I figured out how large the central angle of 10 minutes was

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And then I went to find the intersected arc.

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But not with the formul

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a

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Instead I figured out how much percentage of 360* the central angle was

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Then I applied that to the circumference

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Solved

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(Longer way, more fun, derived this method independently...)

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which made me feel a sense of accomplishment because doing things on my own makes me feel good... u get the same feeling?

acoustic jungle
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Ok

deft imp
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@everyone yo im fucked boys is

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@everyone

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could anyone help?

keen aspen
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Lol

acoustic jungle
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Lol

deft imp
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vc?

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@keen aspen can you help?

keen aspen
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I cant see anything

deft imp
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shit

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can you join vc?

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ill stream

keen aspen
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Damn

verbal gust
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Hello, I just need help on one thing. I have already found (x=3), however I'm unsure on how to find "y". Is there a formula that I need to use to find it?

📎 unknown.png
keen aspen
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How do you relate that vertical height with one of the bases

upper karma
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@verbal gust is 5y-15 the angle of BDA?

verbal gust
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Would "y" be 15 because it would then equal (60). Then you would multiple (60 x 3) having the triangle equal 180?

upper karma
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But why multiplying by 3 if you already have y lol

verbal gust
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to double check if it equals the sum of the triangle (180 degrees

upper karma
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Lol ok

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Wait

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,w (5y-15)+60+30=180

somber coyoteBOT
upper karma
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@verbal gust y is 21.

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Would "y" be 15 because it would then equal (60). Then you would multiple (60 x 3) having the triangle equal 180?
@verbal gust and the first phrase is wrong bc that angle isnt 60, is 90

deep trail
keen aspen
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Whats one of the lengths of the triangle

deep trail
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48

keen aspen
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No

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The perimeter is 48

deep trail
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oh nvm

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16

keen aspen
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Yes

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So make a diagram

deep trail
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k i did

keen aspen
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And construct a triangle with the vertical height, half of the base and the diagonal

deep trail
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oh i get it

obsidian cedar
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Can someone help me with this?

glacial haven
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you know what a unit circle is?

obsidian cedar
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Nope

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I had a ton of other assignments to do so couldn’t really check out math

upper karma
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google it

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"unit circle"

obsidian cedar
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Alright

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Yep watching a video on it rn

glacial haven
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instead of 0<x<360 use 0<x<2pi

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radians>degrees

obsidian cedar
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Oh ok 👍

lunar flower
upper karma
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@lunar flower all of them?

lunar flower
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yes

upper karma
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For the first one, divide the figure into a triangle and a rectangle

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And just apply the corresponding formulas

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For the second, we have again to add the areas of triangles $\triangle{ABC} + \triangle{ACD}$

somber coyoteBOT
upper karma
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And for the third calculate the total area of the big square and substract the areas of the 4 squares on the corners

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@lunar flower

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👍 Thanks

upper karma
acoustic jungle
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angle u and angle u prime

upper karma
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yes, but is there a specific name for angles that are in a similiar location as u and u prim?

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e.g you have adjacent angles for angles next to each other, vert. angles etc

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gonna have to be more specific with your example

upper karma
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How would i solve this

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i am a bit stupid

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i need the area

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well do u see how the 8 and the 9 form a right angle triangle?

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how do u know for sure?

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o wait they bisect

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ye

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so would it be 8*9 over 2

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yes

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then multiply by 4

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because you have 4 triangles

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wait why?

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wdym why

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wait so another way of doing it

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would be 18 * 16 /2

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since its half the line

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(8*9)/2 is just the area of the triangle

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i dont understand

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smh

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so p and q would be 18 and 16 right?

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instead of 9 and 8

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yes

umbral snow
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Yes I think so, but it's not beautifully clear

upper karma
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so the area would be 144

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yes

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which is 36 * 4

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bet thanks ❤️

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oh i get it now

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np

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❤️

upper karma
#

I will be forever indebted to whoever can help me with these

stuck torrent
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Yes you are indebted to me

upper karma
#

Thank you so much

stone falcon
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I'm sorry but I am totally unable to help you.

upper karma
#

Oh I see the pun

stone falcon
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There is no pun.

upper karma
#

@upper karma its a graded exam, it goes against the rules

acoustic jungle
#

forgot to crop 😦
@upper karma

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@upper karma its a graded exam, it goes against the rules

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Now nobody will ever know

upper karma
#

Ok well if nobody here would like to help could you link to a different server

acoustic jungle
#

@upper karma

novel flax
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lmao

upper karma
#

tf is with the 💩 emoji on my heart bruh @acoustic jungle

covert rune
#

Got a problem I’m having trouble, can anyone guide me in the right direction?

#

“Suppose a line segment AB fully contained within a sphere of radius 2 has a length of 4. If an infinitely large two dimensional wall splits the sphere into two pieces whilst intersecting AB at a 90* angle, what would be the volume of the smallest separation be when the intersection between AB & the wall occurs at
A) 1/2 The way of AB
B) 1/3 The way of AB
C) 1/7 The way of AB”

#

A Can be solved with common sense, but I’m having trouble with the other two

rich wolf
#

Draw a picture

#

4 is the length of the diameter

#

Any 4 unit long segment contained entirely within the sphere must therefore be a diameter

covert rune
#

I understand that, but it seems to be a “spherical chord” per se, we haven’t learnt much about these so I’m unsure if there is any formula we can use to find it

brisk holly
#

can someone please explain why this graph is incorrect

#

the midline is -4 and its amplitude is 5 so it should go 5 above and below it

#

and its period is 2pi/pi/16 which is 32

rich wolf
#

What is the amplitude?

brisk holly
#

wdym

rich wolf
#

Look at the parent cosine graph

brisk holly
#

the amplitude of cos is 1

#

or 2pi

rich wolf
#

No look at the shape of the cosine curve

#

It starts at the max

#

But this function in the picture has a negative amplitude

#

So it must start at the min

brisk holly
#

oh I see

sacred tusk
upper karma
#

how do u not know

sacred tusk
#

never been taught

#

is it just rounding?

upper karma
#

idk

sacred tusk
#

do you not know?

upper karma
#

i do not know

#

i assume 24* is 24 degrees on a clock

#

for 13 imma guess the answer is 24*49

#

for 14 imma guess the answe is 39 degrees

#

and for 15 imma guess the answer is

#

35degrees and 86 minutes

sacred tusk
#

ill take a look see if its in any books

lilac gull
#

isnt that just radian stuff

sacred tusk
#

is it i dont know

lilac gull
#

looks like radians to me

#

do you need help with all of them

sacred tusk
#

yes i am stumped

lilac gull
#

there are 60 minutes in a degree and 60 seconds in a minute

#

' is minutes and " is seconds

sacred tusk
#

ok website says its the right step

ancient jacinth
#

please ping me

upper karma
#

what does the arrow on top of the letter mean again?

#

@ancient jacinth

ancient jacinth
#

i got it

#

that just means angle

upper karma
#

ok

#

well if all the sides are 20 cm

#

then we can find out the lenghts on the dotted lines on the square

#

@ancient jacinth

ancient jacinth
#

yea

#

i got it

#

im done lol

#

thanks for trying though

upper karma
#

Is there a way to do this? I’m not sure how to use the ratio

#

Plot M and N

#

I did

barren tide
#

here's a better picture

small gulch
#

Its like algebra

#

Like treat cot x as x

#

Then you should be able to solve it

upper karma
#

do product sum factor

barren tide
#

i got 2cot^(2)x + 5cotx + 6cotx + 15, so how do i get the (cotx) in the end of the question?

versed river
#

factor 2cotx out of 2cot^2(x)+6cotx and 5 out of 5cotx+15

#

you get 2cotx(cotx+3)+5(cotx+3)

upper karma
#

@barren tide do product sum factor

versed river
#

thats whatg

barren tide
#

oh okay I was getting confused cuz the question wasn't showing those factors

versed river
#

he did

#

rent free

upper karma
#

ok

#

tru

barren tide
#

That's what I did, but thank you for your help

upper karma
#

np

versed river
#

you good now?

barren tide
#

I figured it out. You are angels sent from the heavens. Thank you

#

one thing is bugging me tho. Why isn't it set up as (2cotx + )(cotx + )^2

versed river
#

it isnt?

upper karma
#

bc thats not correct

versed river
#

its set up as (2cotx+?)(cotx+?)

#

theres no square

#

2cotx(cotx+3)+5(cotx+3)=(2cotx+5)(cotx+3)

#

do you see how i got that?

barren tide
#

so what happens to the other one?

versed river
#

the "other" cotx+3?

barren tide
#

yeah

versed river
#

a) try expanding (2cotx+5)(cotx+3)

#

you'll see its right

#

b) what do you get if you add something like 4a+3a

#

i know b) doesnt sound related, but i can use it as an example if you dont get it after a)

barren tide
#

oh so it just replaces it, because it is multipying

versed river
#

yeah. The example i use here if that 4a+3a=(4+3)a

barren tide
#

oooohhhh

versed river
#

in the question above, a is cotx+3

barren tide
#

thank you good sir

versed river
#

and obviously its not 4cotx+3cotx, but its analagous

#

no problem

brisk holly
#

What does it mean by give the reference angle?

upper karma
#

the angle given

#

so 115 degreesfor question 13 and350 degrees for question 14

brisk holly
#

Why would it ask for that

barren tide
brisk holly
#

I don't get it

barren tide
#

The reference angle is the angle that is closest to the x axis from the given conterminal angle. For example if you're given 145 degrees, then you would have to subtract 145 from 180 in order to find the reference angle.

brisk holly
#

Do you always just subtract 180?

barren tide
#

No, that's just to find the reference angle for coterminal angles that fall under the second quadrant

#

I don't know the actual formula to find the reference angles, it just seems to be the left over angle for the quadrant that it's in.

brisk holly
#

How do we find trig functions of degrees without a calculator

#

Like sin (60)

upper karma
#

use Taylor series

barren tide
#

Oh I thought it was the unit circle

brisk holly
#

whats that

versed river
#

You dont need taylor series for sin60..

upper karma
#

yeah just draw an equilateral triangle

#

sully lmfao

brisk holly
#

what the fuck is taylor series

barren tide
#

wtf

upper karma
#

do you know calculus? or not?

brisk holly
#

bro im out here doing basic trigonometry

upper karma
#

haha @upper karma

brisk holly
#

no I do not know calculus

upper karma
#

ok then

barren tide
upper karma
#

taylor series is good for approximating functions

#

@brisk holly

barren tide
#

I dunno how to explain what the unit circle is

brisk holly
#

its a circle with radius 1

upper karma
#

circle with centre at (0,0) and radius 1

brisk holly
#

centre

barren tide
#

i don't even fucking know what is even happening anymore. I'm just trynna do my homework

brisk holly
#

same man

upper karma
#

what's the question lol

barren tide
#

😂

versed river
#

exact value of sin60. Thats all the question was

plucky marlin
#

use maclaurin series

upper karma
#

xD

brisk holly
#

lol

plucky marlin
upper karma
#

just draw an equilateral triangle and done

brisk holly
#

how do calculators calculate trig functions

plucky marlin
#

sin(x) = x kongouDerp

upper karma
#

sinx = x and e=pi=3

glacial haven
#

e^x=x^n/n!

rich wolf
#

cos(x)=x

glacial haven
#

sinx=y

plucky marlin
upper karma
#

pi^2 = g

glacial haven
#

y=rsintheta

upper karma
#

Fundamental Theorem of Engineering

rich wolf
#

Original content

glacial haven
#

whats the fundamental theorem of engineering

rich wolf
#

Car go fast

glacial haven
#

sinx is actually x

#

according to the fundamental theorem of engineering

#

tf

barren tide
#

white = black because people are equal

upper karma
#

me = you then

barren tide
#

😳 👉 👈

upper karma
#

All people should be equal

barren tide
#

word(n)=Black and White=Black

#

I think ya'll know where I'm going with this 👀

glacial haven
#

nah criminals

#

shouldnt be equal

upper karma
#

Race isn’t white or black

#

It’s human

#

We’re integrated

pallid cloud
#

$ABC$ is a triangle with elements $a, b, c, \hat{A}, \hat{B}, \hat{C}. \ $ Suppose that $\cosine^2{\hat{A}} + \cosine^2{\hat{B}} + \cosine ^2{\hat{C}}=1. \ $ Show that $ABC$ is a right triangle.

somber coyoteBOT
pallid cloud
#

Ping me if you answer
Thanks

#

<@&286206848099549185>

pallid cloud
#

Solved it
Thanks.

plucky marlin
#

np

pallid cloud
#

My friends' solution
And mine

upper karma
#

hi

pallid cloud
#

Hi

upper karma
#

Anyone on aops

upper karma
#

,rotate -90

somber coyoteBOT
upper karma
#

there we go

candid lily
#

is the rate of degree change between 1 and 2 radians consistent?

dark sparrow
#

what's a "rate of degree change"

candid lily
#

right, sorry, i meant for the cosine function, it seems like between 1 and 2 radians, the rate of degree change, the slope is linear

#

is that the case?

dark sparrow
#

what the fuck is a "rate of degree change"

#

no, the cosine function is not linear between 1 and 2 radians

#

or anywhere for that matter

solid sequoia
#

How would you do "In e^2" to simplify it?

versed river
#

ln(e^2)? ln(e)^2? PARENS

solid sequoia
#

It just says In e^2

#

simplifying it?

#

Logarithms

versed river
#

$\ln(a^b)=a\ln b$

upper karma
#

Taking the exponent as a factor of In e

somber coyoteBOT
upper karma
#

Yes

#

And In e = 1

#

Bc In has base e

versed river
#

do you get what i said mr/mrs goodnight

solid sequoia
#

So would the answer be In 2e

upper karma
#

No

versed river
#

their name is japanese for goodnight

upper karma
#

Oh

#

Cool

versed river
#

2lne

solid sequoia
#

oh mb

versed river
#

but that can be further simplified

#

do you see how?

solid sequoia
#

no?

upper karma
#

And In e = 1
@solid sequoia do you know that 2 times 1 is 2 right? And not 2e

versed river
#

$\lne=?$

somber coyoteBOT
upper karma
#

Lol

versed river
#

bruh

solid sequoia
#

why do all logarithms always equal 1?

upper karma
#

No

versed river
#

lne=log base e of e

#

you can think of it as asking "what number do you raise e to the power of to get e"

upper karma
#

In has base e , so you are asking what exponent has to be to satisfy this

$e^?=e$

somber coyoteBOT
solid sequoia
#

1

upper karma
#

Exactly

#

So

#

2*1=?

#

Lol

solid sequoia
#

2

upper karma
#

Thats it

solid sequoia
#

So the answer is 2? what

#

Im stupid sorry

upper karma
#

Yes

solid sequoia
#

would In e^10 = 10 then

#

yay

versed river
#

good job

solid sequoia
#

In 1/e^3 = ?

#

1/3?

upper karma
#

Almost

versed river
#

not quite

upper karma
#

-3

#

Well yeah its not close lol

solid sequoia
#

oof

versed river
#

1/e^3=e^-3

#

so we have ln(e^-3)

solid sequoia
#

when exponent is negative you flip it to a fraction

upper karma
#

$\frac{1}{a^b}=1*a^{-b}$

solid sequoia
#

ooh thats handy

somber coyoteBOT
upper karma
#

Better

#

And the same the other way around

versed river
#

well you're evaulating ln(1/e^3), so thats equal to ln(e^-3)

solid sequoia
#

In 1/root e would be 1/2 😄

versed river
#

oo

#

close

#

but not quite

solid sequoia
#

-1/2

versed river
#

dont just gues

#

ok yes

#

lmao

upper karma
#

Yeah lol

solid sequoia
#

idek what im doing everything is just become fraction or negatives

upper karma
#

So back to your original question

#

well you're evaulating ln(1/e^3), so thats equal to ln(e^-3)
@solid sequoia

#

Try to continue

solid sequoia
#

^-3 = cube root?

dark sparrow
#

no

upper karma
#

Nope

#

$\sqrt_b{a^c}}=a^{\frac{c}{b}}$

solid sequoia
#

crap, it's 1/cubic root e

#

or idk this is just sad

upper karma
#

Uhh

versed river
#

are you trying

solid sequoia
#

wait why am i doing cubic root

versed river
#

bth root

dark sparrow
#

$\sqrt[n]{a^m} = a^{\frac{m}{n}}$

somber coyoteBOT
upper karma
#

Thanks lol

solid sequoia
#

a lot of math geniuses here

upper karma
#

You too

solid sequoia
#

nope haha

#

negative 3 cubic root e?

upper karma
#

$-3\sqrt[3]{e}$?

somber coyoteBOT
solid sequoia
#

or idk ;-;

upper karma
#

But its a question or what

solid sequoia
#

an answer to the In (e^-3)?

upper karma
#

Ay not quite

#

You got right the part of -3

solid sequoia
#

uuuuuuuu

upper karma
#

So after removing the exponent of e, we have:

$-3\cdot In(e)$

solid sequoia
#

you all are angels ;-;

dark sparrow
#

\ln, not \Ln

upper karma
#

But thats what i had

dark sparrow
#

ln

#

not In

solid sequoia
#

wait ughhhhh I knew that you can just move the exponent behind it ;-; why'd i do that

upper karma
#

I dont have that lol Ann

somber coyoteBOT
upper karma
#

@solid sequoia we are here anyways

#

Remember

#

In has base e , so you are asking what exponent has to be to satisfy this

$e^?=e$

somber coyoteBOT
solid sequoia
#

okay I gotcha

upper karma
#

So the final ans is...

solid sequoia
#

for which one?

dark sparrow
#

you can't type the lowercase L?

#

what?

# 
$ \ln(e) $
somber coyoteBOT
upper karma
#

Oh its that lol, i thought it was another letter

#

Nvm

dark sparrow
#

bruh

upper karma
#

Lol

solid sequoia
#

In 1 what do you even do

#

the answer is 1

upper karma
#

for which one?
@solid sequoia for $-3*\ln{3}$

solid sequoia
#

no matter what you do to the problem your answer is 1 haha

#

wat

upper karma
#

For the question $-3*\ln{e}$

somber coyoteBOT
solid sequoia
#

isnt that the simplified version?

upper karma
#

Thats what you said

#

As -3 is the exponent of e, it goes as a factor

solid sequoia
#

yeah, the answer is 1/3e?

upper karma
#

Why

solid sequoia
#

because you flip the negative and sign stuff

upper karma
#

$\ln {e}=? $

somber coyoteBOT
solid sequoia
#

-1/3?

upper karma
#

you can think of it as asking "what number do you raise e to the power of to get e"
Think of this

solid sequoia
#

to get to -3?

upper karma
#

No, to get e

#

To what power do you raise e to get e as a result?

solid sequoia
#

that sounds like a trick question

upper karma
#

No lol

#

Simpler

#

To what power do you raise 2 to get 2 as a result?

solid sequoia
#

1

upper karma
#

:)

#

So its the same thing

solid sequoia
#

so In e = 1?

upper karma
#

YES

solid sequoia
#

haha i feel like you've strained a lot to get there haha

upper karma
#

Lol

#

So obviously

$-3\cdot \ln{e=?}$

somber coyoteBOT
upper karma
#

Pls think your answer

solid sequoia
#

-3

#

wait

#

ok

upper karma
#

So yeah anything else

#

?

solid sequoia
#

what is In

upper karma
#

Natural log

solid sequoia
#

why not say log

upper karma
#

Havent you been taught this

#

Bc usually ln has base e while log has base 10

solid sequoia
#

i dont really understand the purpose, it's like backwards exponents

upper karma
#

I dont know what to answer to that

solid sequoia
#

ya thats okay ofc

#

In 0 = 0?

upper karma
#

no

#

e^0 = 1 so..

solid sequoia
#

what my calculator said 0, how could it lie to me like this ;-;

acoustic jungle
#

ln 0 is not defined

upper karma
#

Bread is back :)

#

ln 0 is not defined

#

Lol

solid sequoia
#

e^In 5, does it mean 5 In e?

upper karma
#

What

acoustic jungle
#

it means e^ln5 = 5

upper karma
#

e^(ln 5) is 5

solid sequoia
#

what

acoustic jungle
#

dude I just sniped you twice

upper karma
#

I think you should really watch a yt video about log and ln @solid sequoia

solid sequoia
#

when you're the one they're sniping over bc u always need help wew

upper karma
#

think about what log is, log is just opposite of exponentiation

solid sequoia
#

I definitely do

upper karma
#

what's sniping?

solid sequoia
#

khan academy here i come

upper karma
#

Good

acoustic jungle
#

it means you say something before the other person says the same thing

upper karma
#

ahhh ok

#

Ann snipes me everytime lol

grizzled zinc
#

I set the two equal to eachother then got 3.1, then I plugged 3.1 into 12(3.1)+1. Would that be right?

#

i didnt wanna interrupt the other person lol

acoustic jungle
#

if you did 43-2x=12x+1 and 3.1 satisfies this equation you are right.

upper karma
#

Bum sniped again

grizzled zinc
#

Do i not have to plug it in? Or is 3.1 just the answer?

upper karma
#

I didn't say anything coz I knew he was gonna say it too lol

#

Lol

acoustic jungle
#

3.1 is not right.

grizzled zinc
#

me doing math at 7am 👁️ 👄 👁️

upper karma
#

Do i not have to plug it in? Or is 3.1 just the answer?
@grizzled zinc as the sniper said, if you didnt mess it up 3.1 is the answer to x not to the problem

#

Lol

#

lol

#

,w 43-2x=12x+1

somber coyoteBOT
grizzled zinc
#

huh

#

bruh howd i get the .1

upper karma
#

Lol

grizzled zinc
#

???? ig i messed up entering it into the calc lmao

upper karma
#

Might have added 1 to 43 happens to the best of us

grizzled zinc
#

oh u right

#

thank you lmao i am a clown

upper karma
#

@grizzled zinc its a test?

grizzled zinc
#

No review

upper karma
#

Huh