I have no idea how to solve this further
#Trig Help
foggy iron
is that a cotangent? not sure if i'm reading what you meant
astral sable
How would i do that?
jolly verge
Eh, never mind, it's not necessary
foggy iron
you can think of it kinda like this:
you want t such that cot(t) = s(3). s for square root
that's cos(t)/sin(t) = s(3)
construct a right triangle such that (adjacent side to t)/(opposide side to t) = s(3), say
adjacent side = s(3) and opposite side = 1.
Then calculate the hypothenuse with pythagorean thm, it is s(1+3) = 2.
So now you can calculate cos(t) and sin(t), and identify what angle should be t using what you know of sine and cosine
astral sable
you can think of it kinda like this:
you want t such that cot(t) = s(3). s for...
Wait i kinda understand it
The only thing im mixed up on is how i would calculate cos/sin(t) based on the lengths
I kinda cant grasp trig too well
astral sable
Oh and cot is x/y
jolly verge
Yep, and tan would be the slope of the line. It's pretty cool actually how it all relates
foggy iron
The only thing im mixed up on is how i would calculate cos/sin(t) based on the l...
on a right trisngle where t is one of the other two angles, cos(t) is (adjacent side length)/(hypothenuse length) and for sine it is opposite (to t) instead of adjacent
astral sable
on a right trisngle where t is one of the other two angles, cos(t) is (adjacent ...
Ohhhh so in this case its be s(3)/2 for cos
And 1/2 for sine
I think
jolly verge
Yep
jolly verge
From there you use the radians that are associated with the cos(t) and sin(t) values
jolly verge
That's also on the unit circle
astral sable
Ohhh then its pi/6
Omg im supposed to memorize the unit circle thatd be so hard
Actually wait it doesnt seem too hard
Thank you guys
foggy iron
👍
foggy iron
Omg im supposed to memorize the unit circle thatd be so hard
the numbers have a not so hard pattern
there are mnemonics to remember them easily
astral sable
the numbers have a not so hard pattern
The hardest part i have is understanding it
jolly verge
One small note it that you'll sometimes have to consider the domain of the functions. The given problem could also have the solutions -s(3)/2, -1/2, but the domain of arccot is only (pi/2, -pi/2)
From what I can remember, Isma, please correct me if I'm wrong here
foggy iron
yeah i think it's the same as tan's domain, so that interval
astral sable
Because like earlier we said that cos(t) = s(3) but then we also determined that cos(t)= s(3)/2 so like thats weird
jolly verge
cos(t)=adjacent/hypotenuse, and in this case the hypotenuse we found was 2
Giving cos(t)=s(3)/2
astral sable
Oh so basically the cos(t) = s(3) part was like missing the hypo and we had to get it?
jolly verge
The s(3) was one of the legs of a triangle with the ratio indicated by the cot
jolly verge
The entirety of trig is pretty much, when in doubt, draw a triangle
That's what I did at least
astral sable
Alr ima try to do that 😂
Thanks, Last thing do you guys have any tips for what csc sec cot and everything are referring to?
jolly verge
csc(x)=1/sin(x), sec(x)=1/cos(x), and cot(x)=1/tan(x)
astral sable
Is there a good way to memorize it or you kinda just gotta know
jolly verge
And since tan(x)=sin(x)/cos(x), cot(x)=cos(x)/sin(x)
foggy iron
it's because the right triangle with sides cos(t) and sin(t) must have hypothenuse 1, but it's easier to just make the sides first and then calculate cos(t) and sin(t) with the triangle you obtain.
Or alternatively you could rewrite your ratio a/b as (a/r) / (b/r) with r = sqrt(a^2+b^2)
there are different ways to think about it
jolly verge
I remember it as being unintuitive. Cosine is related to secant, even cosecant is the one that starts with cos, and cosecant is related to sine. And then cotangent is related to tangent because, tangent
astral sable
there are different ways to think about it
Ohhh ok i got ya
jolly verge
I remember it as being unintuitive. Cosine is related to secant, even cosecant i...
I dunno, I'm worried this may be more confusing, but that's how I remember it
astral sable
Wait can you guys just verify to what i did was right?
If csc(t) = 2 then sin(t)= 1/2
foggy iron
Is there a good way to memorize it or you kinda just gotta know
for csc and sec i remember they are inverses and that they begin with the opposite letter, so
csc is "inverse" of "s... sin"
sec is "inverse" of "c... cos"
and cotan is inverse of tan
astral sable
I remember it as being unintuitive. Cosine is related to secant, even cosecant i...
The tangent one makes sense to me the only thing was the inverse ones
jolly verge
The tangent one makes sense to me the only thing was the inverse ones
Yeah, sorry, I'm not really good at explaining things
It's probably better to refer to Isma's answer for that
foggy iron
If csc(t) = 2 then sin(t)= 1/2
yeah
astral sable
for csc and sec i remember they are inverses and that they begin with the opposi...
I could match the c woth the s like csc is sec and s with c
foggy iron
yeah it's like, not the same initial
jolly verge
Nah, if there's anything else let us know
astral sable
Ok so like if csc relates to secant or y on a right triangle how can you visualize what csc would be on it
Like sin would be the y part of it
jolly verge
Actually, I can't really think about how it could be visualized on a triangle
It can be graphed though
astral sable
Yeah thats whats kinda hard bc im tryna visualize it but i cant get it
How would you do this one
arccot(1)
So far i know sin(t) =1/2
foggy iron
there's stuff like this but i never bothered knowing them
apart maybe from the tangent
cuz all that thing about the tangent going to infinity when you approach pi/2 from below
becomes visual
jolly verge
So, the blue line is a sine wave, and the red line is the graph of csc. Because it's 1/sin(x), as sin(x) gets smaller, csc(x) gets larger. And when sin(x)=0, there's no value for csc(x)
astral sable
there's stuff like this but i never bothered knowing them
Woah this is so weird
jolly verge
there's stuff like this but i never bothered knowing them
Oh, this is really interesting, thanks for sharing this!
jolly verge
You found the sin(t), so you can use that with the unit circle to find the angle
jolly verge
Yep
astral sable
Pi/6 then
jolly verge
Yeah, that's right
astral sable
Sick thanks
Lemme do one more then if i get it im prob good
Perfect got it
Thanks man
astral sable
there's stuff like this but i never bothered knowing them
Thank you too
jolly verge
Any time, good luck with all of your future studies
foggy iron
have a nice day both of you <3
astral sable
Any time, good luck with all of your future studies
Thank you