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May I leave an evil comment? Just the other day, the algorithm suggested me a video of yours where you talk about your reason why to rationalize the denominator. I'd suggest, though, that in this case, *_not_* rationalizing it is preferable. If you start with -1/√3 instead of -(√3)/3, the tangent triangle has the side lengths 1, √3, and 2. The cosecant will then be 2/-1=-2 which is a more straightforward calculation! 😀
It's fascinating, that you don't even need the result of the term inside the csc brackets. You just need the values, that you figured out on the way to calculate it.
We all know that tan∅=opposite/adjacent. Then it follows that ∅=arctan(opposite/adjacent). Let: -√3 as the opposite, and 3 as the adjacent. By process of elimination, the only possible orientation of -√3 as the right triangle's opposite (y-coordinate) are quadrant III or IV; the only possible orientation of 3 as the right triangle's adjacent (x-coordinate) are quadrant 1 or IV. But, it can't be at quadrant I since the opposite is -√3 which should be below x-axis. Therefore, the right triangle lies at quadrant IV with its angle (∅) set below the x-axis. Since arctan(-√3/3) is basically equal to just ∅, then it follows that csc(arctan(-√3/3)) = csc∅. By pythagorean theorem, let: h as the hypotenuse, it follows that h=√(-√3)²+(3)², so h=√12 or 2√3. Csc∅=hypotenuse/opposite, then csc∅=(2√3)/(-√3). Therefore, csc∅= -2.
Hello sir, I am curious about a complex equation I did and I would love if you made a video on it and pointed out error/s because the answer makes no sense🤣. Here it is: So I calculated i^i and I got e^(-π/2-2nπ) {n E Z}. So I thought, what happens if I calculate the principal value and try to find an interception with x^x; sadly x^x ≠ 0.20788 for any real number x. So then I tried with n = -1: e^(3π/2) ≈ 111.3178. I plotted the functions y = x^x and y=111.3718 on Desmos and I found that the intercept is at ≈ 3.644. So I concluded that i^i (One of the values) ≈ 111.3718 ≈ 3.644^(3.644). Does that mean i ≈ 3.644? I'd really love to hear your thoughts about this! Love your vids btw they are high quality videos that make my day.
Here's a situation where the trigonometry _triangle_ is better than the trigonometry _circle._ Given a right triangle x^2 +y^2 = z^2: arctan (- sqrt(3) / 3) -> arctan (y / x) where x = 3 and y = - sqrt(3) Thus, z^2 = x^2 + y^2 = 12. Cosecant is the reciprocal of a sine: = csc(angle) where "angle" is derived from the above triangle's arctan. = csc(angle) = z/y = sqrt(12) / -sqrt(3) = 2*sqrt(3) / -sqrt(3) = -2
pls tell if we can do like this In bracket tan-root3/3 = tan - 30 = - tan 30 tan inverse and tan get canceled so we are left with -30 in bracket outside the bracket csc - 30 = 1/sin -30 = 1/- sin 30 = - csc 30 = -2
Yes, I'm surprised he doesn't say that you should know the tangent values of the easy angles (0,pi/6,pi/4,pi/3 and pi/2 and the corresponding ones in the other quasrants) by heart. These come up very often I feel like. This question is trivial if you just know the tangent value. What you did was indeed correct and a lot more efficient.
@@RohitKulan There is an English equivalent to that. It took me a few seconds to remember the sequence Sin Cos Tan Cot Sec Csc and then I had what I needed to finish the problem. Not that in the last 73 years I have ever used cosecant for anything but reviewing math!
Because tan^-1 as an operator is only defined to give values either in Q1 or Q4 (because that is the principal branch of the tangent function), and since the input is negative, the output must be in Q4. And cosecant in Q4 will give the same signs as its reciprocal, sine, in Q4: negative values.
@@ronaldking1054 I get it, yeah, but in my trigonometry class we had the sine, cosine, arcsine, arccosine, tangent, cotangent. arctangent and arccotanget. No secant or cosecant. I believe nobody even told us they exist, lol. Though, I hated everything geometry related (still kinda hate it) so I might've missed that part, but we definetly didn't solve anything with sec or csc.
I help students master the basics of math. You can show your support and help me create even better content by becoming a patron on Patreon 👉 www.patreon.com/blackpenredpen. Every bit of support means the world to me and motivates me to keep bringing you the best math lessons! Thank you!
why tf you change your name
May I leave an evil comment?
Just the other day, the algorithm suggested me a video of yours where you talk about your reason why to rationalize the denominator.
I'd suggest, though, that in this case, *_not_* rationalizing it is preferable.
If you start with -1/√3 instead of -(√3)/3, the tangent triangle has the side lengths 1, √3, and 2. The cosecant will then be 2/-1=-2 which is a more straightforward calculation! 😀
Given: cosec(arctan([-√3]/3))
= cosec(-π/6)
= -cosec(π/6)
= -1/sin(30°)
= -1/(1/2)
= -2
Simple method:-
csc(tan^-1(-1/sq.rt.3))
csc((-30)degrees)
-2
It's fascinating, that you don't even need the result of the term inside the csc brackets. You just need the values, that you figured out on the way to calculate it.
csc(arctan(x))
=sec(arctan(x))/tan(arctan(x))
=sqrt(1 + tan(arctan(x))^2)/x
=sqrt(1+x^2)/x
For x =-sqrt(3)/3, ans = -2
We all know that tan∅=opposite/adjacent. Then it follows that ∅=arctan(opposite/adjacent). Let: -√3 as the opposite, and 3 as the adjacent. By process of elimination, the only possible orientation of -√3 as the right triangle's opposite (y-coordinate) are quadrant III or IV; the only possible orientation of 3 as the right triangle's adjacent (x-coordinate) are quadrant 1 or IV. But, it can't be at quadrant I since the opposite is -√3 which should be below x-axis. Therefore, the right triangle lies at quadrant IV with its angle (∅) set below the x-axis. Since arctan(-√3/3) is basically equal to just ∅, then it follows that csc(arctan(-√3/3)) = csc∅. By pythagorean theorem, let: h as the hypotenuse, it follows that h=√(-√3)²+(3)², so h=√12 or 2√3. Csc∅=hypotenuse/opposite, then csc∅=(2√3)/(-√3). Therefore, csc∅= -2.
I think the answer is -2
Yes you are right ✅️
Same bro Mera bhi
Thank you sir😊
Hello sir, I am curious about a complex equation I did and I would love if you made a video on it and pointed out error/s because the answer makes no sense🤣.
Here it is:
So I calculated i^i and I got e^(-π/2-2nπ) {n E Z}.
So I thought, what happens if I calculate the principal value and try to find an interception with x^x; sadly x^x ≠ 0.20788 for any real number x.
So then I tried with n = -1:
e^(3π/2) ≈ 111.3178. I plotted the functions y = x^x and y=111.3718 on Desmos and I found that the intercept is at ≈ 3.644.
So I concluded that i^i (One of the values) ≈ 111.3718 ≈ 3.644^(3.644). Does that mean i ≈ 3.644?
I'd really love to hear your thoughts about this! Love your vids btw they are high quality videos that make my day.
How to find the volume when y=1/x For 0
Here's a situation where the trigonometry _triangle_ is better than the trigonometry _circle._
Given a right triangle x^2 +y^2 = z^2:
arctan (- sqrt(3) / 3)
-> arctan (y / x) where x = 3 and y = - sqrt(3)
Thus, z^2 = x^2 + y^2 = 12.
Cosecant is the reciprocal of a sine:
= csc(angle) where "angle" is derived from the above triangle's arctan.
= csc(angle) = z/y
= sqrt(12) / -sqrt(3)
= 2*sqrt(3) / -sqrt(3)
= -2
pls tell if we can do like this
In bracket
tan-root3/3 = tan - 30 = - tan 30
tan inverse and tan get canceled so we are left with -30 in bracket
outside the bracket
csc - 30 = 1/sin -30 = 1/- sin 30 = - csc 30 = -2
That's correct
I did it like this
Yes, I'm surprised he doesn't say that you should know the tangent values of the easy angles (0,pi/6,pi/4,pi/3 and pi/2 and the corresponding ones in the other quasrants) by heart. These come up very often I feel like. This question is trivial if you just know the tangent value. What you did was indeed correct and a lot more efficient.
@@viktorsmets29 thanks for telling
Sir is the sqrt of -1 considered undefined?
in Real numbers is undefined but in Complex numbers it has a answer its "i"
Is cos(tan)(tan) and cos(tan)^2 has the same value?
I forgot all this thanks
I did have to think for a moment which sides are used for cosecant. r/x or r/y? But I did remember correctly.
Always remember ChoShaCao!!!
@@RohitKulan There is an English equivalent to that. It took me a few seconds to remember the sequence Sin Cos Tan Cot Sec Csc and then I had what I needed to finish the problem. Not that in the last 73 years I have ever used cosecant for anything but reviewing math!
@@rickhole it is English, it's the reciprocal function version of SohCahToa
Just remember sin and csc go together, cos and sec go together.
What you realy want
Cos(120)= cos(-120)=-1/2
1/cos(-120)=-2
Cos(theta) works fine in the range [0,180> degree on the calculator.
sec noko noko noko cos tan tan
why can't this be ±2 and just -2?
Because tan^-1 as an operator is only defined to give values either in Q1 or Q4 (because that is the principal branch of the tangent function), and since the input is negative, the output must be in Q4. And cosecant in Q4 will give the same signs as its reciprocal, sine, in Q4: negative values.
@@stephenbeck7222 okay.... that makes sense...
wrote that on the calculator and gave me 2(sqrt3)/3 ez
2/sqrt(3) is also my answer. I have no problem with sqrt(3) in the denominator.
Thank god, I am not count in many stutends
Well, it's first time in my life I heard about the csc thing, so no wonder that many people can't solve it, lol 😅
csc x = 1/sin x, sec x = 1/cos x, cot x = 1/tan x
@@ronaldking1054 I get it, yeah, but in my trigonometry class we had the sine, cosine, arcsine, arccosine, tangent, cotangent. arctangent and arccotanget. No secant or cosecant.
I believe nobody even told us they exist, lol. Though, I hated everything geometry related (still kinda hate it) so I might've missed that part, but we definetly didn't solve anything with sec or csc.