I'm glad I could find your video on this exact problem I ran into. Wonderfully simple explanation, and it helped me verify that I did my work to find the solution properly!
Thank you so much! You went through the steps very clearly, so it was super easy to follow. I definitely understand this concept now, so thank you for your help!
Asking for clarification. Does it matter that the tangent function is positive in the first and third quadrant? If so, how does one factor that into the answer?
Am I correct in believing that this approach or a similar one won't work for evaluating for example sin(sqrt(2)*arctan(x)) as an algebraic expression? If so, is there such a method?
sin(k arctan(x)) is certainly algebraic in x if k is an integer or a rational with denominator of the form 2^n (as in the integer case the problem can be solved using multiple angle formulae and in the fractional case by using the half angle formula for cosθ/2 derived from cosθ = 2cos^2(θ/2)-1 repeatedly together with the Pythagorean identity), e.g.: For k = 1, sin(arctan(x)): if x > 0, Let θ = arctan(x) (0 < θ < π/2), tan θ = x, sec θ = sqrt(1 + x^2), cos θ = 1/sqrt(1+x^2), sin θ [=sin(arctan(x))] = tan θ×cos θ = x/sqrt(1+x^2). For k = 1/2, sin(1/2 arctan(x)): if x > 0, Let θ = arctan(x) (0 < θ < π/2), as before cos θ = 1/sqrt(1+x^2) = 1 - 2 sin^2(θ/2), sin^2(θ/2) = [1 - 1/sqrt(1+x^2)]/2, sin(θ/2) [= sin(1/2 arctan(x))] = sqrt([1 - 1/sqrt(1+x^2)]/2). However, if k is irrational I think it unlikely the expression is algebraic in x.
I solved it by going through a more complex method of using the sin double angle formula you used, turning the sines into cosines (sin^2+cos^2=1), turning the cosines into secants, turning the secants into tangents (sec^2=tan^1+1), and ataning all of it. That method works too.
sin(2arctanx) Pretty easy let me show you guys 2arctanx=arcsin(2x/√1+x²) Hence answer is directly 2x/√1+x² We are made to remember all this formula as a serious jee aspirant.
sin(2arctanx) Pretty easy let me show you guys 2arctanx=arcsin(2x/√1+x²) Hence answer is directly 2x/√1+x² We are made to remember all this formula as a serious jee aspirant
God bless you. I've been on this exact question for over an hour until I came across your video. I'm so grateful for people like you😁
been integration so much that all I see in the answer is the derivative of ln(x^2 + 1)
Nithesh S. lol!! This is for my trig class
Nithesh S. hahahahha same was hoping someone would say it
I'm glad I could find your video on this exact problem I ran into. Wonderfully simple explanation, and it helped me verify that I did my work to find the solution properly!
this helps a lot, specially with math classes being online and when all i am given to study by my teacher is the textbook. Thank you.
Thank you so much! You went through the steps very clearly, so it was super easy to follow. I definitely understand this concept now, so thank you for your help!
This is exactly what i was looking for thank you for your aid keep up the good work
This video saved me so much trouble. Thank you so much. =)
Thank you for your amazing videos.👍
Thank you very much! Doing Trig on Coursera and the "book/pdf" does not cover this very well.
Obrigado por me ajudar ❤ 🇧🇷
Obs; não sei como cheguei no seu vídeo 🇧🇷
Asking for clarification. Does it matter that the tangent function is positive in the first and third quadrant? If so, how does one factor that into the answer?
Am I correct in believing that this approach or a similar one won't work for evaluating for example sin(sqrt(2)*arctan(x)) as an algebraic expression? If so, is there such a method?
sin(k arctan(x)) is certainly algebraic in x if k is an integer or a rational with denominator of the form 2^n (as in the integer case the problem can be solved using multiple angle formulae and in the fractional case by using the half angle formula for cosθ/2 derived from cosθ = 2cos^2(θ/2)-1 repeatedly together with the Pythagorean identity), e.g.:
For k = 1, sin(arctan(x)):
if x > 0, Let θ = arctan(x) (0 < θ < π/2), tan θ = x, sec θ = sqrt(1 + x^2), cos θ = 1/sqrt(1+x^2), sin θ [=sin(arctan(x))] = tan θ×cos θ = x/sqrt(1+x^2).
For k = 1/2, sin(1/2 arctan(x)):
if x > 0, Let θ = arctan(x) (0 < θ < π/2), as before cos θ = 1/sqrt(1+x^2) = 1 - 2 sin^2(θ/2), sin^2(θ/2) = [1 - 1/sqrt(1+x^2)]/2, sin(θ/2) [= sin(1/2 arctan(x))] = sqrt([1 - 1/sqrt(1+x^2)]/2).
However, if k is irrational I think it unlikely the expression is algebraic in x.
Thank u so much🎉
Thank you for this 😭
Wonderful video!! Thank you so much.
what happens if it is cos(nsin^-1(x))? where n can be any real number
I solved it by going through a more complex method of using the sin double angle formula you used, turning the sines into cosines (sin^2+cos^2=1), turning the cosines into secants, turning the secants into tangents (sec^2=tan^1+1), and ataning all of it.
That method works too.
Where would one see this most likely?I am very intrigued lol.
trigonometry
Thank you dude.
no problemo
Nowhere, it's just a hobby for nerds
can i use this for integratio or to derive?
Thank you for the clear and simple explanation:))
Legend.
sin(2arctanx)
Pretty easy let me show you guys
2arctanx=arcsin(2x/√1+x²)
Hence answer is directly 2x/√1+x²
We are made to remember all this formula as a serious jee aspirant.
Thank you so much! Very helpful!
infinite series of 2x(-x^2)^n, interesting, I'd like to see if you can mess around with Taylor series to achieve this same result
Very good!!
Thanks a lot for this !
Really cool but disappointed Bc I thought it'd be with series. But now I'll try with series
Ah I see. This is for my trig class.
my goat.
2:46 - 2:49 That’s 3 whole seconds of pen switching!
but you figured out the algebraic expression right!?!?
thank you so much
thank you so much that helped me a lot xD
rien compris....
te quiero mucho =)
Wow that intersting❤
thank youuuuuuuuu
amazing!
Dario Fervenza thanks!
Thanks
ahhhhhhhhhhhh. Thank you sir.
OK THANKS,
verigood
시발 존나유용하네
tg^(-1) uwu
sin(2arctanx)
Pretty easy let me show you guys
2arctanx=arcsin(2x/√1+x²)
Hence answer is directly 2x/√1+x²
We are made to remember all this formula as a serious jee aspirant