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Closed form for $sin(narctan(x))$
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Is there a closed form for the function $sin(narctan x)$, perhaps where $n$ is restricted to being an integer, or if not, perhaps some special integers (such as triangular numbers or some other figurate numbers)?
From playing around with a few values, it seems that
$$sinarctan(x)=fracxsqrt1+x^2,~sin(2arctan x)=frac2x1+x^2,~sin(3arctan x)=frac3x-x^3(1+x^2)^3/2,$$
I can see that the denominator is $(1+x^2)^tfrac12n$ but can't quite see the form of the numerator.
Motivation: This is motivated by an inconvenient but necessary change of coordinates from polar to Cartesian when a function involves not $sintheta$ but $sin ntheta$ for some integer $n.$
trigonometry
asked 2 days ago
Hobbyist
579212
add a comment |Â
up vote
3
down vote
favorite
Is there a closed form for the function $sin(narctan x)$, perhaps where $n$ is restricted to being an integer, or if not, perhaps some special integers (such as triangular numbers or some other figurate numbers)?
From playing around with a few values, it seems that
$$sinarctan(x)=fracxsqrt1+x^2,~sin(2arctan x)=frac2x1+x^2,~sin(3arctan x)=frac3x-x^3(1+x^2)^3/2,$$
I can see that the denominator is $(1+x^2)^tfrac12n$ but can't quite see the form of the numerator.
Motivation: This is motivated by an inconvenient but necessary change of coordinates from polar to Cartesian when a function involves not $sintheta$ but $sin ntheta$ for some integer $n.$
trigonometry
asked 2 days ago
Hobbyist
579212
mathworld.wolfram.com/Multiple-AngleFormulas.html
– Sorfosh
2 days ago
add a comment |Â
up vote
3
down vote
favorite
up vote
3
down vote
favorite
Is there a closed form for the function $sin(narctan x)$, perhaps where $n$ is restricted to being an integer, or if not, perhaps some special integers (such as triangular numbers or some other figurate numbers)?
From playing around with a few values, it seems that
$$sinarctan(x)=fracxsqrt1+x^2,~sin(2arctan x)=frac2x1+x^2,~sin(3arctan x)=frac3x-x^3(1+x^2)^3/2,$$
I can see that the denominator is $(1+x^2)^tfrac12n$ but can't quite see the form of the numerator.
Motivation: This is motivated by an inconvenient but necessary change of coordinates from polar to Cartesian when a function involves not $sintheta$ but $sin ntheta$ for some integer $n.$
trigonometry
asked 2 days ago
Hobbyist
579212
Is there a closed form for the function $sin(narctan x)$, perhaps where $n$ is restricted to being an integer, or if not, perhaps some special integers (such as triangular numbers or some other figurate numbers)?
From playing around with a few values, it seems that
$$sinarctan(x)=fracxsqrt1+x^2,~sin(2arctan x)=frac2x1+x^2,~sin(3arctan x)=frac3x-x^3(1+x^2)^3/2,$$
I can see that the denominator is $(1+x^2)^tfrac12n$ but can't quite see the form of the numerator.
Motivation: This is motivated by an inconvenient but necessary change of coordinates from polar to Cartesian when a function involves not $sintheta$ but $sin ntheta$ for some integer $n.$
trigonometry
asked 2 days ago
Hobbyist
579212
asked 2 days ago
Hobbyist
579212
asked 2 days ago
Hobbyist
579212
asked 2 days ago
Hobbyist
579212
579212
mathworld.wolfram.com/Multiple-AngleFormulas.html
– Sorfosh
2 days ago
add a comment |Â
mathworld.wolfram.com/Multiple-AngleFormulas.html
– Sorfosh
2 days ago
mathworld.wolfram.com/Multiple-AngleFormulas.html
– Sorfosh
2 days ago
mathworld.wolfram.com/Multiple-AngleFormulas.html
– Sorfosh
2 days ago
add a comment |Â
3 Answers
3
active
oldest
votes
up vote
5
down vote
The Chebyshev polynomials of the second kind give you
$$
sin(n theta) = U_n-1(cos theta) sin theta
.
$$
In this formula, put $theta = arctan x$ and use $cos theta = dfrac1sqrt1+x^2$ and $sin theta = dfracxsqrt1+x^2$.
answered 2 days ago
Hans Lundmark
32.7k563109
add a comment |Â
up vote
3
down vote
We have that
$$tan^-1(x)=fraci2logleft(frac1-ix1+ixright)$$
And
$$sin(x)=fraci2e^-ix-fraci2e^ix$$
So
$$sin(ntan^-1(x))=fraci2expleft(fraci2nlogleft(frac1-ix1+ixright)right)-fraci2expleft(fraci2nlogleft(frac1+x1-ixright)right)$$
$$sin(ntan^-1(x))=fraci2left(frac(1-ix)^fracn2(1+ix)^fracn2-frac(1+ix)^fracn2(1-ix)^fracn2right)$$
$$sin(ntan^-1(x))=fraci2left(frac(1-ix)^n-(1+ix)^n(x^2+1)^fracn2right)$$
Now we can use the Binomial theorem:
$$sin(ntan^-1(x))=fraci2left(fracsum_k=0^n binomnk(-ix)^k-sum_k=0^n binomnk(ix)^k(x^2+1)^fracn2right)$$
$$sin(ntan^-1(x))=fraci2left(fracsum_k=0^n binomnkx^k((-i)^k-i^k)(x^2+1)^fracn2right)$$
$$sin(ntan^-1(x))=frac1(x^2+1)^fracn2sum_k=0^n binomnkx^kfraci((-i)^k-i^k)2$$
$$sin(ntan^-1(x))=frac1(x^2+1)^fracn2 left(binomn1x^1-binomn3x^3+binomn5x^5+dotsright)$$
$$sin(ntan^-1(x))=frac1(x^2+1)^fracn2 sum_k=0^nbinomnkcosleft((n-1)fracpi2right)x^k$$
answered 2 days ago
Botond
3,8702532
Given that it appears the numerator is always a real-valued polynomial for any $n,$ assuming $x$ is real, is there a convenient way to factor out those $i$s? A form like that would be perfect.
– Hobbyist
2 days ago
@Hobbyist I think I made it pure real, but please check if I didn't mess anything up.
– Botond
2 days ago
add a comment |Â
up vote
0
down vote
Too long for a comment
Assume $x$ is real.
Let $S_n(x)=sin(noperatornamearctan(x))$.
Denote $operatornamearctan$ as $g$.
Let $T_n(x)=e^ing(x)$.
Sloppily, $$T_n(x)=left(e^-ig(x)right)^-n$$
Let $f(x)=e^-ig(x)$.
Then, $$fracdT_n(x)df(x)=-nT_n+1$$
Thus we obtain a recursive relation:
$$T_n+1=-frac incdot e^ioperatornamearctan(x)(1+x^2)cdot T_n’(x)$$
Also, $$S_n(x)=Im T_n(x)$$
answered 2 days ago
Szeto
3,8431421
add a comment |Â
3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
5
down vote
The Chebyshev polynomials of the second kind give you
$$
sin(n theta) = U_n-1(cos theta) sin theta
.
$$
In this formula, put $theta = arctan x$ and use $cos theta = dfrac1sqrt1+x^2$ and $sin theta = dfracxsqrt1+x^2$.
answered 2 days ago
Hans Lundmark
32.7k563109
add a comment |Â
up vote
5
down vote
The Chebyshev polynomials of the second kind give you
$$
sin(n theta) = U_n-1(cos theta) sin theta
.
$$
In this formula, put $theta = arctan x$ and use $cos theta = dfrac1sqrt1+x^2$ and $sin theta = dfracxsqrt1+x^2$.
answered 2 days ago
Hans Lundmark
32.7k563109
add a comment |Â
up vote
5
down vote
up vote
5
down vote
The Chebyshev polynomials of the second kind give you
$$
sin(n theta) = U_n-1(cos theta) sin theta
.
$$
In this formula, put $theta = arctan x$ and use $cos theta = dfrac1sqrt1+x^2$ and $sin theta = dfracxsqrt1+x^2$.
answered 2 days ago
Hans Lundmark
32.7k563109
The Chebyshev polynomials of the second kind give you
$$
sin(n theta) = U_n-1(cos theta) sin theta
.
$$
In this formula, put $theta = arctan x$ and use $cos theta = dfrac1sqrt1+x^2$ and $sin theta = dfracxsqrt1+x^2$.
answered 2 days ago
Hans Lundmark
32.7k563109
answered 2 days ago
Hans Lundmark
32.7k563109
answered 2 days ago
Hans Lundmark
32.7k563109
answered 2 days ago
Hans Lundmark
32.7k563109
32.7k563109
add a comment |Â
add a comment |Â
up vote
3
down vote
We have that
$$tan^-1(x)=fraci2logleft(frac1-ix1+ixright)$$
And
$$sin(x)=fraci2e^-ix-fraci2e^ix$$
So
$$sin(ntan^-1(x))=fraci2expleft(fraci2nlogleft(frac1-ix1+ixright)right)-fraci2expleft(fraci2nlogleft(frac1+x1-ixright)right)$$
$$sin(ntan^-1(x))=fraci2left(frac(1-ix)^fracn2(1+ix)^fracn2-frac(1+ix)^fracn2(1-ix)^fracn2right)$$
$$sin(ntan^-1(x))=fraci2left(frac(1-ix)^n-(1+ix)^n(x^2+1)^fracn2right)$$
Now we can use the Binomial theorem:
$$sin(ntan^-1(x))=fraci2left(fracsum_k=0^n binomnk(-ix)^k-sum_k=0^n binomnk(ix)^k(x^2+1)^fracn2right)$$
$$sin(ntan^-1(x))=fraci2left(fracsum_k=0^n binomnkx^k((-i)^k-i^k)(x^2+1)^fracn2right)$$
$$sin(ntan^-1(x))=frac1(x^2+1)^fracn2sum_k=0^n binomnkx^kfraci((-i)^k-i^k)2$$
$$sin(ntan^-1(x))=frac1(x^2+1)^fracn2 left(binomn1x^1-binomn3x^3+binomn5x^5+dotsright)$$
$$sin(ntan^-1(x))=frac1(x^2+1)^fracn2 sum_k=0^nbinomnkcosleft((n-1)fracpi2right)x^k$$
answered 2 days ago
Botond
3,8702532
Given that it appears the numerator is always a real-valued polynomial for any $n,$ assuming $x$ is real, is there a convenient way to factor out those $i$s? A form like that would be perfect.
– Hobbyist
2 days ago
@Hobbyist I think I made it pure real, but please check if I didn't mess anything up.
– Botond
2 days ago
add a comment |Â
up vote
3
down vote
We have that
$$tan^-1(x)=fraci2logleft(frac1-ix1+ixright)$$
And
$$sin(x)=fraci2e^-ix-fraci2e^ix$$
So
$$sin(ntan^-1(x))=fraci2expleft(fraci2nlogleft(frac1-ix1+ixright)right)-fraci2expleft(fraci2nlogleft(frac1+x1-ixright)right)$$
$$sin(ntan^-1(x))=fraci2left(frac(1-ix)^fracn2(1+ix)^fracn2-frac(1+ix)^fracn2(1-ix)^fracn2right)$$
$$sin(ntan^-1(x))=fraci2left(frac(1-ix)^n-(1+ix)^n(x^2+1)^fracn2right)$$
Now we can use the Binomial theorem:
$$sin(ntan^-1(x))=fraci2left(fracsum_k=0^n binomnk(-ix)^k-sum_k=0^n binomnk(ix)^k(x^2+1)^fracn2right)$$
$$sin(ntan^-1(x))=fraci2left(fracsum_k=0^n binomnkx^k((-i)^k-i^k)(x^2+1)^fracn2right)$$
$$sin(ntan^-1(x))=frac1(x^2+1)^fracn2sum_k=0^n binomnkx^kfraci((-i)^k-i^k)2$$
$$sin(ntan^-1(x))=frac1(x^2+1)^fracn2 left(binomn1x^1-binomn3x^3+binomn5x^5+dotsright)$$
$$sin(ntan^-1(x))=frac1(x^2+1)^fracn2 sum_k=0^nbinomnkcosleft((n-1)fracpi2right)x^k$$
answered 2 days ago
Botond
3,8702532
Given that it appears the numerator is always a real-valued polynomial for any $n,$ assuming $x$ is real, is there a convenient way to factor out those $i$s? A form like that would be perfect.
– Hobbyist
2 days ago
@Hobbyist I think I made it pure real, but please check if I didn't mess anything up.
– Botond
2 days ago
add a comment |Â
up vote
3
down vote
up vote
3
down vote
We have that
$$tan^-1(x)=fraci2logleft(frac1-ix1+ixright)$$
And
$$sin(x)=fraci2e^-ix-fraci2e^ix$$
So
$$sin(ntan^-1(x))=fraci2expleft(fraci2nlogleft(frac1-ix1+ixright)right)-fraci2expleft(fraci2nlogleft(frac1+x1-ixright)right)$$
$$sin(ntan^-1(x))=fraci2left(frac(1-ix)^fracn2(1+ix)^fracn2-frac(1+ix)^fracn2(1-ix)^fracn2right)$$
$$sin(ntan^-1(x))=fraci2left(frac(1-ix)^n-(1+ix)^n(x^2+1)^fracn2right)$$
Now we can use the Binomial theorem:
$$sin(ntan^-1(x))=fraci2left(fracsum_k=0^n binomnk(-ix)^k-sum_k=0^n binomnk(ix)^k(x^2+1)^fracn2right)$$
$$sin(ntan^-1(x))=fraci2left(fracsum_k=0^n binomnkx^k((-i)^k-i^k)(x^2+1)^fracn2right)$$
$$sin(ntan^-1(x))=frac1(x^2+1)^fracn2sum_k=0^n binomnkx^kfraci((-i)^k-i^k)2$$
$$sin(ntan^-1(x))=frac1(x^2+1)^fracn2 left(binomn1x^1-binomn3x^3+binomn5x^5+dotsright)$$
$$sin(ntan^-1(x))=frac1(x^2+1)^fracn2 sum_k=0^nbinomnkcosleft((n-1)fracpi2right)x^k$$
answered 2 days ago
Botond
3,8702532
We have that
$$tan^-1(x)=fraci2logleft(frac1-ix1+ixright)$$
And
$$sin(x)=fraci2e^-ix-fraci2e^ix$$
So
$$sin(ntan^-1(x))=fraci2expleft(fraci2nlogleft(frac1-ix1+ixright)right)-fraci2expleft(fraci2nlogleft(frac1+x1-ixright)right)$$
$$sin(ntan^-1(x))=fraci2left(frac(1-ix)^fracn2(1+ix)^fracn2-frac(1+ix)^fracn2(1-ix)^fracn2right)$$
$$sin(ntan^-1(x))=fraci2left(frac(1-ix)^n-(1+ix)^n(x^2+1)^fracn2right)$$
Now we can use the Binomial theorem:
$$sin(ntan^-1(x))=fraci2left(fracsum_k=0^n binomnk(-ix)^k-sum_k=0^n binomnk(ix)^k(x^2+1)^fracn2right)$$
$$sin(ntan^-1(x))=fraci2left(fracsum_k=0^n binomnkx^k((-i)^k-i^k)(x^2+1)^fracn2right)$$
$$sin(ntan^-1(x))=frac1(x^2+1)^fracn2sum_k=0^n binomnkx^kfraci((-i)^k-i^k)2$$
$$sin(ntan^-1(x))=frac1(x^2+1)^fracn2 left(binomn1x^1-binomn3x^3+binomn5x^5+dotsright)$$
$$sin(ntan^-1(x))=frac1(x^2+1)^fracn2 sum_k=0^nbinomnkcosleft((n-1)fracpi2right)x^k$$
answered 2 days ago
Botond
3,8702532
edited 2 days ago
answered 2 days ago
Botond
3,8702532
answered 2 days ago
Botond
3,8702532
answered 2 days ago
Botond
3,8702532
3,8702532
Given that it appears the numerator is always a real-valued polynomial for any $n,$ assuming $x$ is real, is there a convenient way to factor out those $i$s? A form like that would be perfect.
– Hobbyist
2 days ago
@Hobbyist I think I made it pure real, but please check if I didn't mess anything up.
– Botond
2 days ago
add a comment |Â
Given that it appears the numerator is always a real-valued polynomial for any $n,$ assuming $x$ is real, is there a convenient way to factor out those $i$s? A form like that would be perfect.
– Hobbyist
2 days ago
@Hobbyist I think I made it pure real, but please check if I didn't mess anything up.
– Botond
2 days ago
Given that it appears the numerator is always a real-valued polynomial for any $n,$ assuming $x$ is real, is there a convenient way to factor out those $i$s? A form like that would be perfect.
– Hobbyist
2 days ago
Given that it appears the numerator is always a real-valued polynomial for any $n,$ assuming $x$ is real, is there a convenient way to factor out those $i$s? A form like that would be perfect.
– Hobbyist
2 days ago
@Hobbyist I think I made it pure real, but please check if I didn't mess anything up.
– Botond
2 days ago
@Hobbyist I think I made it pure real, but please check if I didn't mess anything up.
– Botond
2 days ago
add a comment |Â
up vote
0
down vote
Too long for a comment
Assume $x$ is real.
Let $S_n(x)=sin(noperatornamearctan(x))$.
Denote $operatornamearctan$ as $g$.
Let $T_n(x)=e^ing(x)$.
Sloppily, $$T_n(x)=left(e^-ig(x)right)^-n$$
Let $f(x)=e^-ig(x)$.
Then, $$fracdT_n(x)df(x)=-nT_n+1$$
Thus we obtain a recursive relation:
$$T_n+1=-frac incdot e^ioperatornamearctan(x)(1+x^2)cdot T_n’(x)$$
Also, $$S_n(x)=Im T_n(x)$$
answered 2 days ago
Szeto
3,8431421
add a comment |Â
up vote
0
down vote
Too long for a comment
Assume $x$ is real.
Let $S_n(x)=sin(noperatornamearctan(x))$.
Denote $operatornamearctan$ as $g$.
Let $T_n(x)=e^ing(x)$.
Sloppily, $$T_n(x)=left(e^-ig(x)right)^-n$$
Let $f(x)=e^-ig(x)$.
Then, $$fracdT_n(x)df(x)=-nT_n+1$$
Thus we obtain a recursive relation:
$$T_n+1=-frac incdot e^ioperatornamearctan(x)(1+x^2)cdot T_n’(x)$$
Also, $$S_n(x)=Im T_n(x)$$
answered 2 days ago
Szeto
3,8431421
add a comment |Â
up vote
0
down vote
up vote
0
down vote
Too long for a comment
Assume $x$ is real.
Let $S_n(x)=sin(noperatornamearctan(x))$.
Denote $operatornamearctan$ as $g$.
Let $T_n(x)=e^ing(x)$.
Sloppily, $$T_n(x)=left(e^-ig(x)right)^-n$$
Let $f(x)=e^-ig(x)$.
Then, $$fracdT_n(x)df(x)=-nT_n+1$$
Thus we obtain a recursive relation:
$$T_n+1=-frac incdot e^ioperatornamearctan(x)(1+x^2)cdot T_n’(x)$$
Also, $$S_n(x)=Im T_n(x)$$
answered 2 days ago
Szeto
3,8431421
Too long for a comment
Assume $x$ is real.
Let $S_n(x)=sin(noperatornamearctan(x))$.
Denote $operatornamearctan$ as $g$.
Let $T_n(x)=e^ing(x)$.
Sloppily, $$T_n(x)=left(e^-ig(x)right)^-n$$
Let $f(x)=e^-ig(x)$.
Then, $$fracdT_n(x)df(x)=-nT_n+1$$
Thus we obtain a recursive relation:
$$T_n+1=-frac incdot e^ioperatornamearctan(x)(1+x^2)cdot T_n’(x)$$
Also, $$S_n(x)=Im T_n(x)$$
answered 2 days ago
Szeto
3,8431421
edited 2 days ago
answered 2 days ago
Szeto
3,8431421
answered 2 days ago
Szeto
3,8431421
answered 2 days ago
Szeto
3,8431421
3,8431421
add a comment |Â
add a comment |Â
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mathworld.wolfram.com/Multiple-AngleFormulas.html
– Sorfosh
2 days ago