#geometry-and-trigonometry | Mathematics | Page 256

rich wolf Mar 20, 2020, 3:39 AM

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y=tan(1)

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you want the tangent of 1 radian?

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or 1 degree?

golden panther Mar 20, 2020, 3:40 AM

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i guess 1 degree

cinder portal Mar 20, 2020, 3:41 AM

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tan(1) degree?

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holy

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wat

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i mean your answer is going to be

golden panther Mar 20, 2020, 3:41 AM

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Bro I dont know

cinder portal Mar 20, 2020, 3:41 AM

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super SUPER nasty

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it's possible

golden panther Mar 20, 2020, 3:42 AM

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it doesnt say

weary drift Mar 20, 2020, 3:42 AM

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$\tan(1)$ or $\tan\inv(1)$?

cinder portal Mar 20, 2020, 3:42 AM

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best way without calculator is just messing with identities

somber coyoteBOT Mar 20, 2020, 3:42 AM

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RokettoJanpu:

📎 150414892225134592.png

cinder portal Mar 20, 2020, 3:42 AM

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oh fair point

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inverse or regular

golden panther Mar 20, 2020, 3:42 AM

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the right one

cinder portal Mar 20, 2020, 3:42 AM

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oh

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nvm

rich wolf Mar 20, 2020, 3:42 AM

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one of those is super easy

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the other one is super hard

cinder portal Mar 20, 2020, 3:43 AM

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you got this then

golden panther Mar 20, 2020, 3:43 AM

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what is the best way

rich wolf Mar 20, 2020, 3:43 AM

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ok

weary drift Mar 20, 2020, 3:43 AM

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ask yourself what angle between -pi/2 and pi/2 is such that tan of that angle produces 1

rich wolf Mar 20, 2020, 3:43 AM

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you are trying to find tan(x)=1

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what is the definition of tan?

golden panther Mar 20, 2020, 3:44 AM

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sin/cos

rich wolf Mar 20, 2020, 3:44 AM

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so if sin/cos=1, what does that imply

golden panther Mar 20, 2020, 3:44 AM

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@weary drift i dont have the unit circle memorized yet

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that its already 1?

rich wolf Mar 20, 2020, 3:45 AM

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no

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if i have two numbers

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a and b

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and i tell you that a/b=1

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can you solve for a

golden panther Mar 20, 2020, 3:46 AM

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yes

rich wolf Mar 20, 2020, 3:47 AM

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what is a

golden panther Mar 20, 2020, 3:48 AM

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b*1

rich wolf Mar 20, 2020, 3:50 AM

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so a = b

golden panther Mar 20, 2020, 3:50 AM

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yes

rich wolf Mar 20, 2020, 3:50 AM

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so in this scenario

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a is sin(x), and b is cos(x)

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at what angle is sin(x)=cos(x)

golden panther Mar 20, 2020, 3:51 AM

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do i need the unit circle memorized for this

rich wolf Mar 20, 2020, 3:51 AM

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kind of

golden panther Mar 20, 2020, 3:51 AM

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crap

cinder portal Mar 20, 2020, 3:51 AM

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guess and check

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lol

rich wolf Mar 20, 2020, 3:51 AM

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not really tho if you know your special right triangles

cinder portal Mar 20, 2020, 3:51 AM

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hint: it involves pi

golden panther Mar 20, 2020, 3:52 AM

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Nah im just going to open a new tab

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i have the degrees memorized just not the square roots

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Its 45

cinder portal Mar 20, 2020, 3:53 AM

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👏

golden panther Mar 20, 2020, 3:54 AM

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bro dont roast

cinder portal Mar 20, 2020, 3:56 AM

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not roasting

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good job

golden panther Mar 20, 2020, 3:56 AM

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oh ok

cinder portal Mar 20, 2020, 3:57 AM

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i guess the clap does look a bit sarcastic

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didnt mean it that way

golden panther Mar 20, 2020, 4:01 AM

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so i have to memorize the square root part of the unit circle to do these

compact spire Mar 20, 2020, 4:04 AM

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Can anyone help me with this? been trying it out myself but I can't even finish it without getting a super crazy number

📎 trig2.png

cinder portal Mar 20, 2020, 4:04 AM

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try this

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$\sin^{-1}(- \frac{3}{5}) = \alpha \
\tan^{-1}(\frac{2}{7}) = \beta$

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holy shit

somber coyoteBOT Mar 20, 2020, 4:05 AM

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Cytis:

📎 190017172074921986.png

cinder portal Mar 20, 2020, 4:05 AM

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make this substitution

compact spire Mar 20, 2020, 4:06 AM

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I've done that for the difference of sine

cinder portal Mar 20, 2020, 4:10 AM

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what did you get

plucky jacinth Mar 20, 2020, 4:11 AM

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📎 7fe93c093bf8b012c2facb940f33d1ad.png

cinder portal Mar 20, 2020, 4:11 AM

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use tr

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ig

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set up your equation

compact spire Mar 20, 2020, 4:12 AM

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📎 Snapchat-1620467048.jpg

plucky jacinth Mar 20, 2020, 4:13 AM

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i have no clue how to do this

cinder portal Mar 20, 2020, 4:13 AM

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WAT

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when you do what I told you to do, you should just get this

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$\sin{\alpha - \beta}$

somber coyoteBOT Mar 20, 2020, 4:13 AM

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Cytis:

📎 190017172074921986.png

cinder portal Mar 20, 2020, 4:13 AM

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ok just use the substitution

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it's a lot easier

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you don't have to write much

compact spire Mar 20, 2020, 4:14 AM

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okay so sin (a-b) = sin(a)cos(b)-sin(b)cos(a)

cinder portal Mar 20, 2020, 4:14 AM

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use the sin subtraction identity

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yes

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and then evaluate each term individually

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in order to do this

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you have to know the domain of your inverse trigonometry

compact spire Mar 20, 2020, 4:16 AM

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yeah I am lost with finding the domain of the inverse.

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So is this where I find out which quadrant they are in?

cinder portal Mar 20, 2020, 4:23 AM

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yea do that

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figure out which trig function has domain in which quadrants

compact spire Mar 20, 2020, 4:24 AM

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so a = Q2 and b = Q1

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Okay, so I've been trying to figure out the rest of this and im having trouble with the 2/7 part. when completing the triangle I keep getting sqaure root of 53.

rich wolf Mar 20, 2020, 4:32 AM

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And whats the problem with sqrt53

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Is it not pretty enough

compact spire Mar 20, 2020, 4:33 AM

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Im just not sure if im doing it wrong

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so it would be (-3/5)(7sqrt53/53)-(2sqrt53/53)(-4/5)

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I think I know where I messed up.

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Thanks for the help @cinder portal and @rich wolf

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the (-4/5) is supposed to be (4/5)

plucky jacinth Mar 20, 2020, 4:47 AM

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https://cdn.discordapp.com/attachments/326138757474680852/690412071707344956/7fe93c093bf8b012c2facb940f33d1ad.png

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can anyone help me in a way that it will help

dark sparrow Mar 20, 2020, 4:50 AM

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@cinder portal parentheses

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$\sin(\alpha - \beta)$

somber coyoteBOT Mar 20, 2020, 4:50 AM

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Ann:

📎 183668144404037632.png

cinder portal Mar 20, 2020, 4:50 AM

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😦

dark sparrow Mar 20, 2020, 4:50 AM

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@plucky jacinth what is giving you trouble here?

plucky jacinth Mar 20, 2020, 4:50 AM

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all of it

dark sparrow Mar 20, 2020, 4:51 AM

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vague!

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okay, so how familiar are you with the basics of trigonometry

plucky jacinth Mar 20, 2020, 4:51 AM

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0

dark sparrow Mar 20, 2020, 4:51 AM

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does the mnemonic "SOH CAH TOA" ring any bells

cinder portal Mar 20, 2020, 4:51 AM

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lol i thought u hated mnemonic

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1gotem

dark sparrow Mar 20, 2020, 4:52 AM

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it's a definitional mnemonic. the one exception.

plucky jacinth Mar 20, 2020, 4:52 AM

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no i have not been to school for 2 weeks

dark sparrow Mar 20, 2020, 4:52 AM

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https://www.khanacademy.org/math/trigonometry/trigonometry-right-triangles

Khan Academy

Trigonometry with right triangles | Math | Khan Academy

Learn the basics of trigonometry: What are sine, cosine, and tangent? How can we use them to solve for unknown sides and angles in right triangles?

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have this, then.

compact spire Mar 20, 2020, 4:53 AM

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@cinder portal so with a calculator I got -0.79 etc and without using one I got 94.64 for that problem you were helping me with

cinder portal Mar 20, 2020, 4:53 AM

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try to get an exact answer

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calculator is cheating

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lol

compact spire Mar 20, 2020, 4:54 AM

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so 94.64 is better to use

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I hate using calculators but I use them to reference what I should aim for in an answer, should I not be doing that?

cinder portal Mar 20, 2020, 4:56 AM

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actually try to compute each term

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you can do it without a calculator

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and the answer comes out incredibly nice

compact spire Mar 20, 2020, 4:56 AM

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I want to find that answer so bad now

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okay

rich wolf Mar 20, 2020, 4:58 AM

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Use WA step by step

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Or symbolab

dark sparrow Mar 20, 2020, 4:58 AM

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symbolab thonkBan

compact spire Mar 20, 2020, 4:58 AM

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@rich wolf what are those?

dark sparrow Mar 20, 2020, 4:58 AM

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don't bother with symbolab. it's crap.

rich wolf Mar 20, 2020, 4:58 AM

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Say what you want about symbolab but its free

dark sparrow Mar 20, 2020, 4:59 AM

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it's free crap.

rich wolf Mar 20, 2020, 4:59 AM

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Photomath also works

dark sparrow Mar 20, 2020, 4:59 AM

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that's even more crap

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copswing

rich wolf Mar 20, 2020, 5:00 AM

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WA is of course top tier

compact spire Mar 20, 2020, 5:00 AM

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what is WA

rich wolf Mar 20, 2020, 5:00 AM

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But sometimes you dont feel like matching your grouping symbols

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Also kind of expensive

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@compact spire it is an easy way to get answers

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But dont use it unless you're really stuck on something

compact spire Mar 20, 2020, 5:03 AM

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Im really stuck on all my homework im doing because school is cancelled and my professor is leaving us to learn everything on our own. But thank you all for the help so far. I greatly appreciate it.

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OMG

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I think I got it

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(-3/5)(7sqrt53/53)-(2sqrt53/53)(4/5) = -29sqrt53/265

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@cinder portal

cinder portal Mar 20, 2020, 5:12 AM

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looks correct

compact spire Mar 20, 2020, 5:12 AM

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YES!

cinder portal Mar 20, 2020, 5:12 AM

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compare it with ur calculator

compact spire Mar 20, 2020, 5:12 AM

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my calculator says im an idiot

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checking now

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I GOT IT

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OMG I did the math!

cinder portal Mar 20, 2020, 5:13 AM

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nice

compact spire Mar 20, 2020, 5:14 AM

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Thank you so much

cinder portal Mar 20, 2020, 5:15 AM

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no need, u did all the work

compact spire Mar 20, 2020, 5:22 AM

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Is there a limit of help I can ask for?

rich wolf Mar 20, 2020, 5:24 AM

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@compact spire no more than once every minute

compact spire Mar 20, 2020, 5:24 AM

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Okay

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📎 trig3.png

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so im pretty sure I just sub the tan-1x for a and sin-1y for B and do the difference of cosine

cinder portal Mar 20, 2020, 5:26 AM

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yup

compact spire Mar 20, 2020, 5:26 AM

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which has me at cos a cos B - sin a sin B

cinder portal Mar 20, 2020, 5:26 AM

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the rest are easy

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ur formula is wrong though

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it should be a +

compact spire Mar 20, 2020, 5:27 AM

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oh really?

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I thought since it was minus it would be minus, is it the opposite of what is shown?

white cradle Mar 20, 2020, 5:27 AM

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ye

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cos(A+B) = cosAcosB - sinAsinB

compact spire Mar 20, 2020, 5:28 AM

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Okay thanks guys

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so I need to complete the triangle I assume

dark sparrow Mar 20, 2020, 5:30 AM

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well you WILL need to simplify $\cos(\arctan(x))$, $\sin(\arctan(x))$, $\cos(\arcsin(y))$ and $\sin(\arcsin(y))$

somber coyoteBOT Mar 20, 2020, 5:30 AM

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Ann:

📎 183668144404037632.png

compact spire Mar 20, 2020, 5:30 AM

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simplify?

dark sparrow Mar 20, 2020, 5:31 AM

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uh

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yes

cinder portal Mar 20, 2020, 5:31 AM

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do the same thing

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you got this

compact spire Mar 20, 2020, 5:32 AM

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Okay

cinder portal Mar 20, 2020, 5:32 AM

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$x = \frac{x}{1}$

somber coyoteBOT Mar 20, 2020, 5:32 AM

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Cytis:

📎 190017172074921986.png

cinder portal Mar 20, 2020, 5:32 AM

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that's the hint for ya

compact spire Mar 20, 2020, 5:33 AM

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📎 JPEG_20200319_223329.jpg

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so sqrt 1+x^2 = the hyp

dark sparrow Mar 20, 2020, 5:34 AM

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sqrt**(1+x^2)**

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now if only you'd marked the angle arctan(x) on the triangle

compact spire Mar 20, 2020, 5:35 AM

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how do you mean?

dark sparrow Mar 20, 2020, 5:37 AM

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i mean exactly what i said

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if you had marked the angle arctan(x) in your triangle, you'd be able to find its sine and cosine with ease now

compact spire Mar 20, 2020, 5:37 AM

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Im sorry, my brain isnt processing so well. been working like crazy

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Okay, I think I got it

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$frac{sqrt{1-y^2} + xy}{sqrt{1 +x^2}}$

somber coyoteBOT Mar 20, 2020, 5:47 AM

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Pony Boy:

📎 98531807913349120.png

dark sparrow Mar 20, 2020, 5:47 AM

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\ \ \ \ \ \

compact spire Mar 20, 2020, 5:48 AM

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Was trying to learn but didnt workout

dark sparrow Mar 20, 2020, 5:48 AM

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you droped some of these

compact spire Mar 20, 2020, 5:48 AM

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$\frac{sqrt{1-y^2} + xy}{sqrt{1 +x^2}}$

somber coyoteBOT Mar 20, 2020, 5:48 AM

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Pony Boy:

📎 98531807913349120.png

compact spire Mar 20, 2020, 5:48 AM

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omg lol

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$\frac{\sqrt{1-y^2} + xy}{\sqrt{1 +x^2}}$

somber coyoteBOT Mar 20, 2020, 5:49 AM

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Pony Boy:

📎 98531807913349120.png

compact spire Mar 20, 2020, 5:49 AM

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Okay, thats what I got.

dark sparrow Mar 20, 2020, 5:49 AM

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for the final answer?

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looks sus

compact spire Mar 20, 2020, 5:49 AM

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I will look over it again then

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📎 JPEG_20200319_225434.jpg

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Okay so im back to this point

dark sparrow Mar 20, 2020, 5:55 AM

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ok so this actually was alright and it really did only look sus and that's it

compact spire Mar 20, 2020, 5:56 AM

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so it was right

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Im so nervous i feel sick. lol

upper karma Mar 20, 2020, 6:11 AM

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📎 unknown.png

dense trail Mar 20, 2020, 6:16 AM

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Is e the ans

upper karma Mar 20, 2020, 6:18 AM

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Is e the ans
@dense trail 😦

dense trail Mar 20, 2020, 6:18 AM

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Option e

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I am just asking bec of the black mark

upper karma Mar 20, 2020, 6:19 AM

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Is e the ans
@dense trail the ans is c

dense trail Mar 20, 2020, 6:20 AM

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Are there any relation betwwn the radii of the circle

upper karma Mar 20, 2020, 6:27 AM

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Are there any relation betwwn the radii of the circle
@dense trail idk

dense trail Mar 20, 2020, 6:28 AM

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Its not working

compact spire Mar 20, 2020, 6:39 AM

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Okay im back, does anyone know what I should google to learn how to solve this problem? I have 0 clue on how to solve this

📎 trig4.png

rich wolf Mar 20, 2020, 6:40 AM

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@compact spire use a graphing calculator

compact spire Mar 20, 2020, 6:41 AM

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I don't have one.

rich wolf Mar 20, 2020, 6:41 AM

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Uh desmos should work

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Or maybe WA

compact spire Mar 20, 2020, 6:42 AM

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📎 c09d4e10fd38d9e6c5ff490eb3fe17f4.png

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like this?

rich wolf Mar 20, 2020, 6:43 AM

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Sure

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Remember the domain

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2pi is 6.28

compact spire Mar 20, 2020, 6:47 AM

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so .14 and 3.0

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Thanks for the help @rich wolf

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📎 trig5.png

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the formula is 2sinxcosx

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=2(3/5)(4/5)

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=2(12/5)

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am I on the right track?

umbral snow Mar 20, 2020, 7:01 AM

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That's exactly right

compact spire Mar 20, 2020, 7:01 AM

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=24/5 ?

umbral snow Mar 20, 2020, 7:02 AM

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Oh oop. (3/5)(4/5) = 12/25

compact spire Mar 20, 2020, 7:02 AM

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okay

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=2(12/25)

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=24/25

umbral snow Mar 20, 2020, 7:03 AM

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You can also use a calculator to confirm the answer. It is correct here.

compact spire Mar 20, 2020, 7:04 AM

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Awesome, thank you

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📎 trig6.png

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okay so half angle formula would be cosx/2 = +- sqrt 1+cos2x/2

umbral snow Mar 20, 2020, 7:08 AM

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I tried

compact spire Mar 20, 2020, 7:08 AM

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I will try it again

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hopefully it works lol

umbral snow Mar 20, 2020, 7:09 AM

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cos(x/2) = ±√[(1 + cos(x))/2]

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So you want to plug in x = 45 here

somber coyoteBOT Mar 20, 2020, 7:10 AM

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Pony Boy:

$ cos(\frac{x}{2}) = ± \frac{\sqrt{1+cos(x)}{2}$
```Compile error! Output:

! File ended while scanning use of \frac .
<inserted text>
\par
<*> 98531807913349120.tex

I suspect you have forgotten a }', causing me to read past where you wanted me to stop. I'll try to recover; but if the error is serious, you'd better type E' or `X' now and fix your file.

📎 98531807913349120.png

compact spire Mar 20, 2020, 7:11 AM

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nevermind

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lol

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but you get it

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where did you get 45? @umbral snow

umbral snow Mar 20, 2020, 7:12 AM

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If x is 45, then the left side is cos(22.5) which is what we want

compact spire Mar 20, 2020, 7:12 AM

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Okay thats what I was thinking

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I thought I had to plug in 22.5 in for x

umbral snow Mar 20, 2020, 7:12 AM

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You could try it, but you'll find quickly it doesn't work here

compact spire Mar 20, 2020, 7:13 AM

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Okay

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so ±√[(1+cos(45))/2]

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do I just plug this into a calculator?

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wait

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cos of 45 is just √2/2

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so = √[(1+√2/2)/2]

umbral snow Mar 20, 2020, 7:18 AM

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Yep, that's a way you can express it

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You can clean it up a little by multiplying numerator and denominator by 2

compact spire Mar 20, 2020, 7:20 AM

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$=\frac{\sqrt{2+\sqrt{2}}}{2}$

somber coyoteBOT Mar 20, 2020, 7:20 AM

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Pony Boy:

📎 98531807913349120.png

umbral snow Mar 20, 2020, 7:20 AM

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Yus that works

compact spire Mar 20, 2020, 7:21 AM

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Okay

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📎 trig7.png

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Okay, this I was never taught as well

dark sparrow Mar 20, 2020, 7:23 AM

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double angle identities no bueno?

compact spire Mar 20, 2020, 7:23 AM

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Well I was supposed to learn double and half angle this week but schools been cancelled so im trying to learn on my own

umbral snow Mar 20, 2020, 7:23 AM

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Nothing you aren't capable of. That ² should scream "quadratic" at you, so see if you can get rid of the cos(2x)

dark sparrow Mar 20, 2020, 7:23 AM

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

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this is the only identity you need here really

upper karma Mar 20, 2020, 7:24 AM

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Does anyone know this one?

📎 image0.jpg

compact spire Mar 20, 2020, 7:26 AM

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So I should get the cos2x alone?

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cos2x = 2-2cos^2x

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?

dark sparrow Mar 20, 2020, 7:27 AM

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bruh.

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

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this identity. apply it.

compact spire Mar 20, 2020, 7:30 AM

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so far im at 4cos^2(x)-3=0

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am I on the right track so far?

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cos(x)= √(3/2)

umbral snow Mar 20, 2020, 7:37 AM

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Gotem

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Wait you mean
cos(x) = √3 / 2

compact spire Mar 20, 2020, 7:37 AM

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yeah

umbral snow Mar 20, 2020, 7:37 AM

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If you're lit with your unit circle, that should ring a bell

compact spire Mar 20, 2020, 7:38 AM

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pi/6

umbral snow Mar 20, 2020, 7:39 AM

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Also [0, 2π) solution range

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You're correct with π/6

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The other would have been 11π/6

compact spire Mar 20, 2020, 7:40 AM

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11pi/6

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okay

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I need a break, thanks for the help.

viscid ginkgo Mar 20, 2020, 7:49 AM

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can someone please explain to me why these are parallel, coplanar lines? https://i.imgur.com/kBH8GFe.png

Imgur

dark sparrow Mar 20, 2020, 7:51 AM

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@compact spire what about cos(x) = -sqrt(3)/2

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you forgot that pensive_grapes

plucky jacinth Mar 20, 2020, 4:52 PM

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📎 6e41ae6cee0a0a710c8ff15e0ada9d05.png

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this hurts me

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i have 2 hours to take and turn in a test i have no idea on how do this

dark sparrow Mar 20, 2020, 4:54 PM

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this is arccos(13/20)

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simply by SOH CAH TOA

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there is nothing else to this

plucky jacinth Mar 20, 2020, 4:59 PM

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40.5416018735

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@dark sparrow 49.4583981265

dark sparrow Mar 20, 2020, 5:03 PM

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were you asked to round to ten decimal places

plucky jacinth Mar 20, 2020, 5:03 PM

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yes

dark sparrow Mar 20, 2020, 5:03 PM

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to ten decimal places?

stone quiver Mar 20, 2020, 5:03 PM

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📎 IMG_20200320_223244.jpg

plucky jacinth Mar 20, 2020, 5:03 PM

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📎 ec8c395236477c056bbfca8eb4324040.png

stone quiver Mar 20, 2020, 5:03 PM

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I am getting 64/40 can someone check?

dark sparrow Mar 20, 2020, 5:04 PM

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@plucky jacinth ANGLES should be rounded to the NEAREST DEGREE, and even otherwise there's a BIG DIFFERENCE between "TEN DECIMAL PLACES" and "TO THE NEAREST TENTH"

plucky jacinth Mar 20, 2020, 5:04 PM

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so 49.5

dark sparrow Mar 20, 2020, 5:04 PM

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ANGLES.

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should be rounded.

plucky jacinth Mar 20, 2020, 5:04 PM

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49

dark sparrow Mar 20, 2020, 5:04 PM

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to the NEAREST DEGREE.

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there we go

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rounding π to the nearest tenth gives 3.1
while rounding π to ten decimal places gives 3.1415926536

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just so you know

plucky jacinth Mar 20, 2020, 5:06 PM

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📎 7eb7d247148c5391ae4fd5cd86024e2b.png

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27/38

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arctan

stone quiver Mar 20, 2020, 5:07 PM

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📎 IMG_20200320_223244.jpg

plucky jacinth Mar 20, 2020, 5:07 PM

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35

paper mauve Mar 20, 2020, 5:07 PM

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

somber coyoteBOT Mar 20, 2020, 5:07 PM

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

📎 364570129171742732.png

paper mauve Mar 20, 2020, 5:08 PM

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so since we want theta we take the inverse of tangent

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of both sides

plucky jacinth Mar 20, 2020, 5:08 PM

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soo arctan(27/38)

paper mauve Mar 20, 2020, 5:08 PM

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yes

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bingo

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hype

plucky jacinth Mar 20, 2020, 5:09 PM

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35.~~394795845~~

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ya

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so 35

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📎 56ab8d64a0af8e8212ed1705aaf73ff2.png

paper mauve Mar 20, 2020, 5:12 PM

#

so the thing here i want you to do is to draw what is given

#

@plucky jacinth

#

if you can, take a picture of your drawing with the given dimensions in the right places

plucky jacinth Mar 20, 2020, 5:14 PM

#

sin68 x/15

paper mauve Mar 20, 2020, 5:15 PM

#

$$sin(68)=\frac{x}{15}

somber coyoteBOT Mar 20, 2020, 5:15 PM

#

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

📎 364570129171742732.png

paper mauve Mar 20, 2020, 5:15 PM

#

yes

#

very good

#

@plucky jacinth

compact spire Mar 20, 2020, 5:17 PM

#

📎 trig8.png

plucky jacinth Mar 20, 2020, 5:17 PM

#

how would i put that into a a calculator

dark sparrow Mar 20, 2020, 5:18 PM

#

@paper mauve \sin

#

@plucky jacinth you wouldn't, unless you're allergic to algebra you should be able to solve this rather simple equation for x

#

$x = 15 \sin(68)$

somber coyoteBOT Mar 20, 2020, 5:18 PM

#

Ann:

📎 183668144404037632.png

compact spire Mar 20, 2020, 5:18 PM

#

so I googled double angle identities and I have 3 options I guess for cos2(x) so I can use 2cos^2(x)-1

#

is that correct?

paper mauve Mar 20, 2020, 5:19 PM

#

@dark sparrow ty

compact spire Mar 20, 2020, 5:19 PM

#

$2cos^2(x) - 1$

#

my answer from my problem was -7/25 can anyone confirm that for me?

somber coyoteBOT Mar 20, 2020, 5:22 PM

#

Pony Boy:

📎 98531807913349120.png

compact spire Mar 20, 2020, 5:22 PM

#

$2(\frac{3}{5})^2 - 1$

somber coyoteBOT Mar 20, 2020, 5:23 PM

#

Pony Boy:

📎 98531807913349120.png

compact spire Mar 20, 2020, 5:23 PM

#

$2(\frac{9}{25})-1$

somber coyoteBOT Mar 20, 2020, 5:23 PM

#

Pony Boy:

📎 98531807913349120.png

compact spire Mar 20, 2020, 5:24 PM

#

$\frac{18}{25}-\frac{25}{25}$

somber coyoteBOT Mar 20, 2020, 5:24 PM

#

Pony Boy:

📎 98531807913349120.png

compact spire Mar 20, 2020, 5:24 PM

#

$= -\frac{7}{25}$

somber coyoteBOT Mar 20, 2020, 5:24 PM

#

Pony Boy:

📎 98531807913349120.png

compact spire Mar 20, 2020, 5:26 PM

#

Can someone help me with reducing cos^4(x) to the first power?

stone quiver Mar 20, 2020, 5:27 PM

#

Can you reduce cos²x into first power ?

compact spire Mar 20, 2020, 5:29 PM

#

ehhh

#

not really

stone quiver Mar 20, 2020, 5:30 PM

#

wait, you want linear terms with the same angle i.e x and not like 2x or sth?

compact spire Mar 20, 2020, 5:31 PM

#

📎 trig9.png

#

My professor never taught us anything like this so I'm trying to learn on my own since he can't

stone quiver Mar 20, 2020, 5:33 PM

#

Isn't $((1+cos2 x)/2)^2)$ a valid answer then?

somber coyoteBOT Mar 20, 2020, 5:33 PM

#

Vatsal:

📎 327366753619279873.png

compact spire Mar 20, 2020, 5:34 PM

#

How would I get rid of the fraction in that situation

#

because it says to factor out all fractions

#

so $cos^2(x) = \frac{1}{2}[1+cos2(x)]$

somber coyoteBOT Mar 20, 2020, 5:35 PM

#

Pony Boy:

📎 98531807913349120.png

compact spire Mar 20, 2020, 5:35 PM

#

that's correct?

stone quiver Mar 20, 2020, 5:35 PM

#

yes

compact spire Mar 20, 2020, 5:35 PM

#

okay so if it were the fourth power it would just be

#

$cos^4(x) = (\frac{1}{2}[1+cos2(x)])^2$

somber coyoteBOT Mar 20, 2020, 5:36 PM

#

Pony Boy:

📎 98531807913349120.png

stone quiver Mar 20, 2020, 5:36 PM

#

Yea exactly

compact spire Mar 20, 2020, 5:36 PM

#

Okay, so far so good.

#

I have the $\frac{1}{2}$

somber coyoteBOT Mar 20, 2020, 5:37 PM

#

Pony Boy:

📎 98531807913349120.png

compact spire Mar 20, 2020, 5:37 PM

#

which I need to get rid of as it says in the problem

#

how would I go about doing that

#

or am I just overthinking it

gentle dome Mar 20, 2020, 8:10 PM

#

I have a strange question.

#

When I was looking at how Archimedes approximated Pi, I was wondering about a few assumptions.

#

When he was inscribing and circumscribing a circle with polygons, how can he assume that the circumference is between the two perimeters?

📎 image0.png

#

I can also draw a picture like this.

#

📎 image0.png

#

In my 2nd diagram, you can easily see that the white perimeter is way bigger than the outer square.

#

If he were comparing the area of the circle with the area of the two polygons, it would be obvious to understand, but I’m troubled that he used perimeters instead.

#

Can anyone help me in any information or step I’m missing?

upper karma Mar 20, 2020, 9:42 PM

#

Can sum 1 help me with number 4 on this

#

📎 image0.jpg

languid hemlock Mar 20, 2020, 10:20 PM

#

Can someone help me with #17

📎 image0.jpg

#

I dont know how to find AE

silent plank Mar 20, 2020, 11:17 PM

#

bisecting and intersecting chords theorem

rain galleon Mar 21, 2020, 1:24 AM

#

📎 JPEG_20200320_212350.jpg

#

We are given 2 radius's and the distance between centers, no other information, how do I find the area of the intersection?

supple wedge Mar 21, 2020, 1:39 AM

#

can u send the full qn

upper karma Mar 21, 2020, 10:36 AM

#

Hi, can someone help me out with this triangle? I'm given the angle in A and told that the bisectors of B and C intersect at I (incenter). I have to find the angle BIC.

📎 IMG_20200321_073257.jpg

#

,rotate

somber coyoteBOT Mar 21, 2020, 10:38 AM

#

📎 326391424188678144.png

upper karma Mar 21, 2020, 11:15 AM

#

nvm

crisp phoenix Mar 21, 2020, 10:09 PM

#

hey, so i have a quiz on circles coming up, would anyone be able to help me find the answers while im doing it? would highly appreciate it since theres some very smart ppl in here

idle bloom Mar 22, 2020, 12:04 AM

#

no

#

we will help you understand the concepts before the quiz

#

but while you're doing the quiz you're on your own

viscid ginkgo Mar 22, 2020, 2:28 AM

#

https://i.imgur.com/1MATzza.png

Imgur

#

how lis line BC perpendicular to line l?

silent plank Mar 22, 2020, 2:31 AM

#

it's not?

#

who's saying it is

viscid ginkgo Mar 22, 2020, 2:58 AM

#

I drew it wrong my bad

#

someone else answered , thanks though.

golden panther Mar 22, 2020, 4:13 AM

#

YO! guys how do i find sin^2(15°) and cos^2(22.5°)

#

Plz guys?

coral sapphire Mar 22, 2020, 4:19 AM

#

Is there any comments on how you should do it?

golden panther Mar 22, 2020, 4:19 AM

#

No

coral sapphire Mar 22, 2020, 4:19 AM

#

Then, just plug it into the calculator if you don't need an exact form

golden panther Mar 22, 2020, 4:19 AM

#

Wth

silent plank Mar 22, 2020, 4:20 AM

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they probably do want exact form

#

consider half angle identities

golden panther Mar 22, 2020, 4:20 AM

#

Ok i see

#

Thx Fam

ionic bluff Mar 22, 2020, 5:35 AM

#

hello friends

#

i have a kind of dumb question

#

how exactly is multiplication and division defined geometrically

#

i've been thinking about the tangent function

#

and how it is sin/cos

#

i get that sin is vertical

#

and cos is the horizontal part

#

but what does the tangent of that tirangle look like

dark sparrow Mar 22, 2020, 5:53 AM

#

how exactly is multiplication and division defined geometrically
wdym

livid epoch Mar 22, 2020, 6:41 AM

#

https://cdn.discordapp.com/attachments/499124802154659841/691169670622871592/image0.jpg

#

$arg(\frac{z_2-z_1}{z_3-z_2})$ what is argument coudl someone help?

somber coyoteBOT Mar 22, 2020, 6:41 AM

#

Space Pirate Wesley:

📎 255238880494616576.png

livid epoch Mar 22, 2020, 6:41 AM

#

please this is very tricky

#

@umbral snow please this is very tricky

umbral snow Mar 22, 2020, 6:46 AM

#

but what does the tangent of that tirangle look like
@ionic bluff
The tangent is the slope of the terminal arm

#

z2 - z1 is the line AB

livid epoch Mar 22, 2020, 6:47 AM

#

ye

umbral snow Mar 22, 2020, 6:47 AM

#

z3 - z2 is the line BC

#

When you divide one complex number by another, you subtract their angles

#

That is,
Arg(a/b) = Arg(a) - Arg(b)

livid epoch Mar 22, 2020, 6:48 AM

#

yes yes

#

📎 image0.jpg

#

i got marked incorrect i do not know why

umbral snow Mar 22, 2020, 6:50 AM

#

I'm not crazy certain what the arg of either is

reef wren Mar 22, 2020, 8:38 AM

#

Hi. I have some struggles with proofs in coordinate geometry. From the textbooks I saw, it seems to be common practice to only prove statements in the first quadrant. So, for example, say I wanna find general equation for a line that has intercepts with axes of x and y at a and b at the respective axes. Then its equation will be:

#

$\frac{x}{a} + \frac{y}{b} = 1$

somber coyoteBOT Mar 22, 2020, 8:39 AM

#

TimkaSolnze:

📎 655737374755520513.png

reef wren Mar 22, 2020, 8:40 AM

#

And almost always, when authors prove it, they 1. Pick up a line that has positive a and b intercepts, 2. Pick up a point on the line that is in first quadrant.

#

I want to know, why do they do so? Is there's a reason why we don't have to prove statements in other quadrants? Does proof in the first quadrant immediately applies to any other quadrant? Is there a shortcut(s) for proving for all the quadrants and their combinations(i.e., say a line might occupy 1, 2 and 4 quadrants, or 1, 2 and 3.)?

dense trail Mar 22, 2020, 8:46 AM

#

Well may be becasuse of the sense

#

@livid epoch

steady forum Mar 22, 2020, 9:26 AM

#

📎 JPEG_20200322_172729.jpg

#

Is answer cos(90-theta)=sine theta

dark sparrow Mar 22, 2020, 9:27 AM

#

there's like 6 problems here all with multiple subproblems

steady forum Mar 22, 2020, 9:27 AM

#

So its cos(90-35)=sine 55

dark sparrow Mar 22, 2020, 9:27 AM

#

which one are you talking about

steady forum Mar 22, 2020, 9:27 AM

#

3i

#

3i my bad

#

So sine 55= cos 35

#

Right

dark sparrow Mar 22, 2020, 9:28 AM

#

yes

steady forum Mar 22, 2020, 9:28 AM

#

So

#

Its because of formula right thats why i get 35

#

Cos(90-x)=sinx

steady forum Mar 22, 2020, 9:49 AM

#

📎 JPEG_20200322_175015.jpg

#

Question 3ii

silent plank Mar 22, 2020, 9:51 AM

#

zoom in please

#

we aren't hawks

steady forum Mar 22, 2020, 9:51 AM

#

3ii

📎 JPEG_20200322_175234.jpg

#

📎 JPEG_20200322_175318.jpg

#

Im wrong alr

#

All my workings r wrong

#

Anyone

silent plank Mar 22, 2020, 9:57 AM

#

the work is unclear

#

atm i can only see sin(40°) = cos(50°)

steady forum Mar 22, 2020, 9:58 AM

#

My workings r all not relevant to question

silent plank Mar 22, 2020, 9:59 AM

#

can you do anything/ apply any identities to:
cos(310°)?

steady forum Mar 22, 2020, 9:59 AM

#

Not to my knowledge

silent plank Mar 22, 2020, 9:59 AM

#

express in terms of relative angle?

steady forum Mar 22, 2020, 10:00 AM

#

Relative angle?

#

-cos50?

silent plank Mar 22, 2020, 10:00 AM

#

not -

steady forum Mar 22, 2020, 10:00 AM

#

Cos(-50)

#

?

white cradle Mar 22, 2020, 10:00 AM

#

thats just cos50

silent plank Mar 22, 2020, 10:00 AM

#

^

steady forum Mar 22, 2020, 10:00 AM

#

Z

white cradle Mar 22, 2020, 10:00 AM

#

cos is positive in fourth quadrant

steady forum Mar 22, 2020, 10:00 AM

#

Rite

silent plank Mar 22, 2020, 10:01 AM

#

which gets you: $\cos^2(50\deg)$

somber coyoteBOT Mar 22, 2020, 10:01 AM

#

ramonov:

📎 220517274845577216.png

silent plank Mar 22, 2020, 10:01 AM

#

and you should be able to get the required answer now

steady forum Mar 22, 2020, 10:02 AM

#

Idgi

silent plank Mar 22, 2020, 10:02 AM

#

idgi?

steady forum Mar 22, 2020, 10:02 AM

#

I dun gte it

#

Get

white cradle Mar 22, 2020, 10:02 AM

#

i dont get it

#

lol

silent plank Mar 22, 2020, 10:02 AM

#

how ironic

white cradle Mar 22, 2020, 10:02 AM

#

whats the question actually

silent plank Mar 22, 2020, 10:02 AM

#

3ii

steady forum Mar 22, 2020, 10:03 AM

#

3ii

silent plank Mar 22, 2020, 10:03 AM

#

here you can apply something very famous

white cradle Mar 22, 2020, 10:03 AM

#

i cant read it ;-;

silent plank Mar 22, 2020, 10:03 AM

#

those most common trig identity

white cradle Mar 22, 2020, 10:03 AM

#

nvm

#

got it

silent plank Mar 22, 2020, 10:03 AM

#

which is essentially pythagoras

steady forum Mar 22, 2020, 10:03 AM

#

Wtf

#

Idgi

silent plank Mar 22, 2020, 10:04 AM

#

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

somber coyoteBOT Mar 22, 2020, 10:04 AM

#

ramonov:

📎 220517274845577216.png

steady forum Mar 22, 2020, 10:04 AM

#

Why wld i apply that

#

I dont have any squares

white cradle Mar 22, 2020, 10:04 AM

#

because you know sin(50)=b

steady forum Mar 22, 2020, 10:04 AM

#

I dont even know how u got cos^250

#

Cos^2(50)

silent plank Mar 22, 2020, 10:04 AM

#

you had sin(40°) = cos(50°) right?

steady forum Mar 22, 2020, 10:05 AM

#

Yes

white cradle Mar 22, 2020, 10:05 AM

#

oh lmao

#

right

silent plank Mar 22, 2020, 10:05 AM

#

and we established just then that cos(310°) is also equal to cos(50°) right?

steady forum Mar 22, 2020, 10:05 AM

#

Yes

white cradle Mar 22, 2020, 10:05 AM

#

so there is your cos^2(50)

steady forum Mar 22, 2020, 10:05 AM

#

Wait

#

Wait why is it equals again

white cradle Mar 22, 2020, 10:05 AM

#

oof

silent plank Mar 22, 2020, 10:05 AM

#

which gets you
cos(50°) * cos(50°) = cos^2(50°)

white cradle Mar 22, 2020, 10:06 AM

#

cos(310) = cos(-50) = cos(50)

silent plank Mar 22, 2020, 10:06 AM

#

cos is 360° periodic and even

steady forum Mar 22, 2020, 10:06 AM

#

Oh ya cuz in

#

Then 4th quad

#

Everything becomes pos

#

Cos(50) =/= b or a

#

Sine 40 =/= b or a

white cradle Mar 22, 2020, 10:08 AM

#

sin(40)=cos(50)

#

are you even reading ramonov's messages

steady forum Mar 22, 2020, 10:08 AM

#

Sine 40 isnt b or a in the question

white cradle Mar 22, 2020, 10:08 AM

#

sin(40)=cos(50)

#

sin(40)=cos(50)

#

sin(40)=cos(50)

#

sin(40)=cos(50)

steady forum Mar 22, 2020, 10:08 AM

#

Yes i got that

#

In my working

#

From sine 90- theta= cos thea

white cradle Mar 22, 2020, 10:08 AM

#

so the whole expression becomes cos^2(50)

silent plank Mar 22, 2020, 10:08 AM

#

which gets you
cos(50°) * cos(50°) = cos^2(50°)

white cradle Mar 22, 2020, 10:09 AM

#

^

steady forum Mar 22, 2020, 10:09 AM

#

They want me to express the equation in b or a

white cradle Mar 22, 2020, 10:09 AM

#

that is what we're trying to do..

silent plank Mar 22, 2020, 10:09 AM

#

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

somber coyoteBOT Mar 22, 2020, 10:09 AM

#

ramonov:

📎 220517274845577216.png

silent plank Mar 22, 2020, 10:09 AM

#

$\sin^2(50\deg) + \cos^2(50\deg) = 1$

somber coyoteBOT Mar 22, 2020, 10:09 AM

#

ramonov:

📎 220517274845577216.png

#

FlynnXD:

📎 216861413157765120.png

silent plank Mar 22, 2020, 10:10 AM

#

do you see a or b anywhere there?

white cradle Mar 22, 2020, 10:10 AM

#

aw cmon

silent plank Mar 22, 2020, 10:10 AM

#

hah

steady forum Mar 22, 2020, 10:10 AM

#

Sine(50)=b

white cradle Mar 22, 2020, 10:10 AM

#

custom tex 😠

steady forum Mar 22, 2020, 10:10 AM

#

Sine^2(50)= b

white cradle Mar 22, 2020, 10:10 AM

#

no

silent plank Mar 22, 2020, 10:10 AM

#

no

steady forum Mar 22, 2020, 10:10 AM

#

B^2

silent plank Mar 22, 2020, 10:10 AM

#

still questionable

white cradle Mar 22, 2020, 10:10 AM

#

^

silent plank Mar 22, 2020, 10:10 AM

#

math is case senstive

white cradle Mar 22, 2020, 10:10 AM

#

lol

steady forum Mar 22, 2020, 10:10 AM

#

b^2

white cradle Mar 22, 2020, 10:10 AM

#

yes

#

so..

steady forum Mar 22, 2020, 10:11 AM

#

Haiz

silent plank Mar 22, 2020, 10:11 AM

#

and what's your final answer

steady forum Mar 22, 2020, 10:14 AM

#

Cos^2(50)=1-b^2

#

Because its essentially cos(50) × cos (50)

silent plank Mar 22, 2020, 10:15 AM

#

lowercase cos but seems ok

steady forum Mar 22, 2020, 10:16 AM

#

Thanks

vague lance Mar 22, 2020, 1:33 PM

#

📎 image0.jpg

#

This can’t be right

#

?

lament pumice Mar 22, 2020, 1:37 PM

#

Not at all.

vague lance Mar 22, 2020, 1:39 PM

#

do you think you could explain it

lament pumice Mar 22, 2020, 1:44 PM

#

Sure, the formula you wrote for a cosine of an angle works only for angles between cathetus and hypothenuse

supple wedge Mar 22, 2020, 1:55 PM

#

@vague lance why would u choose 90

#

use theta

#

cosθ=AB/1
=AB

golden panther Mar 22, 2020, 9:10 PM

#

YO! do i need to calculator to find the max of y=2cos^2(x/2) in the closed interval from -π/2 to 3π/2?

burnt bolt Mar 22, 2020, 10:14 PM

#

No. Think about the shape of the cos(x/2) graph and where it would be a maximum then work out what the maximum y value for your specific function would be in that interval

rapid pollen Mar 23, 2020, 2:16 AM

#

I don’t understand the law of sine

somber coyoteBOT Mar 23, 2020, 2:22 AM

#

Beef101:

📎 473975912166064140.png

coral sapphire Mar 23, 2020, 2:37 AM

#

What do you not get?

rapid pollen Mar 23, 2020, 2:37 AM

#

I’ll show you

#

📎 image0.jpg

#

📎 image0.jpg

#

I just don’t understand how to use the formula

#

Or law

coral sapphire Mar 23, 2020, 2:42 AM

#

A better way to phrase it in English is that the law of sines states "the ratio of any side and the sine of the opposite angle are the same"

#

Since you already have the triangles labelled, you can just plug values into the equation

#

Lets look at the last triangle you posted

rapid pollen Mar 23, 2020, 2:43 AM

#

Ok

coral sapphire Mar 23, 2020, 2:44 AM

#

Wait, you did it correctly

rapid pollen Mar 23, 2020, 2:44 AM

#

Wait for angles is it b/B sin

#

And for degrees is it B/bsin

#

Or is it the same for both

coral sapphire Mar 23, 2020, 2:45 AM

#

Neither of those statements are valid

#

What is the argument of sin?

rapid pollen Mar 23, 2020, 2:46 AM

#

Opposite over hypotenuse

coral sapphire Mar 23, 2020, 2:47 AM

#

Wait for angles is it b/B sin
@rapid pollen

rapid pollen Mar 23, 2020, 2:47 AM

#

Yes

coral sapphire Mar 23, 2020, 2:47 AM

#

I meant what is this referring to? Sin has no argument

rapid pollen Mar 23, 2020, 2:48 AM

#

I mean for the equation a/sin A= b/Bsin

coral sapphire Mar 23, 2020, 2:49 AM

#

b/sinB?

rapid pollen Mar 23, 2020, 2:49 AM

#

I’ll send

#

📎 image0.jpg

coral sapphire Mar 23, 2020, 2:50 AM

#

Ok, the work is perfect

#

What's wrong thonkzoom

rapid pollen Mar 23, 2020, 2:51 AM

#

Oh it is

coral sapphire Mar 23, 2020, 2:51 AM

#

Yeah, this is correct

rapid pollen Mar 23, 2020, 2:52 AM

#

I’m just tired I should go to sleep lol. I’m studying for pre calculus after trigonometry what’s next?

coral sapphire Mar 23, 2020, 2:52 AM

#

Law of cosines if you didn't learn it yet

#

Then, you will probably learn about graphing trig functions

rapid pollen Mar 23, 2020, 2:53 AM

#

Is that the same as law of sines

coral sapphire Mar 23, 2020, 2:53 AM

#

No

#

Law of cosines and law of sines are different

rapid pollen Mar 23, 2020, 2:53 AM

#

Ok

#

Does graphing and functions use algebra 2

coral sapphire Mar 23, 2020, 2:54 AM

#

Uhhh, yeah

rapid pollen Mar 23, 2020, 2:54 AM

#

Sounds great 👍

crisp phoenix Mar 23, 2020, 3:03 AM

#

hi

#

can someone pls help me w this

#

📎 image0.jpg

coral sapphire Mar 23, 2020, 3:06 AM

#

Is there more info?

crisp phoenix Mar 23, 2020, 3:06 AM

#

no i dont think so

#

i just need to find the measure of angle ABO

coral sapphire Mar 23, 2020, 3:14 AM

#

Not enough info

#

Also, it says correct answer below is 22 degrees

crisp phoenix Mar 23, 2020, 3:15 AM

#

yeah thats what i think it is

coral sapphire Mar 23, 2020, 3:15 AM

#

Wdym think

crisp phoenix Mar 23, 2020, 3:15 AM

#

also poco this one

coral sapphire Mar 23, 2020, 3:15 AM

#

That is the answer

crisp phoenix Mar 23, 2020, 3:16 AM

#

📎 image0.jpg

#

theres

#

the full question so

#

those answers are random not correct

coral sapphire Mar 23, 2020, 3:16 AM

#

Try it yourself first

crisp phoenix Mar 23, 2020, 3:16 AM

#

i did

#

thats why i need help

coral sapphire Mar 23, 2020, 3:17 AM

#

So the stuff entered is what you think is correct

crisp phoenix Mar 23, 2020, 3:17 AM

#

no

#

pretend they arent there

coral sapphire Mar 23, 2020, 3:19 AM

#

Last 3 are wrong

#

The angles are not exactly the arc the encompass

crisp phoenix Mar 23, 2020, 3:22 AM

#

ok poco

#

how would i find angle WVU

coral sapphire Mar 23, 2020, 3:24 AM

#

What is arc WTU

crisp phoenix Mar 23, 2020, 3:24 AM

#

230 degrees

coral sapphire Mar 23, 2020, 3:25 AM

#

So do you know any theorems relating an inscribed angle to the arc it intercepts

crisp phoenix Mar 23, 2020, 3:25 AM

#

ok wvu would be 115 correct?

#

yeah the intercepted arc is half the length of the arc length right?

coral sapphire Mar 23, 2020, 3:27 AM

#

Yes

#

Now do the same for the rest

crisp phoenix Mar 23, 2020, 3:27 AM

#

21 for the last 2 right

#

?

coral sapphire Mar 23, 2020, 3:29 AM

#

Should be

crisp phoenix Mar 23, 2020, 3:35 AM

#

ok poco

#

i have another one

#

📎 image0.png

#

how would i go about finding this

#

nvm i got

#

it

coral sapphire Mar 23, 2020, 3:38 AM

#

Please stop pinging me for every question, expecting me to be there

quiet mason Mar 23, 2020, 3:42 AM

#

what have you tried btw

#

what formulas do you know

coral sapphire Mar 23, 2020, 3:42 AM

#

He already got it

viscid ginkgo Mar 23, 2020, 5:56 AM

#

Would this be a case of alternate interior angles?

📎 JPEG_20200322_225557.jpg

#

And the line is not tranversal right, it would have to be a straight angle

#

So the sum should be 80 + 30?

silent plank Mar 23, 2020, 6:01 AM

#

,rotate

somber coyoteBOT Mar 23, 2020, 6:01 AM

#

📎 220517274845577216.png

viscid ginkgo Mar 23, 2020, 6:02 AM

#

sorry about that, cool bot though

silent plank Mar 23, 2020, 6:02 AM

#

And the line is not tranversal right, it would have to be a straight angle
what's that in reference to

viscid ginkgo Mar 23, 2020, 6:03 AM

#

the line that is in the middle of the parallel lines

#

that breaks into two pieces

silent plank Mar 23, 2020, 6:03 AM

#

that'd be 2 separate lines

viscid ginkgo Mar 23, 2020, 6:03 AM

#

ok

#

so it's just a case of alternate interior angles

silent plank Mar 23, 2020, 6:03 AM

#

yes

viscid ginkgo Mar 23, 2020, 6:04 AM

#

y is 80, x is 30

#

ty

#

there's a theorem here that says alternate interior angles formed by two lines cut by a tranversal are congruent if and only if the lines are parallel

#

so I wonder how does the problem I Showed you relate to this

plain pollen Mar 23, 2020, 8:19 AM

#

Does anyone have a decent proof for this? (I couldn't find out how to place arrows above letters but a and b are vectors in the second half of the equation)

📎 Screen_Shot_2020-03-23_at_9.18.40_AM.png

lament pumice Mar 23, 2020, 8:43 AM

#

Are you asking why half of the length of cross product equals half of the area between two vectors?

plain pollen Mar 23, 2020, 8:46 AM

#

Uh, well. My teacher just showed us that the area formula for a triangle can be rewritten as half of the vector product of a and b

#

And I'm trying to find a mathematical proof for that

lament pumice Mar 23, 2020, 8:50 AM

#

Basically, cross product is defined to have a length of a parallelogram between those two vectors and if you connect any two opposite points of a parallelogram you'd divide it in half(because two those triangles are equal because their sides are equal).

white cradle Mar 23, 2020, 8:51 AM

#

area* not length

lament pumice Mar 23, 2020, 8:51 AM

#

Or that formula could be proven without cross product. Just take a regular formula with triangle height and express that height using sine.

white cradle Mar 23, 2020, 8:51 AM

#

magnitude of cross product = area of parallelogram

plain pollen Mar 23, 2020, 8:52 AM

#

oh and 1/2*parallelogram is a triangle

lament pumice Mar 23, 2020, 8:52 AM

#

Sure, I never said length of a parallelogram, I've said length of a cross product.

white cradle Mar 23, 2020, 8:52 AM

#

Basically, cross product is defined to have a length of a parallelogram between those two vectors

plain pollen Mar 23, 2020, 8:52 AM

#

lol

lament pumice Mar 23, 2020, 8:53 AM

#

Lol, how did I write that, hah

white cradle Mar 23, 2020, 8:53 AM

#

missed a word ig

#

lol

lament pumice Mar 23, 2020, 8:53 AM

#

Nah, what I meant is that a vector of a cross product has length equal to the area of that parallelogram.

plain pollen Mar 23, 2020, 8:54 AM

#

oh and the cross product of two vectors yields a perpendicular vector?

#

is that a given?

lament pumice Mar 23, 2020, 8:54 AM

#

Yep, that's part of a definition

plain pollen Mar 23, 2020, 8:56 AM

#

wait but how is that relevant here? Is it? lol I'm confusing myself

lament pumice Mar 23, 2020, 8:59 AM

#

Nah, in that case we don't care at all where our vectors point.

uncut cradle Mar 23, 2020, 10:02 AM

#

can someone give me SOHCAHTOA problems?

dark sparrow Mar 23, 2020, 10:03 AM

#

gimme a moment.

uncut cradle Mar 23, 2020, 10:03 AM

#

tnks bro

dark sparrow Mar 23, 2020, 10:04 AM

#

i'm not your "bro". and would prefer not to be addressed as such.

#

anyway, here

📎 unknown-35.png

#

make sure to read the instructions carefully.
if you want to check your answers then pm them to me indicating which problem you're giving me the answer to. but do take note that i can and will reject answers which are not properly formatted. you may omit the degree sign in problems that ask you to find the angle.

uncut cradle Mar 23, 2020, 10:05 AM

#

ok ok thank youu

storm cave Mar 23, 2020, 1:20 PM

#

Can anyone help me with my math soon

dark sparrow Mar 23, 2020, 1:24 PM

#

wdym soon

#

why can't you post the thing you need help with now

storm cave Mar 23, 2020, 1:25 PM

#

I took a look at it before, had a lot of trouble with it, took out the dog and I’ll be back soon

steady forum Mar 23, 2020, 1:39 PM

#

whats the diff

#

between

#

things like

#

-tan (theta) and tan(-theta)?

dark sparrow Mar 23, 2020, 1:40 PM

#

wdym by "things like"

steady forum Mar 23, 2020, 1:45 PM

#

the two i stated

dark sparrow Mar 23, 2020, 1:47 PM

#

i mean, tan(-θ) = -tan(θ) for all θ where both sides make sense

#

but if you didn't know anything about tan, then you could not say automatically that negating the input and then passing it through the function would just so happen to produce the same result as passing your input through the function and then negating the output

steady forum Mar 23, 2020, 1:48 PM

#

So its similar but diff

dark sparrow Mar 23, 2020, 1:48 PM

#

...

#

tan(-θ) and -tan(θ) are EQUAL.

#

functions for which this thing works are called "odd functions"

steady forum Mar 23, 2020, 1:49 PM

#

A1bi) if they question asks for -tan(theta) steps wld be same?

📎 JPEG_20200323_214950.jpg

dark sparrow Mar 23, 2020, 1:51 PM

#

"the" steps?

steady forum Mar 23, 2020, 1:51 PM

#

Working

dark sparrow Mar 23, 2020, 1:51 PM

#

if they asked for -tan(θ) then maybe you could cut the step where tan(-θ) is replaced by -tan(θ) out of the solution

#

so maybe no the exact set of steps could differ

steady forum Mar 23, 2020, 1:52 PM

#

What the heck

dark sparrow Mar 23, 2020, 1:52 PM

#

you ask a stupid question you get a stupid answer

steady forum Mar 23, 2020, 1:52 PM

#

:(

dark sparrow Mar 23, 2020, 1:53 PM

#

"what's the difference between these two things that are actually the same"

#

"would The Steps™️ differ for these two superficially different problems even though each problem admits multiple different sequences of steps to solve it so the question as stated just doesn't make any sense and there is no constructive answer to it"

steady forum Mar 23, 2020, 1:56 PM

#

Okay

#

Thanks

storm cave Mar 23, 2020, 2:09 PM

#

https://gyazo.com/b89ed2179a26175208a9b4b0a79df976

Gyazo

#

How am I supposed to get A?

dark sparrow Mar 23, 2020, 2:11 PM

#

consider that you can calculate the lengths of DB and EB

storm cave Mar 23, 2020, 2:11 PM

#

I'll try

#

How would I go about calculating EB?

supple wedge Mar 23, 2020, 2:19 PM

#

EB is the opposite side of triangle BDE

#

and 8 is the hyp

#

so u should use $sin25=\frac{8}{EB}$ and solve for EB

somber coyoteBOT Mar 23, 2020, 2:20 PM

#

Hmm:

📎 460383628157911050.png

supple wedge Mar 23, 2020, 2:20 PM

#

@storm cave

storm cave Mar 23, 2020, 2:20 PM

#

Ohhh I see! Thank you!

supple wedge Mar 23, 2020, 2:21 PM

#

:))

dark sparrow Mar 23, 2020, 2:21 PM

#

uh

#

this is wrong @supple wedge @storm cave

supple wedge Mar 23, 2020, 2:21 PM

#

he said EB

#

😛

dark sparrow Mar 23, 2020, 2:21 PM

#

yes EB

storm cave Mar 23, 2020, 2:21 PM

#

For EB?

supple wedge Mar 23, 2020, 2:21 PM

#

isnt it sin

#

opp/hyp

dark sparrow Mar 23, 2020, 2:22 PM

#

it's $\sin(25) = \frac{EB}{8}$

somber coyoteBOT Mar 23, 2020, 2:22 PM

#

Ann:

📎 183668144404037632.png

supple wedge Mar 23, 2020, 2:22 PM

#

fuck(

dark sparrow Mar 23, 2020, 2:22 PM

#

8 is the hypotenuse.

supple wedge Mar 23, 2020, 2:22 PM

#

REEEEEEEEEEEEEEEEEEEE

#

hahhahhahhahahhahaha i am rart

#

sure

#

thanks st. big brain ann

storm cave Mar 23, 2020, 2:22 PM

#

The answer is -0,016... I doubt thats correct.... Is it my calculator?

supple wedge Mar 23, 2020, 2:22 PM

#

wrong because i am fucking stupid

#

$sin25=\frac {EB}{8}$

somber coyoteBOT Mar 23, 2020, 2:23 PM

#

Hmm:

📎 460383628157911050.png

supple wedge Mar 23, 2020, 2:23 PM

#

hehe

storm cave Mar 23, 2020, 2:23 PM

#

Oh wait

#

The answer for that is -1.05... I think I'm missing something

supple wedge Mar 23, 2020, 2:25 PM

#

,w calculate sin 25

somber coyoteBOT Mar 23, 2020, 2:25 PM

#

📎 wolf.png

Display results online and refine query

supple wedge Mar 23, 2020, 2:26 PM

#

,w calculate 8(sin25)

somber coyoteBOT Mar 23, 2020, 2:26 PM

#

📎 wolf.png

Display results online and refine query

supple wedge Mar 23, 2020, 2:26 PM

#

there u go @storm cave

storm cave Mar 23, 2020, 2:26 PM

#

It's my calculator.... Sin(25) gives me -0.13

#

Any idea how to fix that?

#

TI-84 Plus

#

Also thank you

supple wedge Mar 23, 2020, 2:27 PM

#

did u put in radian

#

or degree

storm cave Mar 23, 2020, 2:27 PM

#

Radian

supple wedge Mar 23, 2020, 2:27 PM

#

its supposed to be in degree

storm cave Mar 23, 2020, 2:27 PM

#

Ohh yes I see, thank you!

supple wedge Mar 23, 2020, 2:27 PM

#

because 25

#

for DB just use some pythagoras theorem

#

same for AB

#

then subtract

storm cave Mar 23, 2020, 2:28 PM

#

Wait do I need DB to get A?

supple wedge Mar 23, 2020, 2:31 PM

#

yes

#

wait i thought u were finding a length mb

#

ignore the subtract part

#

after u find AB use cosA to find A

storm cave Mar 23, 2020, 2:33 PM

#

isnt it tan?

supple wedge Mar 23, 2020, 2:33 PM

#

yes

#

omg

#

my ratios

#

are shit

#

for fucks sake

#

tan

#

yes

storm cave Mar 23, 2020, 2:35 PM

#

You're good, thanks for the help man :)

supple wedge Mar 23, 2020, 2:36 PM

#

sorry for fucking up my ratios so badly omg

storm cave Mar 23, 2020, 2:38 PM

#

It's fine man, ty!

#

Last one

📎 01d10923c49b65bde434674e03518de9.png

#

If C1 has r = 2 and C2 has r = 3, how do you go about calculating PS?

grizzled zinc Mar 23, 2020, 2:50 PM

#

i have literally 0 clue what this is asking me to do

📎 Snapchat-675363948.jpg

golden panther Mar 23, 2020, 2:56 PM

#

lel

grizzled zinc Mar 23, 2020, 3:26 PM

#

Yo nvm had to inverse

quiet mason Mar 23, 2020, 3:59 PM

#

@storm cave what have you tried

storm cave Mar 23, 2020, 4:03 PM

#

nothing much.... I dont understand it at all

quiet mason Mar 23, 2020, 4:05 PM

#

do you know what similar triangles are or trigonometry

rich wolf Mar 23, 2020, 4:08 PM

#

,w calculate sin(25 rad)

somber coyoteBOT Mar 23, 2020, 4:09 PM

#

📎 wolf.png

Display results online and refine query

spice ridge Mar 23, 2020, 4:55 PM

#

📎 unknown.png

#

can someone help me with this question?

#

its 4 right?

rich wolf Mar 23, 2020, 5:23 PM

#

@spice ridge no

#

You are wrong

spice ridge Mar 23, 2020, 5:24 PM

#

then what is it?

#

i graphed both on desmos and its a mirror

#

thats not very helpful by just saying im wrong

rich wolf Mar 23, 2020, 6:13 PM

#

@spice ridge

#

📎 Screenshot_20200323-141342_Desmos.jpg

#

Does this look like something mirrored across the line y=x

spice ridge Mar 23, 2020, 6:18 PM

#

must've put in a different equation my bad

rich wolf Mar 23, 2020, 6:20 PM

#

Lmao

upper karma Mar 23, 2020, 7:33 PM

#

f(x) =1,5x-18
g(x)=0.5x+6

how can i calulate the angle?

wind heart Mar 23, 2020, 7:55 PM

#

📎 image0.jpg

#

Does anyone know how to do this? I got it wrong and I have no idea how to to get -5,7

lyric lichen Mar 23, 2020, 8:10 PM

#

I think I can help, imagine a coordinate grid, and the point (3, -1) on it, if we reflect it across the line y = 3 (a horizontal line that crosses (0, 3)) then we just take the distance the current point is from the line and move it that length away in the other direction

#

That's not a great explanation, but in this case, since the y value of (3, -1) is 4 away from +3, the new y value is 4 away on the opposite side of the line

#

(3 + 4 = 7, so the new point is at (x, 7)) then do the same thing across the vertical line x = -1

#

@wind heart feel free to ping me if you need any further assistance :)

wind heart Mar 23, 2020, 8:14 PM

#

Oh thanks sm!

upper karma Mar 24, 2020, 2:45 AM

#

@loud pike ?

loud pike Mar 24, 2020, 2:46 AM

#

oh wait nvm

#

I misunderstood what it was saying

upper karma Mar 24, 2020, 2:48 AM

#

📎 unknown.png

#

btw here's a geo problem it's kinda competition math

#

Solution:
||So what I did was first make the square into a 3x3 grid. The total area is 9. Then, I drew the diagonal from the top left to the bottom right. That diagonal will divide the square into 1/2, so you know that the non-hashed area divided by that diagonal is 9/2. Then you want to find the area of that trapezoid, so you set up two equations:
y = -x + 2
y = -3x + 3
Solve for the system of equations:
-x + 2 = -3x + 3
2x = 1
x = 1/2
y = 3/2
Now, find the distance from that to (2,0), which is root(2)/2.
The other side is similar, so you subtract the 2(root(2)/2) from 2(root(2)).
You then get root(2). That is the first base of the trapezoid.
You can get the second base, which is just the diagonal, which is 3(root(3)).
If you have the grid drawn out, you can recognize that the height of the trapezoid is root(2)/2.
The average of root(2) and 3(root(2)) is 2(root(2)).
Now, multiply root(2)/2 and 2(root(2) to find the area of the trapezoid.
You get 2.
9/2 + 2 = 13/2
(13/2)/9 = 13/18
But don't forget that we have only calculated for the non-hashed out area!
The final solution is 5/18!||

lyric lichen Mar 24, 2020, 3:47 AM

#

That's awesome, I'll give it a go, wish me luck

lilac isle Mar 24, 2020, 4:50 AM

#

Can anyone help me I’m stuck.

📎 image0.jpg

dark sparrow Mar 24, 2020, 4:58 AM

#

is the tank meant to be cylindrical

lilac isle Mar 24, 2020, 4:59 AM

#

I believe so

dark sparrow Mar 24, 2020, 5:01 AM

#

aight so

#

it's made of half-inch-thick steel

lilac isle Mar 24, 2020, 5:02 AM

#

yup

dark sparrow Mar 24, 2020, 5:02 AM

#

and it stands to reason that 15 ft and 20 ft are the dimensions of the interior of the tank

#

ie the part that actually holds water

lilac isle Mar 24, 2020, 5:03 AM

#

15-ft in diameter and 20-ft tall.

dark sparrow Mar 24, 2020, 5:03 AM

#

so really it's a cylinder with a diameter of 15'1" and a height of 20'1" with a hole inside shaped like a cylinder of diameter 15' and 20'

#

1 inch bc the half-inch of steel contributes to the diameter and height on both sides

#

anyway, with that in mind, you should be able to figure out the volume of steel used in the making of the tank

lilac isle Mar 24, 2020, 5:04 AM

#

yup

dark sparrow Mar 24, 2020, 5:05 AM

#

and once you have the volume, you can get the mass

#

since you have the density

lilac isle Mar 24, 2020, 5:06 AM

#

I think I see what you talking about

#

gimme me a min to calculate it

dark sparrow Mar 24, 2020, 5:06 AM

#

yeah ofc i'm leaving the specific calculations to you

#

in part bc i'm a bit too lazy to do them

lilac isle Mar 24, 2020, 5:06 AM

#

👏🏿

dark sparrow Mar 24, 2020, 5:06 AM

#

so i don't think i can check them if you send them to me

lilac isle Mar 24, 2020, 5:07 AM

#

it fine, thank!

dark sparrow Mar 24, 2020, 7:32 AM

#

alright yeah i guess i'm gonna have to ask for help again lmao

#

let α, β and γ be the angles of a triangle. prove that sin(α)sin(β)sin(γ) ≤ 3sqrt(3)/8 and find all triangles for which this inequality becomes an equality

#

i have a hunch that equality holds iff α=β=γ=π/3 and i've tried naïvely applying am-gm but this didn't really get me anywhere productive

#

i have... this?

📎 unknown.png

#

where a, b and c are the sides of the triangle and S is the area

#

but honestly i have little to no clue how to progress

#

i've got another four of these angle inequality problems and honestly i'm just as stuck on them as on this one

#

hang on

#

i think i have an idea.

#

yeah nvm, i solved it.

#

turns out jensen's inequality did the trick lmfao

viscid ginkgo Mar 24, 2020, 8:09 AM

#

https://i.imgur.com/BV0wXXx.png

Imgur

#

how can I prove that x is 35?

#

the remote interior angles on the left should be 60 + 70 = 130, but that won't work.

#

x = 35.

dark sparrow Mar 24, 2020, 8:10 AM

#

what's known about that angle next to x

viscid ginkgo Mar 24, 2020, 8:11 AM

#

nothing

#

x = 35 is the answer actually

dark sparrow Mar 24, 2020, 8:11 AM

#

nothing? then x need not be 35

#

are you sure you aren't omitting some crucial info

viscid ginkgo Mar 24, 2020, 8:11 AM

#

yeah man

#

the middle line is not even straight, so it doesn't seem like a bisector.

dark sparrow Mar 24, 2020, 8:12 AM

#

the middle line can be swung back and forth as much as you want within that triangle, and x will change but none of the other data will

#

so you MUST be omitting some info

#

can you show the problem exactly as stated

viscid ginkgo Mar 24, 2020, 8:13 AM

#

I'm really not omitting any info

#

but I'll redraw it.

cinder portal Mar 24, 2020, 8:13 AM

#

do what Ann tells you

#

show the problem exactly as stated

viscid ginkgo Mar 24, 2020, 8:13 AM

#

https://i.imgur.com/SIGN1y4.png

Imgur

#

it's even less info lol

#

actually I missed the y

#

https://i.imgur.com/1wXQL7t.png

Imgur

#

but same thing

cinder portal Mar 24, 2020, 8:14 AM

#

ok yea this question is broken

viscid ginkgo Mar 24, 2020, 8:14 AM

#

lol

flint osprey Mar 24, 2020, 8:15 AM

#

who gave you a question in the format of ms paint?

viscid ginkgo Mar 24, 2020, 8:15 AM

#

I copied it from my book.

#

ms paint is just useful as hell for this lol

#

would be nice for a more professional software

flint osprey Mar 24, 2020, 8:17 AM

#

so you drew this, it's not the original problem

#

you really should take the advice that's given to you and take a picture of the problem

viscid ginkgo Mar 24, 2020, 8:17 AM

#

lol I drew it exactly but ok I'll take a picture

cinder portal Mar 24, 2020, 8:18 AM

#

its fine, we cant track where you are when you send a pic

viscid ginkgo Mar 24, 2020, 8:18 AM

#

Number 11

📎 JPEG_20200324_011811.jpg

#

lmao

#

pepega

#

they make a lot of spelling errors in this book, probably rushed

viscid ginkgo Mar 24, 2020, 9:02 AM

#

ok now I'm stuck on #17

#

I don't see same side interior angles, corresponding angles, alternate interior or exterior angles

#

I tried using the remote interior angles trick

#

that doesn't help either

silent plank Mar 24, 2020, 9:08 AM

#

were you able to find x?

#

||x° and 30° are alternate interior||

viscid ginkgo Mar 24, 2020, 9:11 AM

#

my best guess so far is that y is 30 degrees

#

possible alternate interior angle

silent plank Mar 24, 2020, 9:11 AM

#

^ did you get x?

viscid ginkgo Mar 24, 2020, 9:11 AM

#

no

#

I saw the answer in the book but no

#

I don't think y is 30 though, it could easily be 20

#

by same logic

silent plank Mar 24, 2020, 9:13 AM

#

well x is alternate interior to the 30°
are you able to spot it after i tell you?

viscid ginkgo Mar 24, 2020, 9:14 AM

#

ah yeah I see that

silent plank Mar 24, 2020, 9:14 AM

#

and then to find y, consider the external angle theorem

#

and/or angle sum of triangle + angle sum on straight line

viscid ginkgo Mar 24, 2020, 9:18 AM

#

thanks

ruby inlet Mar 24, 2020, 3:54 PM

#

Need some help: is sin(wt-phi) = -sin(wt+phi)?

gritty siren Mar 24, 2020, 3:56 PM

#

it doesn't quite work with phi=0 for example

daring solstice Mar 24, 2020, 4:13 PM

#

Im not sure how to solve this

📎 image0.jpg

#

Bc the radius isnt give.

#

*given

#

Bros plz help

dim sable Mar 24, 2020, 4:18 PM

#

📎 Tues_3.24_Exrtra_Extra_Practice.docx

#

i need help i have a test on this tomorrow

#

can some one help me study'

timber coral Mar 24, 2020, 4:27 PM

#

📎 PlaneGeometryQuestion.JPG

upper karma Mar 24, 2020, 5:26 PM

#

📎 image0.png

lament pumice Mar 24, 2020, 5:46 PM

#

Oh yes, the geometry question

upper karma Mar 24, 2020, 6:56 PM

#

@timber coral how would i go about finding the locus

#

it's just a quartercircle

#

right

timber coral Mar 24, 2020, 8:43 PM

#

yesthank you

#

26quonbarkley i believe it is circle equation

pure pivot Mar 24, 2020, 10:03 PM

#

📎 image0.png

solid sequoia Mar 24, 2020, 10:56 PM

#

I'm trying to find the equation of the hyperbola: Foci (0,-8), (0,8); difference of focal radii is 10

#

I've been looking online for hours on how to solve this problem and all I can muster up is that apparently C = 8, which means that A = 4. But 64 = 16 + B^2 makes B = positive negative 6.92820323028 please halp

#

<@&286206848099549185>

wind heart Mar 25, 2020, 1:27 AM

#

📎 image0.jpg

#

That’s my answer, and this is the correct answer

#

📎 image0.jpg

#

I’m confused. What does it mean to reflect the line over the line of reflection? Like,

silent plank Mar 25, 2020, 1:27 AM

#

read the instructions

#

it seems its indicated by x=6

wind heart Mar 25, 2020, 1:28 AM

#

How would I know to change it over the x axis? I don’t get it

#

Ohhhhh

#

I see

#

So like

#

Your supposed to count the things

#

And the points are equal opposites