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Similarity between graphs of sin and tan inverse

Functional inverse of $sinthetasqrt{tantheta}$Functional inverse of $(a + bsintheta)^2tantheta$- Trigonometry Identities - If $2sin(x-y) = sin(x+y)$, find $frac{tan(x)}{tan(y)}$Show these approximations of $cos$, $sin$ and $tan$ are exact.Prove: $sin (tan x) geq {x}$why $tan x = frac{sin x}{cos x}$? and not $tan x$ = opposite/adjacent?Inverse of $tan^{2}theta$?Finding the sin inversewhat are the function of sin, cos and tan?on the inverse of trigonometric or/ and hyperbolic functions

0

$begingroup$

Why is it that the graphs of tan inverse and sin in the interval $$left[-frac pi 2 , frac pi 2right]$$ are so similar.

Is it just some coincidence or something deeper?

trigonometry inverse-function

share|cite|improve this question

edited Mar 20 at 20:00

clathratus

5,1041439

asked Mar 20 at 19:34

RaghavRaghav

1

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  What are your thoughts on this problem? They are both functions used in trigonometry: of course they are similar.
  $endgroup$
  – clathratus
  Mar 20 at 20:01
  

  
  
  
  

add a comment | 

0

$begingroup$

Why is it that the graphs of tan inverse and sin in the interval $$left[-frac pi 2 , frac pi 2right]$$ are so similar.

Is it just some coincidence or something deeper?

trigonometry inverse-function

share|cite|improve this question

edited Mar 20 at 20:00

clathratus

5,1041439

asked Mar 20 at 19:34

RaghavRaghav

1

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  What are your thoughts on this problem? They are both functions used in trigonometry: of course they are similar.
  $endgroup$
  – clathratus
  Mar 20 at 20:01
  

  
  
  
  

add a comment | 

$begingroup$

Why is it that the graphs of tan inverse and sin in the interval $$left[-frac pi 2 , frac pi 2right]$$ are so similar.

Is it just some coincidence or something deeper?

trigonometry inverse-function

share|cite|improve this question

edited Mar 20 at 20:00

clathratus

5,1041439

asked Mar 20 at 19:34

RaghavRaghav

1

$endgroup$

share|cite|improve this question

edited Mar 20 at 20:00

clathratus

5,1041439

asked Mar 20 at 19:34

RaghavRaghav

1

share|cite|improve this question

edited Mar 20 at 20:00

clathratus

5,1041439

asked Mar 20 at 19:34

RaghavRaghav

1

edited Mar 20 at 20:00

clathratus

5,1041439

edited Mar 20 at 20:00

clathratus

5,1041439

asked Mar 20 at 19:34

RaghavRaghav

1

asked Mar 20 at 19:34

RaghavRaghav

1

2 Answers
2

active

oldest

votes

0

$begingroup$

It's nothing deep.

Both $f(x) = arctan x$ and $g(x) = sin x$ are increasing on $(-pi/2,pi/2)$ and both have an inflection point at $x = 0$. Their somewhat similar appearance is due to those properties and the fact that $arctan dfrac pi 2$ is remarkably close to $1$.

share|cite|improve this answer

answered Mar 20 at 20:07

Umberto P.Umberto P.

40.3k13370

$endgroup$

add a comment | 

0

$begingroup$

The first few terms of their Taylor series around $0$ are $x-x^3/6+x^5/120$ and $x-x^3/3+x^5/5$, respectively, so it’s not terribly surprising that they look similar in a relatively small interval around the origin. I doubt that there’s any deep connection beyond that, though.

share|cite|improve this answer

answered Mar 20 at 20:12

amdamd

31.6k21052

$endgroup$

add a comment | 

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0

$begingroup$

It's nothing deep.

Both $f(x) = arctan x$ and $g(x) = sin x$ are increasing on $(-pi/2,pi/2)$ and both have an inflection point at $x = 0$. Their somewhat similar appearance is due to those properties and the fact that $arctan dfrac pi 2$ is remarkably close to $1$.

share|cite|improve this answer

answered Mar 20 at 20:07

Umberto P.Umberto P.

40.3k13370

$endgroup$

add a comment | 

0

$begingroup$

It's nothing deep.

Both $f(x) = arctan x$ and $g(x) = sin x$ are increasing on $(-pi/2,pi/2)$ and both have an inflection point at $x = 0$. Their somewhat similar appearance is due to those properties and the fact that $arctan dfrac pi 2$ is remarkably close to $1$.

share|cite|improve this answer

answered Mar 20 at 20:07

Umberto P.Umberto P.

40.3k13370

$endgroup$

add a comment | 

$begingroup$

It's nothing deep.

Both $f(x) = arctan x$ and $g(x) = sin x$ are increasing on $(-pi/2,pi/2)$ and both have an inflection point at $x = 0$. Their somewhat similar appearance is due to those properties and the fact that $arctan dfrac pi 2$ is remarkably close to $1$.

share|cite|improve this answer

answered Mar 20 at 20:07

Umberto P.Umberto P.

40.3k13370

$endgroup$

share|cite|improve this answer

answered Mar 20 at 20:07

Umberto P.Umberto P.

40.3k13370

answered Mar 20 at 20:07

Umberto P.Umberto P.

40.3k13370

answered Mar 20 at 20:07

Umberto P.Umberto P.

40.3k13370

0

$begingroup$

The first few terms of their Taylor series around $0$ are $x-x^3/6+x^5/120$ and $x-x^3/3+x^5/5$, respectively, so it’s not terribly surprising that they look similar in a relatively small interval around the origin. I doubt that there’s any deep connection beyond that, though.

share|cite|improve this answer

answered Mar 20 at 20:12

amdamd

31.6k21052

$endgroup$

add a comment | 

0

$begingroup$

The first few terms of their Taylor series around $0$ are $x-x^3/6+x^5/120$ and $x-x^3/3+x^5/5$, respectively, so it’s not terribly surprising that they look similar in a relatively small interval around the origin. I doubt that there’s any deep connection beyond that, though.

share|cite|improve this answer

answered Mar 20 at 20:12

amdamd

31.6k21052

$endgroup$

add a comment | 

$begingroup$

The first few terms of their Taylor series around $0$ are $x-x^3/6+x^5/120$ and $x-x^3/3+x^5/5$, respectively, so it’s not terribly surprising that they look similar in a relatively small interval around the origin. I doubt that there’s any deep connection beyond that, though.

share|cite|improve this answer

answered Mar 20 at 20:12

amdamd

31.6k21052

$endgroup$

share|cite|improve this answer

answered Mar 20 at 20:12

amdamd

31.6k21052

answered Mar 20 at 20:12

amdamd

31.6k21052

answered Mar 20 at 20:12

amdamd

31.6k21052