precalculus

[gilded marten]

pi/4

ok so 3cos(pi/4x-2)?

or is it supposed to be stated as a sine function?

[viscid thistle]

Looks like your amplitude is 3 and the period is 8 units (wut???)

Would have figured the x would be in terms of π

[gilded marten]

yea me too this is really confusing

[viscid thistle]

Oh I see

Let me roll back my precalc notes real quick

Ok so this actually looks like it is a cosine function

And its phase shift is -2 units

Since cosine usually has a maximum at x = 0

So let's set it up

y = a cos (bx - c) + d

a = amplitude = 3

[gilded marten]

yes

[viscid thistle]

b = period = 8?

I believe so

Hang on

No the period of a cosine curve is pi

Ok so the formula for finding the period of a cosine curve is 2π/|b|

Your period is 8 so

8 = 2π/b

b = 2π/8

b = π/4

[gilded marten]

right

[viscid thistle]

And the phase shift is -2

c = -2

[gilded marten]

because it moves right 2 units?

[viscid thistle]

Or is it 2?

Sorry had to play around with desmos to visualize

It "moves right" when you subtract a positive value

So since we moved left we subtracted a negative value

y = a cos (bx - c) + d

a = 3

b = π/4

c = -2

[gilded marten]

but the equation makes it positive?

-(-2)

[viscid thistle]

And the graph looks like its vertical shift is 2 units as well

d = 2

y = 3 cos (π/4x + 2) + 2

[gilded marten]

ah right

*correct

[viscid thistle]

Yeah that phase shift part confused me a bit

Dammit nope that's wrong

When you plot that in desmos we got everything right except for the phase shift (c)

[gilded marten]

correct answer was -3sin(pi/4x)+2

[viscid thistle]

Hmmm

[gilded marten]

that is ok

i can try again

[viscid thistle]

Yeah I'm rusty with that area of precalc

Gotta figure out that phase shift

[gilded marten]

that is quite alright thank you for answering my questions!

[viscid thistle]

Np

[gilded marten]

would these problems just b easier if i memorized the unit circle, or is there some way to actually work out the problem

[dense zealot]

Yes

They would be much easier

But

it's commonly known root2/2 = pi/4

so if the height is negative

it's -pi/4

[viscid thistle]

Hmmm

[gilded marten]

right but how does that help

like saying that it was tan^-1 of sqrt2/2

how does that differ

[viscid thistle]

Ok so this requires the Unit Circle

You've seen it right?

[gilded marten]

yes

[viscid thistle]

And you know how x and y correspond to cos and sin?

[hexed ermine]

@gilded marten sorry for leaving had to do something

[viscid thistle]

Yes ^

[gilded marten]

its perfectly fine pjs

[viscid thistle]

So basically they're asking what is the value of theta when y = -sqrt(2)/2

@gilded marten I'll make a visual

Hopefully you can see what I'm trying to show you

Because it's inverse, we're given our value of sine, in this case, y = -(rad 2)/2 on the unit circle

And it appears twice

[gilded marten]

it appears twice?

[viscid thistle]

Yes

[gilded marten]

ok sure i see that

[viscid thistle]

You can see the two points I plotted right there

So it's a matter of deciding which one to use

[gilded marten]

in quadrant 3 and 4

[viscid thistle]

yep

For some odd reason they want us to pick the one in quadrant 4

So in this case, solve for theta base 2

And ignore theta base 1

[gilded marten]

oh i remember that you never use quadrant 3

[viscid thistle]

yes

I don't know why, I'll have to look it up

But either way, we solve for theta base 2

In this case, what do you think the angle measurement is there?

[gilded marten]

45

id guess

[viscid thistle]

Close, it's 45 degrees in relation to the positive x-axis but since we go counterclockwise, it's 360 - 45

= 315 degrees

You see why?

If not I can make a visual representation

[gilded marten]

no im sorry i dont

[viscid thistle]

Ok

You know how the Unit Circle is counterclockwise right?

[gilded marten]

yea

[viscid thistle]

Ok then you'll understand this

[dense zealot]

Arcsins range is pi/2 to -pi/2

315° isn't an acceptable answer

[viscid thistle]

@gilded marten hopefully this shows it

[gilded marten]

okay yes

[viscid thistle]

Also what baka says is right

So first we need to convert this into radians

315 degrees into radians is what?

@gilded marten do you know how to convert degrees to radians?

[gilded marten]

315/1*pi/180

correct?

[viscid thistle]

So basically (π/180) x 315

= 315π/180

= 7π/4

Just asking, does that make sense?

Those formulas are something you should memorize

[gilded marten]

so let me try to explain this

i can do the radians to degree stuff

but i still need to memorize the unit circle

but anyways

[viscid thistle]

So we have 7π/4 as our answer (it's not correct though, we have one final step)

[gilded marten]

to find the exact value we need to find theta, saying that theta is the angle of sin^-1

[viscid thistle]

yes

Do you know how inverse functions work?

[gilded marten]

in this situation no

[viscid thistle]

Ok

[gilded marten]

man i thought i understood this chapter

[viscid thistle]

So if we have a function y = x + 1

Then the inverse function is x = y + 1 right?

Just switch the x and y value

[gilded marten]

yea

i get that

[viscid thistle]

So let's say we had sin (x) = y

Then the inverse is sin (y) = x

[grizzled hull]

Mmm...

[viscid thistle]

So sin (θ) = y

And sin^-1 (y) = θ

[gilded marten]

yea

[grizzled hull]

There's the rigor. I can leave.

[viscid thistle]

Does that make sense?

[gilded marten]

yes

[viscid thistle]

kk

So back to our answer, 7π/4, we have one final step

As Bacca said earlier, it's not in the domain since inverse sine only goes from -π/2 to π/2

So we have to subtract from it

What can we add or subtract to radians and still get the same values for our trig functions?

If you don't know I can help so don't worry

Just let me know

[gilded marten]

well dont you have to get the base to be the same

[viscid thistle]

It's basically 2π

So for example sin (π) = sin (3π)

sin (π/2) = sin (5π/2)

Do you want me to show you why?

[gilded marten]

yea

[viscid thistle]

kk

So as you can see, π/4 is the same as 9π/4 in radians

[gilded marten]

yea

[viscid thistle]

This is because we did one full revolution around the Unit Circle and the value is still on the same point

So for example sin (π) = sin (3π) = sin (5π) etc

[gilded marten]

ok yea

[viscid thistle]

Yeah so now we do 7π/4 plus or minus 2π

Since we're too high we subtract

(7π/4) - 2π

= 1.75π - 2π

= -0.25π

[gilded marten]

ok so its -pi/4?

[viscid thistle]

= -π/4

Yep

[gilded marten]

wow

cool

[viscid thistle]

Sorry it's hard to teach this stuff without visual representation lol

[gilded marten]

thank you so much

[viscid thistle]

Would have been much faster if we had a whiteboard or something

[gilded marten]

i dont understand the beginning part

[viscid thistle]

Oh

I can still go over it

[gilded marten]

sure if you can give me one second

[viscid thistle]

kk

Basically going back to inverse functions, since it's sine, we're inputting our y value and our output is theta

[gilded marten]

ok

im back

[viscid thistle]

If sin (θ) = y

Then sin^-1 (y) = θ

[gilded marten]

yea i understand that

thanks

[viscid thistle]

And in your problem, y = - (sqrt 2)/2

So we find where y equals that on the unit circle and find where theta satisfies that

[gilded marten]

is "rad" the same as sqrt?

[viscid thistle]

Picture should demonstrate

Yep

[gilded marten]

ah

[viscid thistle]

I should say sqrt lol

[gilded marten]

no its fine

[viscid thistle]

Hopefully the picture shows what I'm trying to say

[gilded marten]

so because there are two locations on the unit circle it must be one of those

[viscid thistle]

Yep

And like you said, can never be in Quadrant III

[gilded marten]

but it can only be in 4 because of some math law

[viscid thistle]

So it's in Quadrant IV

[gilded marten]

yes

but where does the sin^-1 come in

like what if it was cos^-1

how does it differ

[viscid thistle]

If it was inverse cosine then you'd plug in the given x value

What they're basically giving you is the y-value of the point that exists on the unit circle

And you're finding angle theta

So for example at 30 degrees, sin(30) = 1/2, since the y value is 1/2

So sin^-1 (1/2) = 30 degrees, because they gave you the value of y, which is 1/2, so you're finding theta, which is 30 degrees

[gilded marten]

uh

[viscid thistle]

?

sin (30) = 1/2 is basically saying "When angle theta equals 30 degrees, y = 1/2 units"

sin^-1 (1/2) = 30 is basically saying "When y = 1/2, angle theta equals 30 degrees"

[gilded marten]

ok i see that now

[viscid thistle]

Yeah

[gilded marten]

sorry took me a minute

[viscid thistle]

So they're giving you the y coordinate, asking you to find the angle

No worries lol

And as for cos^-1, they'll be giving you the x coordinate

Asking you to still yet find theta

[gilded marten]

OH

wow

[viscid thistle]

Did it click?

[gilded marten]

thanks

yea lol

[viscid thistle]

Good

[gilded marten]

i have a quiz tomorrow and none of my friends are online so you have been a huge help

[viscid thistle]

No problem dude lol

[gilded marten]

one more thing

[viscid thistle]

sure

?

[gilded marten]

sorry one sec

[viscid thistle]

0_0

[gilded marten]

i got it wrong entirely oops

[viscid thistle]

Oh

[gilded marten]

i know that this is basically saying

[viscid thistle]

The answer is 7π/4

[gilded marten]

the answer is ^

yea

but

that doesnt fit

[viscid thistle]

?

[gilded marten]

oh wow one second again my apologies

[viscid thistle]

np

[viscid thistle]

Wait a second now I'm confused

Also the answer is just 5π/7 for that one

[gilded marten]

yea

my question is

5pi/7 doesnt fit

right?

[viscid thistle]

It does fit, it's just not a perfect number in the degree

Converting 5π/7 to degrees = 900/7 degrees = 128 4/7 degrees

Or 128.5714285714 degrees

[gilded marten]

i meant like what baka said

[viscid thistle]

Yeah there was a problem with that one we spent quite a while on

The answer was actually π/4

Not -π/4

I'm still trying to figure out why

[gilded marten]

wait you mean the one i posted above is pi/4

[viscid thistle]

No no no

What the hell????

Ok my Windows 10 calculator is saying the answer to the very first one (the inverse sine one)

Is π/4

But a scientific calculator online is saying it's -π/4

[gilded marten]

uhm

[viscid thistle]

Actually it can't be π/4 because sin (π/4) is positive

So no, we were right

[gilded marten]

cool

[viscid thistle]

-π/4 is the answer to the very first problem we did

[gilded marten]

nice

[viscid thistle]

Screw windows calculators, they're defective

So what was the second problem?

cos^-1(cos 7π/4)?

[gilded marten]

yea

[viscid thistle]

I see

This one might be difficult to explain

[gilded marten]

they cancel each other out so its just 7pi/4

[viscid thistle]

Yeah that's where they get to you

[gilded marten]

and you have to use the thing baka said

[viscid thistle]

7π/4 has the correct x-value, not the right y-value

The correct answer is actually π/4 for that one

I'll make a visual

See the problem is, we can't just cancel out

You remember what cos (7π/4) is right?

[gilded marten]

?

what do you mean

[viscid thistle]

We can't just cancel them out

Inverse sine can't exist in Quadrant III

Inverse cosine can't exist in Quadrant IV

So do you remember what cos (7π/4) is?

[gilded marten]

i thought invers sin was quad 4 and inverse cos was quad 2

or no..?

[viscid thistle]

Hold on

[gilded marten]

i have no idea

[viscid thistle]

Ok inverse sine can only exist in Quadrants I and IV

Inverse cosine can only exist in Quadrants I and II

And inverse tangent can only exist in Quadrants I and IV (because it's sine over cosine, duh)

So with that said, our first answer is right because our inverse sine landed in quadrant IV

But for inverse cosine it has to be in Quadrants I and II

[gilded marten]

ok cool

[viscid thistle]

And because 7π/4 is in Quadrant IV it doesn't work

So what is cos (7π/4)?

e.g. what is the x value at 7π/4 radians?

[gilded marten]

i get it!

ok

ok

thanks so much

thanks for spending so much time with me

i understand this now

[viscid thistle]

Great!

Let me just send one last visual

Ok so since inverse cosine can't exist in the shaded area, our only other option is π/4

Which is our answer

^

So cos^-1(cos 7π/4) = π/4

gg dude

Good luck!

[gilded marten]

thanks man

appreciate it

[viscid thistle]

Could someone help me with #9?

Ok so first find inverse cosine of 3/5

arccos (3/5) = 2θ

@viscid thistle you might need your calculator for this one

I got .927 the ÷ by 2 and got .463

Good, now convert that to degrees

I got 83.34°, would I now just plug that into the rest of the functions?

No, you wouldn't

Also, 0.463 in radians is actually 26.53 degrees

Ah I see, you forgot to divide by pi

83.34/pi = 26.53 degrees

But if you look it says that 90 < θ < 180, and 26 doesn't fit in that domain

So there's no answer for theta

Hate to be that guy but the book has answers for all 6 trig functions in the back I just have no idea how they got it

Hmmm?

What did the answer say on the back?

#9

Huh

I don’t get it

Oh of course yes

I thought you were only solving for theta

Ok

We found theta

Just find the 6 trig functions of that

Theta was .463 or 26.53° right?

I believe so but it seems like their answers were written exactly, not to a rounded decimal so we’ll have to back up

arccos (3/5) = 2θ

Let me see if I can find my trig reference sheet

For now, (arccos (3/5))/2 = θ

@viscid thistle you still here?

I found something

Try applying some of these properties

In particular, the top 2 interest me the most

Especially the second one. HINT: cos cancels arccos

I got it, thanks (:

Good luck!

[gilded marten]

hey sirius if you are still there i want to thank you for all your help

i have no idea how it will go tomorrow but thank you for assisting me

[grand axle]

So for cos and sin, we have this periodic rule of sin/cos alpha= sin/cos(alpha+k*360)

But if i know that sin is 0 or 180 and cos is 90 or 270 can i write it as sin/cos (alpha+k*180)?

Or that would be incorrect

[grizzled hull]

Those special cases are true if I'm interpreting the way you wrote it correctly.

It doesn't usually help to note that in that way, though.

[grand axle]

But if a question asks me to get the cos/sin values without giving any limits, wont i miss out on half the answers by using k*360 in those cases?

[grizzled hull]

Yes and no. If you give the k*360 in those cases, you can specify both angles that would give the value as separate answers.

[grand axle]

Ah good point

[grizzled hull]

More often than not, this is the way it's done. The presentation generalizes.

It's kind of the same as thinking about $$\sin(30^\circ)=\sin(150^\circ)$$. You wouldn't come up with the special case that links these in general. You'd state the two angle cases in general separately.

[grand axle]

Ah, true

Thank you

[grizzled hull]

No problem.

[grand axle]

One more trig question, there is another rule:
Cos(beta)=-cos(beta-180)

Now if cos is for e.g 200 then that means cos200= cos-20

But 200 is in the 3. Quadrant, -20 is in the 4. So they cant be equal

Because cos is the x coordinate

Or am i missing something?

[grizzled hull]

Careful.

Yeah, there's a couple of things.

$$\cos(200^\circ) = {\color{red} -}\cos(20^\circ)$$

[grizzled hull]

The quadrant change that you're noting is applied here.

You can go a step further and say $$\cos(200^\circ) = {\color{red} -}\cos(20^\circ)={\color{red} -}\cos(-20^\circ)$$

[grand axle]

Ahh yeah, its not cos-20 but -cos20

My bad

[grizzled hull]

^^

[grand axle]

Thank you :D

[grizzled hull]

No problem!

[viscid thistle]

Also @gilded marten if you’re still here, there’s a reason why sine and cosine can’t exist on some quadrants

Take for example sine

sin (y) = x is the equation here

However that’s not a function, therefore the range is restricted

Graph of y = sin^-1 (x)

As you can see inverse sine’s range is only -pi/2 < y < pi/2

(Change those inequalities to “or equal to”, I’m on mobile)

[eternal solstice]

Can anyone explain this to me? I don't really understand the concept.

Or anyone know a good video about it?

[hexed ermine]

Ask me that in two days and I will :v

We are about to learn Polar graphs

[viscid thistle]

The heck is 2pik

[hexed ermine]

2pik

Is used for solutions involving trignometric functions

since trig functions are periodic

Lets say the answer is pi and you are asked to solve in the interval from (-inf,inf) you would add 2pik; where k is an integer

Because pi+2pi=3pi+2pi=5pi

pi;3pi; and 5pi are all the same values

if you understand :v

[viscid thistle]

I think I get it friend v: I might be back for some help tho

[viscid thistle]

Hi, is anyone here willing to help me with a question?

thanks ❤

Am I doing this wrong

the part where I put the ln?

oh crap

sorry

second line should be +16

I moved the 16 to the other side

the first starting line is the base problem

okay lol

what should I do after the second then

it seems like a quadratic function

but I don't know what to do exactly or how to move the numbers

so u^2-8u+16?

😮

okay I can figure that one what about

Do I have to reduce the 4^x to 2^2 or something

[sinful cedar]

yup 😄

and then its the same as the one you did above

[dense zealot]

Ur creepy

[gilded marten]

.________.

Hello I would like some assistance with a previous problem if someone would be so kind to look at it. Thank you for your time.

this one here

[hexed ermine]

Where is sine negative?

[gilded marten]

in 2 and 3 i think

[hexed ermine]

Not quite

sin, not cosine

[gilded marten]

3 and 4 then?

[hexed ermine]

yes

and where is tan negative

[gilded marten]

no idea

4?

i cant remember

[hexed ermine]

positive I mean

[gilded marten]

1?

[hexed ermine]

and?

[gilded marten]

im not sure

[hexed ermine]

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

when tan is positive; both sin and cos have to be both positive or negative

so when is cosine and sine both negative

[gilded marten]

4

no

3

right?

[hexed ermine]

Yes 3

[gilded marten]

thats a big help thank you

[hexed ermine]

Np

[eternal solstice]

Can anyone help me set this up? Would it have one or two variables?

[bleak fractal]

after 1 year, she will have an interest of r% on her first investment and (r+2)% on her second

1200·r% + 1400·(r+2)% = 236

solve for r basically

[gilded marten]

any help?

[hexed ermine]

Recall that circumference is 2pir

[gilded marten]

ok

[hexed ermine]

So 64pi right?

As the circumference

[gilded marten]

i got 128pi

oh no nvrmnd

you are right

so what do you do with that

sorry i know nothing about this

[hexed ermine]

Okay so it turns 14 times in one hour

So 64pi*14

[gilded marten]

alright

896pi

[hexed ermine]

Actually no I believe for angular speed you don't account for the radius yet

What you would do is 2pi*14

So 28pi

That's yours angular speed

And then multiply that by your radius to get your linear speed in feet per hour

[gilded marten]

oh

unless im wrong the second answer says its incorrect

rounded to the nearest whole number i got 3

any ideas?

[hexed ermine]

2813 for b right?

[gilded marten]

?

no

i got 3

my mistake

you said multiply

i divided

Thank you for your help and time. It was much appreciated

[hexed ermine]

Np

[pure tree]

hey all.
i had a question about filling a table with trig functions.
how do i go about finding the rest?

this is from a mock exam, if anyone was wondering.

[hexed ermine]

Well you know that cot(x) = cos(x)/sin(x)

And tangent is the inverse

[pure tree]

so tan(θ) would be -4/3

but im not sure how you get the sin and cos

it wouldnt be just cos(θ) = -3

or sin(θ) = 4

right?

[hexed ermine]

Flipped

[pure tree]

ah sorry

edited, but it's not as simple as that right?

[hexed ermine]

No

Now its in quadrant IV

So is cosine positive or negative

[pure tree]

Positive

[hexed ermine]

Yes

So cos should be positive 3

[pure tree]

cos(θ) = 3? instead of cos(θ) = -3

[hexed ermine]

And sin is -4

Yep

[pure tree]

because it is in quadrant IV

[hexed ermine]

Yes

[pure tree]

ahhhh

[candid vine]

It could also be positive 1/4 👀

[hexed ermine]

If it were quadrant II it would be right

[candid vine]

And you have to divide by the norm too

[hexed ermine]

So now csc and sec

[pure tree]

yes. how do we find those without the denominator?

[viscid thistle]

Q1: A Q2: S Q3: T Q4: C
A S T C
All Students Take Calculus
All Functions positive:Q1
Sine positive:Q2
Tangent positive:Q3
Cosine postivie: Q4
All + :Q1
Sine + :Q2
Tan +: Q3
Cosine+ :Q4

[pure tree]

or is it cos(θ)= 3/1

and flipped is csc(θ)=1/3

[hexed ermine]

Yes

No

Sec

Not csc

Secant is the inverse of cosine

[pure tree]

cos(θ)=sec(θ)?

ok

i always get those confused

[hexed ermine]

Yep its dumb

[pure tree]

i hear co in csc

[viscid thistle]

^

[hexed ermine]

Yep exactly

[pure tree]

so thats why i get those confused

ok

thats great

[hexed ermine]

That's how I decipher

[viscid thistle]

I had to muscle that function in my head and it wasn't easy

[hexed ermine]

Bc of how stupid it is

[pure tree]

my midterm is in 30 minutes and i think i can remember it

maybe

llo

[viscid thistle]

Cosine is secant
Sine is csc

Cos = sec
Sin = csc
Co=Se
Si=Cs

[dense zealot]

Lol

do u wanna know how to remember

it's the opposite

Cosecant - sin

[pure tree]

yep... thats how im gonna remember it...

[dense zealot]

(cos)ecant - sin

[viscid thistle]

You really gotta know this is you plan to take calculus

Write them 100 times each if you have to lol

[pure tree]

yeah idk if this is related but i have to take phy 211 by winter 2019

[viscid thistle]

I went through the whol difference quotient to get the derivative of cosecant, and I wrote secant *-cot and it's csc * -cot

You'll make it. No worries

[pure tree]

im trying to decide if i should take 251

over the summer

[hexed ermine]

Physics?

[pure tree]

mth 251

calc 1

[hexed ermine]

O I'm taking it fall

[pure tree]

how does it compare to 112

i want to take summer classes this year so i finish winter 2019, lol.

i mean right now, mth112 is just a lot of things to take in at once, but i think it's the repetitive part that gets me. so would mth251 just be applying the lessons from mth112?

[dense zealot]

Well math 251 is just 2(math112) +math27

[viscid thistle]

lol

[pure tree]

looool

alright time for my midterm exam. thx for all the help!

[viscid thistle]

whats the subject of m251 ?

[dense zealot]

Bai

[wind igloo]

Course inflation. Calculus I was just Math 1 where I went.

[pure tree]

thank god

exam cancelled lmao

just extra practice

but can someone tell me why this is coming out as a line

[rocky bison]

Because it's constant

You've not got a value of x in that graph

So it's just returning the same value over and over

Making y a straight line at that value

[pure tree]

ah okay thank you

[eternal solstice]

@bleak fractal Just saw your reply. Thanks for the help. That was it.

[haughty thicket]

That’s a precalc question?

[eternal solstice]

I'm currently in precalc and it's in my homework.

Idk is it an algebra two problem?

[grand axle]

what happened here at the end?

[calm thicket]

Multiplied by x^-1/x^-1

[grand axle]

why do i need to do that to get the lim tho?

[calm thicket]

Dunno

It seems pretty obvious at step two that you have a larger degree on the bottom so the limit is 0

but maybe this'll go to something different later

I'm not an expert in finding limits unfortunately

[grand axle]

nah it only asked for the lim, which is indeed 0

but this way to get it seems weird

[calm thicket]

Agreed

[wide frost]

its probably to get something/x which approaches 0

so u dont have an inf/inf situation anymore

if u "plug in" inf ont he right side, you get 2/inf which gets to 0

[calm thicket]

Right

[grand axle]

ah true

[calm thicket]

You can't do that on the second step

but at the same time, you can just sprinkle in some intuition and see that it's like that anyways

=tex \frac{2\Omega+3}{\Omega^2+1}\approx\frac{1}{\Omega}\approx 0

[calm thicket]

where Omega is a "big" number

[wide frost]

you cant do what on the second step?

[calm thicket]

plug in infinity, you'd get inf/inf

[wide frost]

on the left side yes

[calm thicket]

but if you plug in a (very) big number, you get ~2/that number, 0

[wide frost]

on the right side u get 2/inf

[calm thicket]

mhm

[thick raptor]

where Omega is a "big" number

👀

why do i need to do that to get the lim tho?
It seems pretty obvious at step two that you have a larger degree on the bottom so the limit is 0
Dividing everything by the highest power in the numerator (or denominator) is the proof that you can just check it based on the "higher power" rule.

And I would imagine that's where Macro was heading

@calm thicket

[wide frost]

or you just use L'Hôpitals rule a few times

[thick raptor]

if you had gotten to the point where you proved LH for limits as x → ∞, you shouldn't be doing this problem

[wide frost]

xD

sometimes one has to do simple stuff to feel omniscient

[odd lichen]

@wide frost you all good? Can you see how it approaches 0 without L'Hôspital's rule?

[wide frost]

ye ofc

denominator degree is higher

[odd lichen]

Oh sorry, I tagged the wrong person.
Also do you mind being a little more pleasant around here?

[thick raptor]

lol

[wide frost]

im pleasant

[dense zealot]

Lol

[candid vine]

👀

[viscid thistle]

👀 ya i found the emoji

[dense zealot]

[viscid thistle]

tfw sister is in algebra 2 in hs

her teacher is a cunt so she gets more work and more retarded mathematical modelling questions than I ever had while doing college algebra 2 at 16

[eternal solstice]

Is it possible to simplify this more? Any idea why Im getting it wrong?

[warped cape]

I think you have to compute cos 7pi/4 and sin7pi/4

[eternal solstice]

@warped cape like this?

[warped cape]

That's not right..

There shouldn't be cos(sqrt2 / 2) and sin (-sqrt2 /2

Nevermind, your signs are right.

[eternal solstice]

Youre right. I forgot to remove the sin/cos when I computed it.

Thank you.

[simple vault]

http://prntscr.com/jbvq0e

im not sure how im supposed to be formatting my answer

[rocky bison]

looks like numbers might be wrong if part b is also wrong?

[simple vault]

I mean its a 2 part question

so you can get one partially right

thats for a

[rocky bison]

Yeah but it's showing both are wrong?

Or will it show b is wrong is a is wrong

[simple vault]

yeah but one should be right

they are seperate

so I can get either wrong while getting the other right

[rocky bison]

but they're on the same thing

So if b is wrong

Then the angles are wrong

which means wrong angles for part a

yes is wrong

[simple vault]

thats the answer for a though

its the provided example

[rocky bison]

=tex 2\cos(360)-\sqrt3\neq0

[rocky bison]

So

That means part a isn't formatted wrong

it's just wrong

because you're using 360 in that

oh nvm that's your period

[simple vault]

http://prntscr.com/jbvrrt

this is the entire problem

[rocky bison]

That looks right then

Well it is right

[simple vault]

it marked it as wrong though

[rocky bison]

Part b is wrong

you put 360 instead of 330

but part a is right

So it's just a formatting error

So I don't think it really matters

Just explain to your teacher or something

[simple vault]

ok

[tacit jewel]

Help a boy out :(

I can't figure out how to find s and t

[simple vault]

http://prntscr.com/jcmu94

[dense terrace]

@tacit jewel nope u dont find s and t because they are just variables that can take values of real numbers

[rocky bison]

@simple vault Still stuck?

[rocky bison]

😦

[simple vault]

http://prntscr.com/jcodr3

I cant figure out this one

[rocky bison]

ok

=tex 2\sin(\theta)+3=-\csc(\theta)

[rocky bison]

@simple vault You about?

[simple vault]

im here

[rocky bison]

awesome

So

Let's make it equals 0

because easier

=tex 2\sin(\theta)+3+\csc(\theta)=0

[rocky bison]

Yes?

Now what actually is cosec

[simple vault]

1/sin(theta)

[rocky bison]

=tex \csc(\theta)=\frac{1}{\sin(\theta)

[granite stirrup]

Rendering failed. Check your code. You can edit your existing message if needed.

[rocky bison]

Exactly

So let's sub that in

=tex 2\sin(\theta)+3+\frac{1}{\sin(\theta)}=0

[rocky bison]

See anythig yet?

[simple vault]

ok

[rocky bison]

Do you see how to do it?

[simple vault]

um

get rid of the sines?

[rocky bison]

nah

Make more of them

Multiply everything by sin

=tex 2\sin^2(\theta)+3\sin(\theta)+1=0

[simple vault]

wait why did you multiply sin

[rocky bison]

Just to get rid of the one in the denominator

That one was a problem

Doing that gave us a quadratic of sin

Which we can solve

[simple vault]

2sin(theta)+1 & sin(theta) + 1

its factorable

[rocky bison]

yep

=tex (2\sin(\theta)+1)(\sin(\theta)+1)=0

[rocky bison]

So

=tex 2\sin(\theta)+1=0\text{ and }\sin(\theta)+1=0

[rocky bison]

Now you can solve each of those for your values of theta

[simple vault]

theta = 7pi/6, 11pi/6?

[rocky bison]

no idea

try it out

[simple vault]

wait supposed to be in degrees

210 degrees, 330 degrees

[rocky bison]

=pup calculate 2\sin(\frac{7\pi}{6})+3+\frac{1}{\sin(\frac{7\pi}{6}}

[granite stirrup]

Wolfram|Alpha didn't send a result back.
Maybe your query was malformed?

[rocky bison]

😦

=pup calc 2\sin(\frac{7\pi}{6})+3+\frac{1}{\sin(\frac{7\pi}{6}}

[granite stirrup]

Query made by @rocky bison
Data sourced from Wolfram|Alpha: http://www.wolframalpha.com/input/?i=calc+2\sin(\frac{7\pi}{6})%2B3%2B\frac{1}{\sin(\frac{7\pi}{6}}
Do more with Wolfram|Alpha Pro: http://www.wolframalpha.com/pro/

[rocky bison]

There we go

Correct

[simple vault]

its not saying that its correct?

http://prntscr.com/jcoj9f

wait a sec

im missing a value

270 degrees

I have a problem with one more can you help me with that one?

@rocky bison

http://prntscr.com/jcojzo I dont know what to do with the cosine

[rocky bison]

sure

ok

This one

Same ide

Make a quadratic in cos

So convert that sin^2 to cos^2

[simple vault]

2 * cos^2 - 1?

[rocky bison]

not quite

=tex \cos^2(\theta)+\sin^2(\theta)=1\implies\sin^2(\theta)=1-\cos^2(\theta)

[rocky bison]

So then you get a quadratic of cos

=tex 2\left(1-\cos^2(x)\right)-\cos(x)-1=0\implies2-2\cos^2(x)-\cos(x)-1=0

[rocky bison]

=tex \cos^2(x)+\cos(x)-1=0

[simple vault]

@rocky bison Thank you

wait

what do I do next

I blanked out

lol

[rocky bison]

It's now a quadratic

factorise it

and then solve for

x

[simple vault]

(cosx - 1) (cosx +1)?

is it factorable??

[rocky bison]

Lemme check it real quick

Ok it should be

=tex 2\cos^2(x)+\cos(x)-1=0

[rocky bison]

I must have misplaced a minus somewhere

[simple vault]

(2cosx - 1)(cosx + 1)

[rocky bison]

=tex 2\sin^2(x)-\cos(x)-1=0\\sin^2(x)=1-\cos^2(x)\to2\sin^2(x)-\cos(x)-1=0\2\left(1-\cos^2(x)\right)-\cos(x)-1=0\2-2\cos^2(x)-\cos(x)-1=0\-2\cos^2(x)-\cos(x)+1=0\2\cos^2(x)+\cos(x)-1=0

[rocky bison]

yep

Then solve each of those factors equal to 0

[simple vault]

got it

also, would it be easy to find the zeros

if I used a graphic calculator?

graphing*

[rocky bison]

You mean plugging in the original function?

[simple vault]

yeah

into a graphing calculator

[rocky bison]

Well it's not as such a functio

Like there's only one variable

So it'll just be a line

So it won't really work

[simple vault]

idk because I have a problem like this

[rocky bison]

Doing it algebraically?

[simple vault]

http://prntscr.com/jcpku9

its kind of similar to that one

and it says use your graphing calculator

so wouldnt it work for that too?

[rocky bison]

I mean technically

Like if you enter the expression

It'll graph vertical lines for all solutions

Like for example

That problem we just did

Each of those lines is a solution

[simple vault]

what if we simplified it

to the 2cos2(x) + cos(x) - 1

[rocky bison]

It'll give the same thing

They're the exact same solutions

[simple vault]

what about the other one then?

what makes those two different

[rocky bison]

What other one?

[simple vault]

http://prntscr.com/jcpku9

[rocky bison]

This is the same idea

Except this one is already in quadratic form

Making it ez pz

[simple vault]

ok

how would I do it?

[rocky bison]

Just factorise it

[simple vault]

I dont think its factorable

[rocky bison]

If you can't factor it

Solve using quad equation

=tex ax^2+bx+c=0\implies x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}

[simple vault]

but this says to use my graphing calculator

[rocky bison]

Do that then

[simple vault]

how can I do that on my graphing calculator?

[rocky bison]

Graph the function

[simple vault]

im graphing it

but im not getting the right answer

like im trying to trace the 0s

[rocky bison]

What do you mean?

[simple vault]

so I graphed it

and got a sin wave graph looking thing

im trying to get the radian solutions

[rocky bison]

I don't think graphing equal to y is good

If you type the expression exactly as you see it

You'll get a bunch of vertical lines which are the solutions

[simple vault]

what if you dont put the =0

and is it in radians?

[rocky bison]

Then it sets it to y

I've no idea

Depends on your calculator

You can normally change the mode

[simple vault]

did you use desmos?

[rocky bison]

yes

[simple vault]

hmm

im trying it on my own graphing calculator too

but its really confusing

[tacit jewel]

@cheeze weird, the markscheme shows us finding them. Is there another way of solving the problem?

[dusk solstice]

Hey guys, need urgent help with a problem

PLs help

[viscid thistle]

@dusk solstice 4sin(theta)

[dusk solstice]

ah

and b?

[dusk solstice]

w8 im confused

[mortal basin]

can someone explain to me descartes rule?

[young lintel]

uh

@mortal basin (wow nice name) The amount of real 0s in a polynomical function is either equal to or less than by even numbers of the number of changes in signs

but that seems pretty simple so what u need explained..

[hexed ermine]

a. sin(theta)=opposite/4

The opposite being that height you are asked for, so to solve for the height multiply both sides by 4, 4sin(theta)

[mortal basin]

how do you find the bounds to the zeros of a function

[young lintel]

https://www.mathsisfun.com/algebra/polynomials-bounds-zeros.html

idont know

xd

[mortal basin]

our teacher didnt teach it to us but gave us homework on it

like the fuck

[trim coyote]

Is this a good place to ask questions about Linear Algebra? Specifically matrices and vectors

[wide frost]

[trim coyote]

I'm confused as to what exactly the dimensionality of a vector like [x, y] is. It should just be two dimensions, but it only has 1 column, whereas a matrix with 3 rows and 3 columns has a 3 x 3 dimensionality

So why is this?

3 x 3 dimensionality is also two dimensions right?

[wind igloo]

A 3x3 matrix is rank 3, I think.

You may need to be more precise about what you mean by dimensionality.

The term is used differently in different fields.

[trim coyote]

I really do. I'm somewhat of a novice to math and just jumping into Linear Algebra with little experience.

A given point in a matrix, say a(sub)3, 4 is a single point in a two dimensional space.

How can you write a 3-dimensional matrix (Tensor if my belief is right)

A vector with 6 components has 6 dimensions, whereas the number of columns and rows in a matrix represent their dimensions right? Because matrices are purposed for indexing quantities? Is this right?

[wind igloo]

In mathematics we usually talk about vectors and scalars. An element of a matrix is generally a scalar.

Not a point.

[trim coyote]

It's a scalar?

[wind igloo]

For a matrix, we usually think of rank, rather than dimensionality.

[trim coyote]

Okay, I'll read into that

[wind igloo]

I assumed that's what you meant.

I tend to think of matrices as transformations

They operate on vectors and send them to other vectors.

[trim coyote]

In machine learning, something I was reading referred to the input vector as having 13,002 dimensions because it had 13,002 components, whereas the values of the matrices for the hidden layers were treated as single points in this 13,002-dimensional space. I'm very likely far off base here.

Again, total novice just trying to get a footing.

[wind igloo]

I don't follow that analogy

[trim coyote]

Likely because I'm getting it completely wrong.

I'll start reading and come back with more understanding.

[wind igloo]

Ok. Good luck

[signal fable]

@dusk solstice MGH

[subtle narwhal]

hey guys

any tips to memorise trignometric identities ?

[rocky bison]

Use them

Or write them out at the begining of every question

I remembered the equations of motion like that

at evvery question I'd write all 5 of them out

[wide frost]

@subtle narwhal complex exponentials

imo best way to remember

[swift heath]

does anyone know how to solve this

[clever inlet]

hmm

that looks like a common denominator already

except the 1 at the start

convert everything on the left to a common denominator

see where that brings you

[swift heath]

i was able to cancel out the cos^2 x by turning the 1 into positive sin^2 x + cos^2 x

[clever inlet]

yeah

now notice the RHS in the original has no tans

so it would make sense to convert tan(x) to sin(x)/cos(x)

it's gonna get very gross

so i recommend rewriting the fraction as division

(tan(x) + sin(x)^2 - sin(x)^2tan(x)) ÷ (tan(x) + 1)

[swift heath]

ok

[clever inlet]

it's just it would be a complex fraction otherwise

it doesn't actually change anything mathematically to do that, it just makes it less confusing

[swift heath]

would it be easier to get say, both sides of the equation to equal something else which is also the same thing?

like say, to prove 2+5 = 1+6, i confirm both sides equal 3+4

[viscid thistle]

Could someone help me with this? )':

#51

[prime prawn]

@viscid thistle my dude 110/8.3=180/x

[viscid thistle]

That's using laws of sine right? o:

This is what I have so far

[storm stirrup]

$$c^2 = a^2 + b^2 -2abcos(C)$$

[viscid thistle]

This is the chapter before Law of cos tho

[storm stirrup]

Oh

Unlucky

[viscid thistle]

Is it possible to solve it using law of sines?

[storm stirrup]

Yes

[viscid thistle]

Tell me your secret duck

How would you do that? o:

[storm stirrup]

O

[viscid thistle]

I'm assuming that it's not centered

[storm stirrup]

It is centered

Impossible to solve without being centered

[viscid thistle]

;_;

[storm stirrup]

It is centered u goober user

Ok so ramchaves

[viscid thistle]

Alright my b 😤 👏

[storm stirrup]

Ignore this goober

Because he a bingus

[viscid thistle]

I can't even see him tbh

[storm stirrup]

You know the triangle is isosceles

Because it is centred

Therefore the two unknown side lengths are the same

Call them "r"

And

The two base angles are equal

So because they're equal

And the sum of all angles in a triangle add to 180

O

Um

Ye

35°

So use sine rule

[viscid thistle]

You are the smartest man I know

Thank you for your kindness sir

[storm stirrup]

$$\frac{r}{sin(35)} = \frac{8.3}{sin(110)}$$

[storm stirrup]

U wanna fight @pine kindle

Wat

Oh well yeah

But thats gay

Sine rule is easier thi

Tho

Less um