-12/5
#precalculus
viscid thistle
willow bear
yes
that is your slope
and now you need the y-intercept
luckily, one of the points you were given is the y-intercept
willow bear
yes
viscid thistle
i wasn't given any points I was assuming y=4 and x=-8
-8*
so if the y intercept is 4
willow bear
what do you mean you weren't given any points? then what are those green points there?
the ones for which it's clear that their coordinates are (0, 4) and (5, -12)?
willow bear
(-2 2/5)x + 4, if you're insisting on writing that as a mixed number
willow bear
i'd leave it as -12/5 honestly
viscid thistle
So I think mixed number?
willow bear
i mean honestly
-2 2/5 and -12/5 both refer to the same thing and are imo equally "simplified"
viscid thistle
yeah I will leave it at -12/5
willow bear
except that writing mixed numbers like that kinda clashes with juxtaposition meaning multiplication in other contexts
nimble sand
viscid thistle
yes
willow bear
so
which ones of the graphs you're shown include the part of that graph that corresponds to x > 1?
willow bear
no
neither D nor C contain even part of y = x+1
you picked precisely the wrong answers to my question
willow bear
yes
viscid thistle
Is the domain (1,1)
willow bear
the domain is the set of all x values for which f(x) is defined
viscid thistle
(-2,2)?
willow bear
answer my question, please
viscid thistle
i'm not sure
willow bear
is f(x) defined when x ≤ 1?
can you calculate f(x) when x ≤ 1?
don't actually do it
just say if you can
viscid thistle
no
willow bear
yes you can.
viscid thistle
how?
willow bear
what do you mean, "how do you get that"? it's literally written there!
willow bear
you can calculate f(x) when x ≤ 1
and when x > 1
so you can calculate f(x) for all values of x
so the domain is all real numbers
viscid thistle
is there supposed to be one or two sets of numbers for the domain?
willow bear
so the domain is all real numbers
all real numbers
do you know how to write that in interval notation?
viscid thistle
no
willow bear
do you know how interval notation works?
viscid thistle
inf is a real number?!
viscid thistle
oh
willow bear
(a, +∞) refers to the set of all real numbers greater than a
(-∞, b) refers to the set of all real numbers smaller than b
(-∞, +∞) refers to the entire real number line
willow bear
yes
willow bear
what were you asked to do?
viscid thistle
yes that's what I wrote sorry in the interface I have the option to select the exponent
willow bear
anyway, yes, that equation corresponds to that graph
viscid thistle
wooo okay thanks
and the graph of y = |x| is reflected in the x-axis, stretched by a factor of 3, then translated 2 units to the left and up 7 units
would be y = 3|x-2|+7 right?
viscid thistle
hmm
that's what I thought, but it's not an option
this one is multiple choice, these are my options:
willow bear
??
viscid thistle
willow bear
you did the composition right, but the functions are definitely not inverses
if they were, then by definition f(g(x)) = g(f(x)) = x would be true
which
it isn't
willow bear
3 - b^2 factors out to (sqrt(3) - b)(sqrt(3) + b)
so if your input window has a square root option it's probably implied that you should do that
past veldt
willow bear
but yes x + iy and x - iy are conjugates
severe anchor
I have a shiton of trig questions
but Ill ask them later, I'm busy now
thanks for the channel
I'm in Calculus, but pre-calc help could be handy too
smoky lance
What was the answer for the question (7 +6i)(7-6i)? 49 +42i - 42i - 36i^2?
calm thicket
First thing's first, 42i - 42i = 0, so you're left with 49 + 36i^2
*-
i^2 is -1, so it's 49 - 36
==49-36
granite stirrupBOT
13
smoky lance
well I was just foiling it out right quick, but the initial approach was just to multiply the 7 +6i by the conj right?
willow bear
49 + 36 @calm thicket
smoky lance
Just making sure, because I never have seen i or dealt with it before
calm thicket
one typo and I'm backwards for the whole question
calm thicket
=tex i := \sqrt{-1}
granite stirrupBOT
willow bear
bad definition
calm thicket
=tex i = \sqrt{-1}
granite stirrupBOT
willow bear
complex numbers are a thing
to construct them, you introduce a new "magic" number, called i, whose defining property is i^2 = -1
calm thicket
(They're also my favorite numbers)
willow bear
and then you just pretend you can do with it anything you can with a real number
calm thicket
Is it possible to real-ify an imaginary number similar to rationalizing the denominator in radical fractions?
willow bear
imaginary as in complex non-real?
"imaginary" or "pure imaginary" typically refers to iy for y in R
calm thicket
For sake of simplicity, just bi
willow bear
iy can be multiplied by i
willow bear
in general though, multiplication by the conjugate is a thing
=tex \frac{1}{x+iy} = \frac{x-iy}{(x+iy)(x-iy)} = \frac{x-iy}{x^2 + y^2}
granite stirrupBOT
The renderer took too long to respond.
faint acorn
willow bear
faint acorn
:)
calm thicket
Mhm
willow bear
feat. the slightly fancy i that i use in handwriting for the imaginary unit
calm thicket
Fancy.
faint acorn
gotta use those fancy letters for maths
calm thicket
"Carl bought two plots of land for a total of $120,000. When he sold the first plot he made a profit of 15%. When he sold the second he lost 10%. His total profit was 5500. How much did he pay for each piece of land?" On attempt four, every attempt I've gotten either impossible solutions (350k for one of the plots) or extremely unlikely ones (99 cents vs 129k, etc), confirm I set it up properly?
Assuming plots are a and b, the cost of a + b = 120000. On plot A, he made 15%, and on plot b, he lost 10 %, so the difference (5500) is applied in a manner of 1.15a + .9b = 125500, and from there I solve the system, correct?
=tex b = \frac{125500 - 1.15a}{.9}
granite stirrupBOT
calm thicket
=tex a + \frac{125500 - 1.15a}{.9} = 100000
granite stirrupBOT
calm thicket
And I just go from here?
But then, we get a = 142000, which means that b has a negative cost, which is impossible
Oh wait, =120000
Then I get a = 70000, and therefore b = 50000
Correct?
willow bear
guess so
smoky lance
so
f(x)= mx + b can be written as f(x) = a(1)x +a(0)
with the (1 and 0) being subscript
can anyone exlplain this subscript stuff?
viscid thistle
u mean
smoky lance
because my book does not and I don't understand it
granite stirrupBOT
viscid thistle
well thats the slope intercept form of a line
$$m = a_1 $$ being the slope of that line
granite stirrupBOT
viscid thistle
$$b = a_0$$ being the Y-intercept
granite stirrupBOT
viscid thistle
oh
i think they just mean it in the way
like you know how polynomials are ax^2 + bx + c
smoky lance
ye
viscid thistle
i think they mean to do that to show that the a_0 term without the variable
in this case x
and the a_1 being a linear (or degree 1) term with the variable in it
x
i would consult with your professor / someone else here just to make sure, but it is the point-intercept form of the equation of a line tho
smoky lance
yea I'll go to the help room on monday because it just doesn't seem to make any sense
like what ever it's written like that for
indigo briar
could someone give me an interesting question I could give to some grade 11's?
dusk kettle
what level of math are these 11th graders?
indigo briar
they completed ib math 10 so they know ~1/3 of the 11 curriculum
dusk kettle
I have no idea what ib math 10 is.
Do they know calculus?
or trig?
"integrate the product rule" is a fun one if they do
indigo briar
they know linear algebra, quadratics, trig up to sin law, linear equations, stuff like that
willow bear
linear algebra?!
dusk kettle
one of these is not like the others
"What is the difference between bijection and isomorphism?"
granite stirrupBOT
indigo briar
thanks I think that'll do
last week was a complete blunder
I told the other leader 'bring a chocolate bar as a reward'
he forgot and the reward was a fruit roll up snack from his lunch -_-
they guys face was like a sad puppy the moment he got his reward
untold wave
Give them 300 years lol
buoyant knot
300 years is too optimistic
viscid thistle
So with row eschelon form thingy, how do you turn them into zeros?
I dont get that
dusk kettle
Type 1: Swap the positions of two rows.
Type 2: Multiply a row by a nonzero scalar.
Type 3: Add to one row a scalar multiple of another. ```^ you do any number of those things, any number of times you want to
and try to fnaggle the matrix into having zeros where you want them
=tex \left[ \begin{array}{ccc|c} 1 & 2 & 3 & 4 \ 5 & 6 & 7 & 8 \ 9 & 10 & 11 & 12 \end{array} \right]
granite stirrupBOT
dusk kettle
now let's turn the 5 into a 0 by adding (5 times the first row) to the second row:
=tex \left[ \begin{array}{ccc|c} 1 & 2 & 3 & 4 \ 0 & -4 & -8 & -12 \ 9 & 10 & 11 & 12 \end{array} \right]
granite stirrupBOT
dusk kettle
@viscid thistle did that make sense to you?
hollow cobalt
can someone expl,ain why log3(x)=2 can be turned into x=3^2
calm thicket
This is just how logs work, they're another way of writing exponents.
=tex \log_a(b)=c \text{ is the same as } a^c=b
granite stirrupBOT
calm thicket
As for why, log_a(b) is basically saying, a to the power of what equals b?
That what is c.
And, if you look at the formula, this is true.
So in your example, log3(x) = 2 is the same as saying x = 3^2, or that x is 9.
== log3(9)
granite stirrupBOT
Error: Unknown name: log3
granite stirrupBOT
2
calm thicket
^ Calculator confirms this is true
Is this sufficient? @hollow cobalt
I'm actually just learning logs myself, so anything you can ask to throw me off so I learn more is appreciated.
viscid thistle
@dusk kettle You say turn 5 into 0 by adding 5 times the entire first row? That doesnt make sense
How would that make it a zero
dusk kettle
If you look at the first row in the matrix, the leftmost number is 1. If we subtract that five times from the second row, where the leftmost number is 5, the 5 will be 0 because 5 - 5*1 = 0
you do that for every number in the row at once, though
and you keep doing things like that, or swapping rows around, or multiplying a single row by a constant, until you can get 1s down the main diagonal and 0's everywhere else.
viscid thistle
Can someone help with 13? We cannot figure out this proof
manic root
induction? $$a_0=1$$ $$a_{n+1}=a_n+3^{n+1}$$
so if $$a_n=\frac{3^{n+1}-1}{2}$$ then $$a_{n+1}=\frac{3^{n+1}-1}{2}+3^{n+1}=\frac{3*3^{n+1}-1}{2}=\frac{3^{n+2}-1}{2}$$
granite stirrupBOT
viscid thistle
Hey it’s much nicer than what we were getting when trying to get the proof of the induction
tepid topaz
g(x) = -3 sin ( 4 ( x - pi/4 ) + 2 Are these correct key points for the function(max,min, x-intercept)? pi/4, 3pi/8, pi/2, 5pi/8, 3pi/4
willow bear
=tex g(x) = 2 -3\sin\left(4\left(x - \frac{\pi}{4}\right)\right)
granite stirrupBOT
willow bear
did you mean this?
@tepid topaz
if by key points you mean maxima, minima, and points where the function crosses its midline rather than the x axis (i.e. when the sin or cos term vanishes), then yes, those are indeed spaced pi/8 units apart
if you want to graph this, though, i'd rewrite 4(x-pi/4) as 4x - pi, and then sin(4x - pi) as -sin(x), thus getting g(x) = 2 + 3 sin(4x)
tepid topaz
thank you
crimson gate
@timid jacinth if it's even, then f(-x)=f(x), if it's odd f(-x)=-f(x), for trig functions you could just try plugging in maybe pi/4 and -pi/4 (-pi/4 is 7pi/4)
Just comparing those should be enough I'd think
timid jacinth
That's the only way to do it without the graph of the functions? @crimson gate
crimson gate
That's just the first way that came into my head based on the definition of even and odd functions
cos and sec are even and sin, csc, tan, and cot are all odd
If it's a composition of functions like f(g(x)) then if g(x) is even f(g(x)) will be even too
If g(x) is odd and f(x) is even, then f(g(x)) is even
f(g(x)) will only be odd if f and g are both odd
gleaming schooner
Hey everyone, where can I learn about quadratic functions? I wanna know how to derive the vertex form from the standard form, and how to identify the components of a parabola represented by a quadratic function
timid jacinth
@crimson gate can you explain why cos is even and sin is odd?
crimson gate
An even function is always mirrored across the y axis, that's why f(-x)=f(x)
Odd functions are rotated 180 degrees about the origin
If you look at sin, you can see how if you rotated it 180 degrees it'd be the same and how if you mirrored cos it would be the same
@timid jacinth
smoky lance
holy ship
timid jacinth
@crimson gatewhat do you mean if you look at sin?
timid jacinth
How do I know this without the knowledge of the graph of sin though?
I'm supposed to find this out without the graphs of the functions because we haven't been introduced to them yet
Only the unit circle
crimson gate
then my prior explanation should have sufficed
if f(-x)=f(x) it's even, if f(-x)=-f(x) it'd be odd
just compare some values
jolly turret
quick question does cos have to do with 2pi and sin with pi? in any sort
timid jacinth
I don't understand this at all; https://gyazo.com/61429b02ed777ee8eecd71b108a8c165
Figure 35 = unit circle
clever inlet
since unit circle radius is 1
x and y coordinates can never been greater than 1
or less than -1
timid jacinth
I understand for y
but why are we even talking about x in regards to range?
I thought x-values corresponds to domain?
timid jacinth
Ohhhhh shit
crimson gate
it's saying that on the unit circle the cos(theta) is just the x coordinate
I think
and since x is between -1 and 1, that means the range of cos is -1 and 1
timid jacinth
cos(theta) = x is actually an output, so technically a y-value, but is represented as an x-value on the unit circle?
crimson gate
yeah
clever inlet
yes
timid jacinth
so I'm not looking at the -1 and 1 on the x axis for -1 <= x <= 1, I'm still getting the -1 and 1 values from the y-axis right?
since it's an output?
timid jacinth
Well, I'm operating from the assumption that the graphs are non-existant right now
because I haven't been taught it
so I'm supposed to determine all of this with only the unit circle
clever inlet
well it's hard without it
timid jacinth
what's hilarious is, this is 2.3 homework. Look at what I learn in 2.4 and 2.5. https://gyazo.com/43ccae0e1a90d0e476e0f4a6eb1c3cbe
crimson gate
what an awful structure
timid jacinth
Hold on Steel, it gets better
my teacher, for some reason, before starting trig (which I've never been introduced to) thought it'd be best if we did 4.1 before starting chapter 2, which is the intro to trig.
This is 4.1 https://gyazo.com/41eaa7c818a3366211a6c7514f4f8b6e
clever inlet
rip
timid jacinth
I did
he said he thinks we ought to know right angle trig before unit circle trig
Literally, I had never seen sin, cos, tan, cotan, sec, csc before, ever
and suddenly I'm asked to find the exact measurements of right angled triangles using trig functions only
needless to say, I was fucking lost, and I'm still lost now
timid jacinth
I'm kind of getting it, very, very slowly
I mean, I see the logic behind it
right angles are formed on the unit circle
but
For someone who had never been introduced to trig
it was massively overwhelming to go from basic functions f(x), f(g(x)) etc
to fucking that
clever inlet
rip
timid jacinth
Also, I'm not understanding how for example, the domain of sin is all real numbers
but it's range is limited to [-1,1]
shouldn't the domain be the exact same?
smallest x-value on the unit circle is -1
clever inlet
it can accept any input
timid jacinth
largest x-value is 1
But for it to be on the unit circle, doesn't the input need to be within x = (-1,0) to (1,0)
timid jacinth
oh
fucking shit
yeah
cause
input = angle
input can be 5000pi
just means we go around the unit circle 2500 times
point will still be the same as 2pi right?
crimson gate
when you were doing stuff in chapter 4 did your teacher explain soh cah toa?
or in this chapter I guess
also yes it's just the same as 2pi, which is just the same as 0
timid jacinth
Yes, he mentioned sohcahtoa in passing. All he said was sin = o/h, cos = a/h, tan = o/a, he didn't explain what any of that meant, he assumed we knew the basic geometry etc
clever inlet
so this is basically back to right angled trig
timid jacinth
I have no clue
right now, I'm just trying to understand the range of all trig functions
through my limited knowledge of the unit circle
I don't want to jump into the graphs of the functions yet
crimson gate
it's related to the unit circle
timid jacinth
because we study that next, after the test for 2.3
crimson gate
I'll see if I can find an image that explains it
clever inlet
have you done inverse functions as well?
timid jacinth
yeah, didn't really understand it too well
crimson gate
so the o = the side opposite to the angle, the a = the side adjacent to the angle, and the h = the hypotenuse of the right triangle.
timid jacinth
isn't it like, the output of f^-1(x) = the input of f(x)?
crimson gate
soh stands for sin(theta) = opposite/hypotenuse, cah is for cos(theta)=adjacent/hypotenuse, toa is for tan(theta)=opposite/adjacent.
timid jacinth
Yeah, this is all sounding familiar, which means I've retained it
always nice
: D
crimson gate
So if you have two right triangles and you know that one of the non 90 degree angles in both of them are the same, then you know that they have to be similar, because the third angle has to add up with the other two to get 180.
That means the two triangles have to be similar and the ratios of their sides have to be the same
Here, y is the opposite side and x is the adjacent side
theta would be the angle between the x side and the hypotenuse (which is 1)
the sin(theta) is opposite/hypotenuse, right? so it's y/1, which gets you y, just like the it says in the point
make sense?
timid jacinth
Yeah, which is why we call sin(theta) = y on the unit circle, as in sin is the y coordindate when the angle is in standard position
crimson gate
and that's just because the unit circle is unit because it has a radius of one. if you take any right triangle, just imagine laying it on top of the unit circle
timid jacinth
Okay, I'm understanding; how does this help me find the range of these functions?
crimson gate
lets say the angle was pi/6, and the hypotenuse was 8, you'd be able to find the length of the side that's adjacent to the angle right?
oops sorry this isn't related to the range of the functions
crimson gate
this is just trying to help you visualize what the functions represent
timid jacinth
it's helping me cement it
so you're saying
the angle is pi/6, the hypotenuse is 8, I'd need to find the length of the adjacent leg?
crimson gate
yeah
timid jacinth
wouldn't I need at least one of the sides?
cause right now, I'm looking at x^2 + y^2 = 8
crimson gate
well you know that cos(theta)=adjacent/hypotenuse
so try using that instead of the pythagorean theorem
crimson gate
This is what the trig functions are great for, because they let you find out the side length using the angle
crimson gate
fill in both sides of the equation
timid jacinth
I genuinely had no idea what their purpose was
so hang on
I only know the constant value of the hypotenuse
at this moment right?
crimson gate
you also know the angle theta=pi/6
timid jacinth
ok
crimson gate
theta is used for angles, generally, I assumed it would have been used in chapter 4, sorry if I made it more confusing
timid jacinth
So I have angle = pi/6, hypotenuse = 8
crimson gate
yep
timid jacinth
it was
So I can find cos by assinging a variable for the adjacent angle? like x/8?
or is that worng
crimson gate
I would just use "x"
you don't need to divide it by anything to start with
wait no sorry
misread your question
what you know:
cos(angle)=adjacent/hypotenuse
angle=pi/6
hypotenuse=8
what you're looking for:
adjacent
timid jacinth
adjacent / 8 then right?
crimson gate
(the formula comes from sohcahtoa)
fill in the formula with the numbers you know first
without simplifying
crimson gate
that's right
so this is where the unit circle comes in handy
what are the sin(pi/6) and cos(pi/6)
(is that image showing up right?)
timid jacinth
Yeah, it's fine
crimson gate
so you can simplify the sin(pi/6) and cos(pi/6) and plug those into the two formulas you've got
timid jacinth
So here is where I got lost in class; cos(pi/6)= sqrt3/2, sin(pi/6)=1/2
Why?
It's something to do with the ratio of a right angle right?
crimson gate
it's because on the unit circle, where the hypotenuse is 1, that's what the ratio is
for sin, it's the ratio of the opposite side to the hypotenuse, and for cos, it's the ratio of adjacent to hypotenuse
timid jacinth
But I'm not understanding where these values are coming from
like the sqrt3/2
I was told it was the ratio of a triangle with 30 degrees
and this was all I was told
let me draw it
timid jacinth
so from that, sin = o/h, x/2x = 1/2
then I was told that sin = y coordinate in (x,y), so (x,1/2)
then I'd figure out cos(30)
cos = a/h, x(sqrt3)/2x, simplifies to sqrt3/2
so now I have my point, (sqrt3/2, 1/2)
I don't know why it works
only how it works
crimson gate
so I had a lot of trouble understanding the idea of "where" the numbers come from, but it was only until I realized that the numbers were so intricately related
Every right triangle with a 30 degree angle is similar, right?
timid jacinth
what does similiar mean again?
timid jacinth
so same ratio of sides?
crimson gate
yes exactly
they all have the same angles and they all have the same ratio of sides
timid jacinth
ok
crimson gate
that's why sin(30) is 1/2, because in EVERY right triangle with a 30 degree angle, the ratio is ALWAYS 1/2 between the opposite angle and the hypotenuse
timid jacinth
1/2 between the opposite angle and the hypotenuse
what does this mean?
1/2 of what?
crimson gate
opposite/hypotenuse = 1/2 when you simplify it
because the ratio is always the same in the similar triangles
and the triangles are always similar because they all are 30 degrees
timid jacinth
whatcha mean by this and the triangles are always similar because they all are 30 degrees
well
I know what tha means
that means*
but, are we applying that to get hte values of the rest of the unit circle?
like 45degrees etc
crimson gate
every right triangle with a 45 degree angle is similar to every other right triangle with a 45 degree angle
so every right triangle with a 45 degree angle has the same ratio between the sides
so that's how you'd get the sin and cos
crimson gate
yes, the unit circle is something you memorize
timid jacinth
hyp = 2x, legs = x & x(sqrt3)
I need to memorize that ratio for triangles with an angle of 30 degrees?
crimson gate
with the unit circle you just memorize that an angle of 30 corresponds to the point (sqrt(3)/2,1/2)
and you can see how you can get to the hyp = 2x, legs = x & x(sqrt3) from there, right?
timid jacinth
hmm
not really
I can see how I got the point
from that ratio
not the reverse
crimson gate
so when you have the 30/60/90 triangle lets say you're told that the hypotenuse is 2, you can find the other two side lengths using the ratio right?
crimson gate
solve for the other two sides
timid jacinth
how would I do that?
crimson gate
well first you'd solve for x
then you know that the other two sides are x and x*sqrt(3), right?
timid jacinth
alright, so lets say I'm told the hypotenuse = 4, I know from my ratio of triangles with 30 degrees, that the hypotenuse = 2x, therefore 4=2x, divide both sides by 2, I get x=2, so now I can plug in 2 for x for one leg, and 2(sqrt3) for the other?
crimson gate
precisely
so if you wanted to do that with trigonometry you'd start with sohcahtoa
sin(30)=opposite/hypotenuse
cos(30)=adjacent/hypotenuse
sin(30)=opposite/4
cos(30)=adjacent/4
1/2=opposite/4
sqrt(3)/2=adjacent/4
2=opposite
2*sqrt(3)=adjacent
it gets the exact same results
timid jacinth
Okay
crimson gate
because trigonometry and geometry are fundamentally linked together
You have to have one to have the other
"Where does the ratio come from?" The Angle
"Where does the angle come from?" The Ratio
so that's how you can use the angle of a right triangle with the hypotenuse to find the leg lengths, (or if you had a leg length you can use tangent to find the hypotenuse and other leg)
timid jacinth
Can you give me an example question for me to work through and tell me if I'm doing it correctly?
crimson gate
lets say you have a right triangle with angle pi/3 and the opposite leg is 6*sqrt(3), find the adjacent leg and hypotenuse
you should be able to do this one, but you'll have to use tangent
just use sohcahtoa and fill in what you know (angle and opposite)
willow bear
👀
crimson gate
lets start with finding the hypotenuse
willow bear
sin(π/3) = o/h
o = 6 sqrt(3)
timid jacinth
Okay
willow bear
h = ?
crimson gate
soh, sin(angle)=opposite/hypotenuse
we know the angle = pi/3, and the opposite side is 6*sqrt(3)
just plug and chug
solve for h
timid jacinth
I'm not sure what to plug in though, for opposite I can obviosuly plug in 6(sqrt(3)), but what am I plugging in for hypotenuse?
willow bear
you're finding the hypotenuse
crimson gate
you're solving for the hypotenuse, so you can't plug anything in for it, just use "x" for now because it's the unknown
timid jacinth
okay
crimson gate
you can also plug in the angle, because you know that
willow bear
your work should look like this
crimson gate
yeah, so you see how she plugged in pi/3 for the angle and 6*sqrt(3) for the opposite?
timid jacinth
yeah
crimson gate
so first things first you use the unit circle to find sin(pi/3)
This is where the domain and range of sin comes in, actually, the domain of sin is all real numbers, but you know that the range is between -1 and 1
so you know that when you plug pi/3 into sin, you'll just get a number between -1 and 1
the unit circle will tell you what that number is
willow bear
which is another way of saying sin accepts any input, but it only ever gives outputs between -1 and 1
timid jacinth
So, at this point exactly, I got lost in class, mostly because we were asked to do these questions before we were introduced to the unit circle
timid jacinth
Yes, from the ratio of the triangle with 30 degrees
granite stirrupBOT
willow bear
x = ?
crimson gate
wait no pause
@timid jacinth You know that sin(pi/3) = sqrt(3)/2 because of the unit circle specifically, right? The point is that you're learning it based on the unit circle, rather than the ratios of the special right triangle of hypotenuse 2x and side lengths x and x*sqrt(3)
forget the 2x, x, x*sqrt(3) for now if that's what you were basing it on and just use the unit circle
timid jacinth
Yes, I know that now in retrospect, but when I was introduced to these types of questions in 4.1, I hadn't been introduced to the unit circle, so I was using the ratios of the special right triangle of hypotenuse 2x and side lengths x and x*sqrt(3)
Okay
so I can just look at my unit circle, go up to the angle pi/3/ find that sin= sqrt(3)/2
crimson gate
forget about your teacher's awful choice of doing 4.1 before 2.1
yes exactly
that's how you're supposed to do it
so sin(pi/3)=sqrt(3)/2
now continue with
=tex \frac{\sqrt{3}}{2} = \frac{6\sqrt{3}}{x}
granite stirrupBOT
crimson gate
and remember that x is the length of the hypotenuse
just reminding you because we kinda got a little off track
timid jacinth
right
willow bear
so, can you solve for x there?
timid jacinth
so would I just multiply the 2 from sqrt(3)/2 by 6 to find x?
crimson gate
Just solve it like you would solve two ratios
if that's how you'd do it then go for it
timid jacinth
Yeah, I don't think I know how to >.>
crimson gate
I'm not sure if I fully picture what you're saying but it's also 1am and I'm a little foggy
timid jacinth
How would I solve for x here?
maybe I'm missing it because it's late for me also
or maybe I just don't have a clue
crimson gate
lets simplify it to a ratio like
=tex \frac{1}{2}=\frac{2}{a}
how would you solve that one
willow bear
...why not use a different letter for that
crimson gate
oh good point
willow bear
just to minimize confusion
granite stirrupBOT
timid jacinth
1/2 = 2/4
willow bear
willow bear
i wrote that one out in as much detail as i possibly could
crimson gate
and remember cos=a/h
willow bear
x/12 = 1/2, yes
crimson gate
yep
timid jacinth
x = 6
crimson gate
there you have it!
timid jacinth
so now I have my adjacent = 6, opposite = 6(sqrt(3)), hypotenuse = 12, angle = pi/3
so I've solved the right triangle at this point?
crimson gate
you have solved the triangle
timid jacinth
So the minimum I need in order to solve a right*trinagle is an angle and 1 side?
willow bear
yeah
to solve a triangle in general, you need any 3 elements, provided at least one of them is a side
where an element is a side or an angle
crimson gate
and because it's a right triangle, you actually know two angles, since one of them is 90
timid jacinth
yeah, one of them is given
okay, gotcha
that's really cleared things up for me, thanks
crimson gate
it might be good, after you finish chapter 2, to review chapter 4 on your own
timid jacinth
I'm going to be reviewing everything through khan academy
willow bear
do that
timid jacinth
because my class isn't teaching me anything
so, going back to understanding the range and domains of the functions
I'm understanding why the domain for sin for example is all real numbers, but I'm not understanding why the output is limited
willow bear
well the output of the sine function is the y-coordinate of the input's point on the unit circle
and the unit circle lies between the lines y = 1 and y = -1
is this clear?
@timid jacinth
willow bear
...did i not draw just that?
crimson gate
remember, you're inputting the angle
willow bear
yeah
timid jacinth
oh fuck yeah
crimson gate
you can input any angle
willow bear
x isn't the input here!
crimson gate
I'm not saying this in a "give up" way, but in a "your class is horribly designed" way, it's great that it makes sense now, but as soon as you start to look closely at the graphs of the functions it'll make WAYYYYY more sense
willow bear
this isn't the graph of a function!
that's the reason we used θ for the inputs of sin, cos and tan last time we talked
rather than x
timid jacinth
Okay
since the input is the angle, and the angle is just going to revolve around the unit circle until it finally stops at a point
that's why the domain is all real numbers?
willow bear
correct conclusion, bad wording
the angle doesn't need to ever stop increasing
you can go around the circle as many times as you want, in both directions
crimson gate
the angle can go around forever and ever
that's why it's all real numbers
because it never has to stop at a point
that's exactly what you said golly gee I'm tired
crimson gate
oh Ann are you a math teacher professionally? That's cool
willow bear
i'm not
i'm a first year uni student who hopes to find some income at some point by tutoring
crimson gate
is it a high level algebra class then?
crimson gate
ooooh now that's fuckin sick
willow bear
which part would you like explained?
timid jacinth
basically
crimson gate
so the first part is just saying that the range goes from -1 to 1 and that can be represented by saying |sin(theta)|≤1, because when you do inequalities with absolute values they become -1≤sin(theta)≤1
timid jacinth
everything from the using absolute value part
crimson gate
ok
timid jacinth
the entire thing more or less
crimson gate
so when theta is a multiple of pi sin is equal to 0. csc is equal to 1/sin, so we know that we can't have 1/0
timid jacinth
right
crimson gate
that means the domain of csc is every angle that isn't a multiple of pi
willow bear
and likewise for the domain of cot for that matter
crimson gate
so think about how sin(theta) is either 1, -1, or a fraction (when theta isn't a multiple of pi)
that means csc(theta) is gonna be 1/1, 1/-1, or 1/fraction
and as those fractions get very very close to zero, 1/fraction gets very very big
it gets infinitely close to 0 and it gets infinitely high
that means that the range of csc is from -infinity to -1 and 1 to infinity, but nothing inbetween -1 and 1
does that make sense?
just different wording than the textbook
timid jacinth
Right, so since the csc(theta) = 1/sin, and sin is either 1, -1, or between as a fraction
so from here
is where I get confused
willow bear
=tex \frac{1}{\langle \text{something smaller than 1} \rangle} = \langle \text{something bigger than 1} \rangle
granite stirrupBOT
timid jacinth
right
so since the range can be anything from 0.9999 to basically an infinitely tiny fraction that is just slightly greater than 0
the 1/0^+ will = a massive number potentially
willow bear
i mean the range of sin includes 1 and -1 themselves
crimson gate
the range can be 1
willow bear
if sin(θ) = 1, then csc(θ) = 1, yes
timid jacinth
so where does the -inf in the (-inf, -1]U[-1, 1]U[1, inf) come from?
timid jacinth
did I get that right?
willow bear
were you trying to state the range of csc?
willow bear
yeah no
(-∞, -1] U [1, +∞)
ok so let's consider the function f: [-1, 0) U (0, 1] -> R defined by
f(t) = 1/t
what we're trying to ascertain is that the range of f is (-∞, -1] U [1, ∞)
crimson gate
not sure what you're going for here
willow bear
well csc is just that function composed with sin
willow bear
i will have to depart at some point
crimson gate
@timid jacinth the reason you know that csc can't be a fraction on the interval (-1,1) is because to get a fraction you'd have to do 1/(not a fraction)
but as we said earlier, when you're talking about csc, sin can only be 1, -1, or a fraction
willow bear
"not a fraction" meaning "something above 1 in absolute value", of course
crimson gate
yeah I should clarify we're not talking about improper fractions here
timid jacinth
right
crimson gate
so yeah, csc(theta) can never be 1/2 because that would require sin(theta) to equal 2, and sin only returns values from [-1,1]
just as an example
crimson gate
yeah that
crimson gate
csc(-.5)=1/sin(-.5)
crimson gate
maybe pull something from the unit circle
timid jacinth
like pi/6
timid jacinth
or if we did -pi/6 the output would be -2 right?
crimson gate
sin(-pi/6) is -.5 yeah
timid jacinth
Ok
and if we say, sin=1
then the output would be 1
and that's the absolute highest value sin can be
smoky lance
in Descartes' rule of signs it says number of pos real zeros is determined by the number of variations in signs through the polynm
and f(-x) same thing for negative zeros
soo what if the polynom has no variations, all numbers are positive
0 positive zeros?
plucky locust
granite stirrupBOT
willow bear
by parts
gleaming schooner
Hi
If I have the following quadratic equation in vertex form:
f(x) = -(x-4)² + 1
How do I find the roots of that?
I set f(x) = 0, right?
crimson gate
yep
viscid thistle
Okay, so we need to expand it
or, I would first expand it, then complete the square
crimson gate
The roots of a function is just when y=0, so if you set 0=-(x-4)² + 1 you just solve for x
viscid thistle
Or that too :P
since yeah, its already complete
one of the roots should be 5, and its easy to solve
gleaming schooner
When I do the algebra, I get up to:
-1 = -1(x-4)²
Do I divide both sides by -1 there?
crimson gate
you can divide both sides by -1 yeah
do you know the FOIL method?
for expanding the (x-4)^2
crimson gate
what's -1/-1
crimson gate
yeah so it's 1 = (x-4)(x-4)
crimson gate
you foil the (x-4)
(you don't need to use foil though, because when you have (a+b)² you can use the shortcut of a²+2ab+b²)
gleaming schooner
So I'm left with 1 = x² - 8x + 16
Do I take 1 away from both sides
To get x² - 8x + 15 ?
And then factor that to get my roots?
Though I don't see how that can be factored.
The first one can be.
willow bear
x^2 - 8x + 15 = (x-3)(x-5)
gleaming schooner
Thanks 😛
Need to brush up on my algebra.
I've got another quadratic equation in vertex form.
f(x) = 2(x+1)² - 8
Based on my knowledge, I believe the vertex is (1, -8)
But the answer sheet I have tells me (-1, -8)
How can this be?
Ah
Nvm
Answered my own question
😛
timid jacinth
Trig functions really catch me out
because I'm so use to regular functions always = y on a graph
their output = y
but cos(theta) = x catches me off guard for some reason
since its output is x on the unit circle
gleaming schooner
I have another quadratic equation
-1/5(x + 4)² + 1
I need to find the roots, so I set that to 0.
And now I'm up to -1 = -1/5(x + 4)²
How do I solve it from this point?
Do I divide both sides by -1/5 ?
jolly turret
hey i asked the question earlier but no one was able to get to it so heres another repost.
I have another problem with that phrase above coming from the link: https://www.khanacademy.org/math/precalculus/trig-equations-and-identities-precalc/solving-sinusoidal-models-precalc/a/trigonometric-equations-review . I dont understand how he comes up with the identity you need to use for the problem
gleaming schooner
Can anyone help? 😃 ❤
clever inlet
?
gleaming schooner
I have another quadratic equation
0 = -1/5(x + 4)² + 1
I need to find the roots, so I set that to 0.
And now I'm up to -1 = -1/5(x + 4)²
How do I solve it from this point?
Do I divide both sides by -1/5 ?
clever inlet
Yeah
You can multiply through by -5
You'll get
(x+4)^2 = 5
then
x + 4 = sqrt5
x + 4 = -sqrt5
x = -4+sqrt5 and x = -4-sqrt5
clever inlet
yeah
timid jacinth
How do I find what cos(theta) is exactly?
sin(theta)=4/5 but doesn't tell me what cos(theta) could be
clever inlet
you can figure out the sides
timid jacinth
Ah
one sec
just looking over my notes, I have two options in order to find the values of all other trig functions when given 1 function and an angle which lies in one of the 4 quadrants
I can either use the concept of a circle or I can use identities
boreal shell
why not both 🤷
timid jacinth
hmm
which is easier I wonder
I'll try using identities for this one
the pythagorean identity states that sin^2(theta) + cos^2(theta) = 1
or r
but I'll use the unit circle and call it 1
timid jacinth
well I'd need to solve for cos^2(theta) first right?
clever inlet
i'm just going straight from the original question
timid jacinth
(4/5)^2 +cos^2(theta) = 1
timid jacinth
cos^2(theta) = 1 - (4/5)^2
I'll take the sqrt of each side
cos(theta) = + or - sqrt(1-(4/5)
since I know theta is in quadrant 1, so it's positive, I'll take the positive sign
so
cos(theta) = sqrt(1-(4/5))
did I arrive at the correct conclusion?
or did I mess up anywhere?
clever inlet
i was thinking more of
you know the opposite, you know the hyptoeneuse
you can just pythagoras the opposite
tired nest
No, you can't just square root like that.
timid jacinth
Yeah, that's one of the methods of using the concept of a circle
I wanted to try to use the concept of identities first
I've been taught there's two methods to figure a problem like this out, through identities, which is what I tried
and the other is the circle
tired nest
Okay, try it like this. Before doing the square root, calculate what the RHS actually is.
As a fraction.
timid jacinth
RHS?
timid jacinth
so I have cos^2(theta) = 1-(4/5)^2
so you're saying find (4/5)^2 first?
before taking the square root?
tired nest
Yeah.
timid jacinth
okay
tired nest
Because you can just get rid of the square root by taking the square root of it.
tired nest
Yeah. And what's 1 - 16/25?
tired nest
Yes! Perfect.
tired nest
Yeah.