#precalculus
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@upper smelt it sounds like to me you don't know the unit circle very well
Yes
<@&286206848099549185> Can i get some help on parametric equations?
please
Show it !!!
yeah just ask!
rip
not sure of this would be in calc or precalc but does anyone know how to help me with this
i know that the sum from k=0 to n of k^2 = (n(n+1)(2n+1))/6
but how would you do it for any power like 5 or 1/2
Math steps unclear. Now stuck in the 5th dimension. What do?
Hi. Where can I find hard questions that are tricky and so regarding a certain subject?
what ^
Can someone help me with this? I am not sure how to follow
I got -1 for Horizontal asymptote, and -2 & 2 for vertical asymptote but it
'is incorrect
where the top part intercepts athe the infinite curve
I thought that was the y-intercept asymptote
?
Don't all horizontal asymptotes have some form of y intercept?
Indeed so

@clever inlet you still need help?
7?
yep
Okay
So you started off a bit correct
You want to determine the amplitude and and the vertical shift
which so far would be 4cos(x)-1
now what we need is the b and c value
@queen ibex
ok how do i find that? thats where im lost on
Alright
so first find the period
What you want to do is subtract the x coordinates
$$\frac{\pi}{6}-\frac{-\pi}{3}$$
yea
Okay so that's the distance from the minimum to the maximum
that means its a half a period
so multiply that by 2 to get the full period
Which is just pi
so pi is the period
Now you know that 2pi/b is the period right?
yep
Okay so we do 2pi/b=pi
since pi is the period
and we can say that b is 2
2pi/2 = pu
pi*
Wait dont u do 2pi/b=7pi/6?
ok
Okay so we have so far
4cos(2x)-1
now lastly we need to find the phase shift
Which is defined as c/b
Now I prefer to use cosine first bc it's easier for this problem imo
I notice that the maximum is pi/6 to the right
So what I can do is c/b=pi/6
and we said that b is 2 already
$$\frac{c}{2}=\frac{\pi}{6}$$
Cross multiply 2pi=6c
c=pi/3
There you have it
and since it's pi/6 to the right you do -pi/6
So we have as a final answer..
$$4\cos(2x-\frac{\pi}{3})-1$$
got it?
yep
Alright and finding the sin is answer
i didnt know u can find the shift like that
is it the same thing but different shift value??
ok thank you so much
No problem
@hexed ermine can C/B to find phase shift can be used for sine equations as well???
Indeed
the only thing you need to do is add pi/2
since cosine and sine waves are 90 degrees out of sync
ok thank you
Draw a force body diagram first @upper smelt
So just draw a diagram first showing the respective forces
Yes
Weird that forces are in kg not N
Oh I see
Anyways, you could decompose each vector into their x and y components then add them up for resultant force
Assume one of the forces is parallel to the x axis for convenience
How does your pic look like, I'm on android so can't draw sorry
If you draw a diagram, the x component is just cos theta multiplied by the magnitude of the vector
Y is sin theta nultiplied by mag
Add together the components, use pythag to find magnitude: sqrt (x^2 + y^2)
Angle is just arctan y/x
Nice, there's an answer on homework help server too for your q @upper smelt
Same method
Llama's method is probably best way
That's what I would've done
can anyone help me with parametric equations
Ok
@viscid thistle How do I graph this stuff
What are the eqns
Add them
x+y=4cos3 (t)+2sin3 (t)
Try get rid of the t
Then you are left with only x and y then you can graph
yeah
What have you tried?
Try solving this
umm
=pup sinx from 0 to 2pi
Query made by @fringe stream
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All values of x greater than that for which sin(x) is -1/sqrt2, and all values of x less than that for which sin(x) is 1/sqrt2
Between 0 to 2pi
???
=pup solve -1/sqrt(2) <= sin(x) <= 1/sqrt(2)
Query made by @fringe stream
Data sourced from Wolfram|Alpha: http://www.wolframalpha.com/input/?i=solve+-1%2Fsqrt(2)+<%3D+sin(x)+<%3D+1%2Fsqrt(2)
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=pup solve -1/sqrt(2) <= sin(x) <= 1/sqrt(2), x between 0 and 2pi
Query made by @fringe stream
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=pup solve -1/sqrt(2) <= sin(x) <= 1/sqrt(2), 0 <= x < 2pi
Query made by @fringe stream
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Lol
Make time stamps and plot it on a graph
If it makes 1.5revolutions per minute/ that means it makes 1 revolution every 1/1.5minutes
so 1 revolution = 40sec
You know i've been working through this problem and im not sure if you can do it
This is the closest thing I have to it
you need a function where it goes up and down in a linear fashion and im not sure if there is a way
<@&286206848099549185>
Maybe they might know
Gut says $$150\sin\left(1.5x\right)+156$$
I dont understand how you got that @grizzled hull
I'm still editing it.
ok
Final answer lol
it doesnt say its right on the website
huh okay I guess ill wait for Shenzao
I hate webassign lol
wouldnt the amplitude be 156?
since its 150 feet + the wheel(6)
but mine is in seconds
We translate it up by radius + height above ground
So we have at least $$-75\cos(_)+81$$
I just need to figure out how to make periodicity of -cos(x) 1.
Mm. Didn't see"minute" there.
oops
still, 1.5 per minute is 3 per two minutes.
I get the period from the revs/min?
It makes one revolution in 40seconds
or 2/3 minutes
so put 3pi as b????
2pi/b=2/3
$$-75\cos\left(3\pi x\right)+81$$
Yes
Looks correct to me
@simple vault $$-75\cos\left(3\pi x\right)+81$$
I dunno, I like $$-r\cos(2\pi\omega x)+r+h$$
Let omega be the number of revolutions in a minute.
r is the radius of the ferris wheel.
h is the distance off the ground from the lowest point.
the negative starts it at the bottom.
Let it be positive if it starts at the top.
Use sin if it starts in the middle.
This make sense @simple vault ?
I was on the right track :v
Lol savage
Did we miss a small detail?
It doesn't say if t is in minutes or seconds
wrote it in terms of h and t, yes?
Try seconds maybe
Something friendlier to the website may be $$h=-75\cos(3\pi t)+81$$
Since it specified that it needed to be an equation in terms of h and t.
If you already did that, we'd just need to turn 3 into 180 to get it into seconds, I think.
Try that @simple vault
Graphed that.
It looks way off.
No
It looks perfect.
Kill me.
I did that right, then edited it 6 times.
It seems I'm no longer confident.
Where's the h?
Do you need it written out for you?
Which, btw, has already been done, multiple times
If that doesn't work, replace pi/20 with 3pi
3pi worked
:+1:
so I have these set of problems here but I don't know how to go about approaching them. The textbook doesn't have good examples of it either
Yes
It's the number
Yes
The arcsin function does something to the number
And then the sine function undoes that
Giving back your original number
okay now what is the inverse is outside of the parenthesis?
if*
if the inverse is outside and the sine/cosine/tan is inside
so theyd still cancel eachother out?
Yeah
mostly try to find the equation of the parabola
put the coordinate system with (0,0) at the vertex
Where is the law of tangents ๐ค
what is the symmetry of sin and cos
Could you elaborate a bit more on that?
Sorry
Im reviewing my test
And the question is what type of symmetry does f(x)=sin(x) have?
I said neither
But that because i compeltely forgot
Oh, well, cos has reflective symmetry over the y axis then
But its asking for odd or even
For sin(x) at 0; it's odd
Yes x(x-2)(x-4)
For f(x)>0
I presumed that the area in between is (2,0)โฉ(4,0)
but I got that wrong
I think it's actually negative
but you can't go below two 0's in one place
so I am not sure how to write this
That does not look right
Go from left to right
It's positive from negative infinity to 0
And again from 2 to 4
Hello, is anyone available to help me understand how future and present values work? I was trying to answer the following question : "How does the interest rate affect the present and future values of an annuity?" This is my answer but after thinking about it more I don't think it is correct : "Assuming the time it takes to pay off a loan stays the same, the present value of an annuity with higher interests rates will have to pay larger annuities while the present value of an annuity with lower interest rates will have to play smaller annuities. Assuming the time given to future value annuity and the outcome is the same, higher interest rates correlate to lower annuities and lower interest rates correlate to higher annuities." It seems the more I think about this I start to get it and then I confuse myself.
Could you answer the question "how does the interest rate affect the present and future values of an annuity?" by saying...
Higher interest rates increase the future value, and lower interest rates increase the present value?
can someone help me simplify a trig expression?
If you send the trig identity sheet
well I can't find a good trig identity for sin(sq)x-1
there are a bunch for 1-sin(sq)x
here is the right one then
naturally it should work
got something
@rotund musk
just found : -1+sin(s*q)*x
that doesn't work. I just checked.
yeah checked too
website is shitty
SHIT
oh
lol
Simplify a Trigonometric Expression - powered by WebMath
had also this one in mind
hope i couldve help you a bit
๐
first it was like ( dosent support multiple functions ) and then you just had to adjust it a bit
I believe that the internet/ computing all these huge ressources made possible a lot of complex / boring tasks such as these calculations and made our lives just simpler and eaiser. If you were to go on math , go on computational / symbolic algebra
machines do the hard work for you
Like imagine a huge repertoire where hunders and thousands of devolopers contribute their efforts in calculating stuff like a huge dictionnary and compile all of that
main intrest left is just to learn new concepts on and on
back then it was just as boring as it might seem
Calculatorsbare best
๐
@candid vine did u simplify it
It's the imaginary part of (cosx + isinx)^17 divided by i
I can complexify it
U can use binomial theorem
Regarding my question yesterday, this is what I needed:
And from then I look up the value in radians and solve for x. Sorry for the mess
do you guys have any tips on remembering reduction, half-angle, etc. formulas other than just writing it a ton.
half angle formulas are dumb
you can get all of them from sin though
by noting cos(x) = sin(x+pi/2)
lright thank you
Ok this problem is driving me crazy.
I know that the Amp = 2 , Midline = 0 and the Period is 6 which is 2pi/6 so pi/3
It is a cosine function shifted to the right one unit so that would be -1
Therefore the answer should be -2cos(pi/3x - 1 )
It looks almost identical on desmos but I cant get it to line up perfectly.
Query made by @rocky bison
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Ok
You want 2 and -2
So you're going to just multiply by 2
So currently we have 2cos(x)
Amplitude should = 2
Which lines up in y
yeah
=pup graph 2cos(x)
Query made by @rocky bison
Data sourced from Wolfram|Alpha: http://www.wolframalpha.com/input/?i=graph+2cos(x)
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Looks good
Yep.
divide 2pi by 6 and simplify
Sure thanks
nice
Oooh
btw this is using radians hopefully that's no issue
I think that might be where I messed up?
Hold up
We want the inverse of that function
Because we'll be making a transformation
Of type
f(ax)
Which is a "squeeze" of a
Wait no that's good, 6 is less than pi
nvm
Let's graph it and see if period is correct though
=pup plot 2\cos(\frac{3x}{\pi})
Query made by @rocky bison
Data sourced from Wolfram|Alpha: http://www.wolframalpha.com/input/?i=plot+2\cos(\frac{3x}{\pi})
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I think pi/3 might be the correct period
Yep
It is
It is the reciprocal
1/a
stupid mistake
whatever though
=pup graph 2cos(\frac{\pi x}{3})
Query made by @rocky bison
Data sourced from Wolfram|Alpha: http://www.wolframalpha.com/input/?i=graph+2cos(\frac{\pi+x}{3})
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ok
perfect
now last
we want the first peak
at -2
you can choose any point on the graph
so -A
just make it consistent
not -a
+a
It's of type
f(x+a)
which shifts everything left
by a
So from 0 to 2
2
=pup graph 2\cos(\frac{x\pi}{3}+2)
Query made by @rocky bison
Data sourced from Wolfram|Alpha: http://www.wolframalpha.com/input/?i=graph+2\cos(\frac{x\pi}{3}%2B2)
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Hopefully that was useful to you
It was
But the software doesn't like that answer either.
But it does look very similar to it on desmos
myopenmath
Ok. So the amplitude and the period and the midline will be the same
Not sure what midline is
So i just need to find the horizontal shift
Ye
splits*
=tex \sin(x)=\cos\left(x-\frac{\pi}{2}\right)
Although might not like that
Might be easier starting agian
Also tests your understanding
like a new problem completely?
yeah
I can reset it. Ok
I'm heading to sleep though, so goodluck ๐
Where am i going wrong? Z is supposed to equal .8 but i got 8/11 which is 0.73
Hi
hi
have you tried row reductions
not sure what that is
inverses?
do you mean like get zeros in the other rows?
No
If u wanna solve a 3x3
U find the inverse
Then transpose
Well I guess there r faster ways
gaussian elim to get 4/5 for z
your last step is wrong
if you do r1+r3 you get 1 in the first column third row
but you set it to 0
oh ok yeah that last step really messed me up
I don't understand what you are doing, what is that 4th columnsupposed to represent?
its a system of equations and the 4th column is the = to part like 4x +5y + 6z = 7 ... the 7 would be on the 4th column
Oh you write your systems like that, that's weird :o
its just something in math i guess

theyre called matrices
its a linear algebra concept
of gaussian elimination in matrices
look up augmented matrices
i shall be back when i fail this next practice example lol

got the next one right 
Im trying to do 37 but no matter what series of changes i keep getting y as 1.5 but the answers says 3.5 for y
im trying to use the matrices though
it does
Why don't you show us what you've done?
It's just that you wrote 2 instead of -2 in the first matrice (last line, last column)
You made a mistake copying down the right hand column.
One of those values should be -2
Those sorts of mistakes are the worst, you never see them when you make them
Sign errors are some of the hardest errors to catch.
on which set?
If you look at problem 37, you have ... = 5, ... = 2 and ... = -2
But your right hand columns in the top line do not contain -2.
oh my god
i made the same mistake on my first attempt when i wrote the second set smh
double check when you copy stuff
its funny because i usually get more errors on that kind of stuff than anything else
by the way gator would you if you add line 1 to line 3 get 2x + 2y = 7?
2x+2y = 3
oh wait nvm i see now
Err. 2x + 2z = 3
its the negative of the 3rd line
Yes
do you guys know if there are solutions when a row in a matrix is 0 = SomeNumber? Normally if its all zeros id say z=z but with this is it just no solution?
inconsistent system
Agreed. There are no solutions.


sounds good. i was making sure there wasnt something like for dependent systems
do you know the identity?
hmm, does this actually require the identity?
not sure
i'm pretty sure it does
i dont
cos(a+b) = cos(a)cos(b) - sin(a)sin(b)
do i need to make the triangle and find the missing side for each one
yea, given hypotenuse = 1, sin(a) = 15/17, you can find cos(a)
etc
but you have to be careful about the signs of the cosine values
for example
if sin(a) = 15/17, and a is in quadrant I
will cos(a) be positive or negative?
so if the missing side is 8, what is cos(a)?
all are positive in 1st quadrant
yea
8/17
well, to find cos(a+b)
you use the formula
that
cos(a+b) = cos(a)cos(b) - sin(a)sin(b)
so you can substitute the values you have for cos(a), sin(a), cos(b), and sin(b)
remember
cos(a+b) does not equal cos(a) + cos(b)
8/17 x 7/25 - 15/17 x 24/25
yep
200/425 119/425 -
is that
right so far ?
i havent solved something like this before so
I think you multiplied the numerators incorrectly
&& \frac{8}{17} \cdot \frac{7}{25} = \frac{56}{425} and -\frac{15}{17} \cdot \frac{24}{25} = -\frac{360}{425} $$
is it just 8 x 7 ?
[(8 x 7) - (15*24)]/425
$$ \frac{8}{17} \cdot \frac{7}{25} = \frac{56}{425} and -\frac{15}{17} \cdot \frac{24}{25} = -\frac{360}{425} $$
-304
so some important identities to know are sin(a+b)=sin(a)cos(b) + sin(b)cos(a)
and the one for cosine
for cos(a+b)?
yeah
you typed only 304, right? not -304 (since it puts the negative there for you)
i tested both
what does that do
so the cos(b) should be negative
so its 416/425 ?
alright sweet
what is the identity for sin now ?
sin ( a + b)
sin(a+b)=sin(a)cos(b) + sin(b)cos(a)
ok
thx
i saw it up there
if you ever wanted to find them online, you can search angle addition formulas
also
since you already know cos(a+b), you could also try to complete the triangle
but if you did it by completing the triangle, you would have to figure out which quadrant a+b is in
you could always just use the sin(a+b) formula
so that you don't have to figure out the quadrant
the only reason finding the quadrant is necessary
yeah im gonna use that
oh alright
but i draw the triangles so i can figure out the sin and cosine
didn't you already figure them out?
yeah but i did it all in my head
oh, do you need to show work?
yea that's fine
better if i can just look at it than try to remember it
lol
is that the quickest way to solve these
with the indentity ?
and what is the identity for tan(a + b )
The identity is going to be the fastest
for tan(a+b)
well
I never remember the identity for this one XD
because usually
oh it gave it to me
oh
just multiply by one
:D
for gnarly fractions like that
you can use the property that anything times one equals itself
like
2 x 1 = 2
fancy
but
you also know that, for example
2/2 = 1
$$ \frac{\frac{-416}{425}}{\frac{87}{425}} $$
so -416 / 87 ?
yea
just multiplying numerator and denominator by 425 (aka multiplying the entire thing by 1)
it says its wrong
what do u mean
i did the identity for each one
yes i got those correct
tan(a+b) = sin(a+b) / cos(a+b)
so it should be
$$ \frac{\frac{87}{425}}{\frac{-416}{425}} $$
is that 87/-416
yea
remember
you can take the negative sign up to the numerator
by multiplying numerator and denominator by -1
just a by the way sorta thing
so it should really be -87/416 ?
yea, they're the same thing, but it's more common to put the negative sign in the numerator
did everything make sense?
alright
aw
well
I mean one problem at a time
you got dis
I'm going to go now though, sorry :/
good luck though
alright
I'm not sure what I'm doing wrong for this problem. My understanding of the definition of the secant line is that it is just the line between two points of a function.(edited)
My understanding to find the solution to this problem is y = 42 * ((heartbeats at t - 2936)/t - 42)
but the website doesn't like my answers
That is the slope of the tangent line
Which is the same as the slope of the secant line
Huh, that's what I have and it doesn't accept it. 1
The numbers shown in the image are not what @haughty thicket has calculated.
The calculation above gives, approximately, 66 beats/min
can anyone suggest me good website for learning function sketching?
Iโm on 4. The answer is going to be in fractions like is for the example problem at the top of the pic. I can also show an example problem if that helps. I am confused when sliding everything to one side and then the factoring.
Hey guys. I've been grinding out some homework all day and got stuck on this problem. Can anyone point me in right direction?
ayyyy general solutions
I know that I can use the double- angle formula to turn it into 16sin(x)cos(x)+15sin(x)=0
Do i combine like terms?
Thanks
Oh cool it's asking for non exact answers
Phew

So yeah getting to 16sin(x)cos(x) + 15sin(x) = 0 is correct
You just have to factor out sin(x)
So you get sinx(16cosx + 15) = 0
This is true when either sin(x) = 0
Ok
or 16cos(x) + 15 = 0
Go from there, if you get stuck just shoot. List your solutions here before you input them
So i need to solve for x for the last solution?
In both
Ok. Thanks for pointing me in the right direction. Let me try and figure the rest out.
Make sure you find ALL solutions between 0 and 2pi, you should have 4
@storm stirrup Ok so I know that sin is 0 at 0, and pi
but I'm blanking on how to find the other two values. cos(x)=-15/16
So for the other two values
The first one you can find by simply putting in your calculator arccos(-15/16)
Make sure you're in radians
yeah
Just a different way of writing it
Do you have or know a formula for general solutions?
+2pi n ?
Because you're dealing with cos(x)
You have cos(x) = cos(2.786)
And because trigonometric functions are cyclic, you can use the general solutions formula for cosine
so x = 2n(pi) +- 2.786
You just have to find the integer(s) "n" that make x between 0 and 2pi
So i could subtract?
yeah
yeah
Well the only two values of n that allow for an x between 0 and 2pi are n=1 and n=0
I thought i could do 360- the value
Yeah stay in radians
but i had to do 2pi-2.786
yep
Great. Thanks for the help.
So what are your 4 solutions?
Very helpful. Thank you.
initial speed of a function?
Mmmmmm are you talking about integrals and that such?
@dense trellis show us the complete problem
its literally that function and it was in a test involving rate of change and instataneous and stuff
and it asked for the initial speed
i agree with PJS in any case ๐
@lucid spindle thats only the vertical speed
Isnt that what they asked for?
they asked for the initial speed
they equation they gave is y=f(x)
you would initially have to find the 2 equations x = f(t) and y = f(t) and then sub one into the other to get y = f(x)
How is that just vertical speed?
He took a differential of the graph which I assumed was based on the idea it's a displacement function
@earnest sable
Meh. It's a poorly-framed question.
The form of the equation, f(x) = -4.9x^2 + 13.4x + 7 -suggests- that it ought to be displacement as a function of time.
Because -4.9 is -9.8/2, which happens to be the acceleration due to gravity in m/s^2
That said, Dusty isn't wrong in that it only accounts for vertical displacement.
There is, however, no information given on what, if any, horizontal displacement/velocity/acceleration there is.
A particle is moving along the curve y^2 โ 6x^4 = y. At the moment when x = โ1 the x-coordinate is increasing at the rate of 5 cm/sec. If the y-coordiante is negative at this moment, is y increasing or decreasing?How ast?
anyone know how to solve this
all i know is i need to take the derivative implicitly
nvm found a khan academy vid on how to do it
What is a "Trace" of a matrix? When and why would we need it?
I can google pretty easily to find it's the sum of the diagonals, but that doesn't tell me what it represents geometrically or answer the second question.
Hi guys. Anyone available for some quick trig help?
I know how to replace the sin^2 with cos^2, and I know how to factor the equation. I get cos=-3/4 and cos = 1/2
I also know the values of cos = 1/2 between 0 and 2pi. But i struggle understanding how to figure out the values of cos=-3/4 over two pi.
can't you just do arccos(-3/4) forthat?
Yeah I can gives me the correct value in the QII
Now I need to find the value in QIII
QIII?
quadrant 3
Nvm. I think I figured it out. I was over thinking it.
take the value and subtract it from 2pi
Yep
or add pi/2
For anyone interested, I spent some time on Desmos making an interactive Unit Circle
You can play around with it if you want
Damn. That is pretty cool.
Ty
๐๐ผ
Okay basic question. I need to distribute 3(2sinxcosx)
would it be 6sinxcosx or 6sinx3cosx
Treat sin x and cos x as if they were just variables
It would be 6 sin (x) cos (x)
okay thanks
np
Okay. Why can't I get this right? Dont i just do arcsin(-.45) and add pi/2?
arcsin(-.45) = -.4668
=tex sin\bigg(\frac{\pi}{2} + x\bigg) \neq sin(x)
Eh.
Oh I see
It has to do with the Unit Circle
The graph I sent up in chat has some explanation
Do you have it open?
yeah
Ok so do you know how sine and cosine correspond with the Unit Circle?
yes, cos in the x value, sin is the y value
yes
Try and visualize that
ok sin is negative in the 3rd and 4th quadrant
Yep
So since it's a circle, what's significant about these two values?
Hint: I'm talking about the x-coords
I'm not sure
They're opposites of each other
oh yeah
For example in another problem one would be x = 1/2 and the other would be at x = -1/2
ok yes
Hold on, no it's asking for the angle in radians not the x coordinate.
Damnit
Ok so you want to find arcsine (-0.45)
And that's one of your answers
You still here?
Yeah I am
Ok good, but since we want our answer as a positive, what can we add to it?
pi
gg
That answer is correct
yes
now i just need to find the other value
so 5.8164 is in the QIV right?
can't i just subtract pi/2 from our original answer
Anything greater than 3ฯ/2 is in quadrant IV
because i need QIII
No, you could only subtract ฯ/2 if theta was halfway through Quadrant IV
(315 degrees)
hmm
I'll make a visual
The shaded areas are the same area, just mirrored across the y-axis
ok
So what we want to do to find the question mark is subtract 5.8164 from 360 degrees or 2ฯ to get the shaded angle
And then add that value to ฯ
Can you visualize why that works?
Yeah that helps
Ok so
np
I really need to get into the habit of drawing
I'm realizing im much more of a visual learner haha
so soon as I saw that picture it clicked.
Visual representations always help, especially when you get into physics
Yeah me too
I need a video or a whiteboard in front of me lol
No problem man
I could use some tips if anyone has any
Some tips would be to determine the amplitude
and the period of the function
and if there is any vertical shift or phase shift
Well so far you have 3cos(x)
maybe cos -1