Proof of range of tan $theta$ [duplicate]












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  • On The Range Of The Tangent

    7 answers




I need someone to review the following proof to find the range of tan $theta$, whether it is acceptable or not:

Starting with |cos$theta|le$ 1

Square both sides: |cos$theta|^2le 1^2$

Simplify: $cos^2thetale 1$

Use reciprocal property for inequality: $frac{1}{cos^2theta}gefrac{1}{1}$


Use reciprocal identity: $sec^2thetage1$


Use Pythagorean identity: $1+tan^2thetage1$

Simplify: $tan^2thetage 0$

Since the square of any real number is always greater or equal to 0, it follows that the range of tan $theta$ is the set of all real numbers.
***************************************************************************

I need someone to review the following proof whether it is acceptable or not.

Using the rectangular coordinates definition of $tan theta=frac{y}{x}$ where $theta$ is in standard position and (x,y) is a point on the terminal side,

tan $theta=frac{y}{x}$

=$frac{y-0}{x-0}$

= Slope of the terminal side (The terminal side starts from the origin)

Since the slope of a straight line is the set of all real numbers, it follows that the range of tan θ is the set of all real numbers.










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marked as duplicate by N. F. Taussig, Cesareo, Vidyanshu Mishra, Leila, user159517 Dec 4 '18 at 14:53


This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.















  • $begingroup$
    $(x^2)^2=x^4ge0$ but this does not imply that the range of $x^2$ is the whole real line. Morover, $cos^2thetage 0$ but the range of $costheta$ is $[-1,1]$.
    $endgroup$
    – Tito Eliatron
    Dec 4 '18 at 9:05












  • $begingroup$
    In fact, the range of tangent is R. This is the basis of a textbook example of a symmetric distribution with undefined mean called the Cauchy distribution, whose cumulative distribution function is the arctangent function with domain R.
    $endgroup$
    – GNUSupporter 8964民主女神 地下教會
    Dec 4 '18 at 9:19
















1












$begingroup$



This question already has an answer here:




  • On The Range Of The Tangent

    7 answers




I need someone to review the following proof to find the range of tan $theta$, whether it is acceptable or not:

Starting with |cos$theta|le$ 1

Square both sides: |cos$theta|^2le 1^2$

Simplify: $cos^2thetale 1$

Use reciprocal property for inequality: $frac{1}{cos^2theta}gefrac{1}{1}$


Use reciprocal identity: $sec^2thetage1$


Use Pythagorean identity: $1+tan^2thetage1$

Simplify: $tan^2thetage 0$

Since the square of any real number is always greater or equal to 0, it follows that the range of tan $theta$ is the set of all real numbers.
***************************************************************************

I need someone to review the following proof whether it is acceptable or not.

Using the rectangular coordinates definition of $tan theta=frac{y}{x}$ where $theta$ is in standard position and (x,y) is a point on the terminal side,

tan $theta=frac{y}{x}$

=$frac{y-0}{x-0}$

= Slope of the terminal side (The terminal side starts from the origin)

Since the slope of a straight line is the set of all real numbers, it follows that the range of tan θ is the set of all real numbers.










share|cite|improve this question











$endgroup$



marked as duplicate by N. F. Taussig, Cesareo, Vidyanshu Mishra, Leila, user159517 Dec 4 '18 at 14:53


This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.















  • $begingroup$
    $(x^2)^2=x^4ge0$ but this does not imply that the range of $x^2$ is the whole real line. Morover, $cos^2thetage 0$ but the range of $costheta$ is $[-1,1]$.
    $endgroup$
    – Tito Eliatron
    Dec 4 '18 at 9:05












  • $begingroup$
    In fact, the range of tangent is R. This is the basis of a textbook example of a symmetric distribution with undefined mean called the Cauchy distribution, whose cumulative distribution function is the arctangent function with domain R.
    $endgroup$
    – GNUSupporter 8964民主女神 地下教會
    Dec 4 '18 at 9:19














1












1








1





$begingroup$



This question already has an answer here:




  • On The Range Of The Tangent

    7 answers




I need someone to review the following proof to find the range of tan $theta$, whether it is acceptable or not:

Starting with |cos$theta|le$ 1

Square both sides: |cos$theta|^2le 1^2$

Simplify: $cos^2thetale 1$

Use reciprocal property for inequality: $frac{1}{cos^2theta}gefrac{1}{1}$


Use reciprocal identity: $sec^2thetage1$


Use Pythagorean identity: $1+tan^2thetage1$

Simplify: $tan^2thetage 0$

Since the square of any real number is always greater or equal to 0, it follows that the range of tan $theta$ is the set of all real numbers.
***************************************************************************

I need someone to review the following proof whether it is acceptable or not.

Using the rectangular coordinates definition of $tan theta=frac{y}{x}$ where $theta$ is in standard position and (x,y) is a point on the terminal side,

tan $theta=frac{y}{x}$

=$frac{y-0}{x-0}$

= Slope of the terminal side (The terminal side starts from the origin)

Since the slope of a straight line is the set of all real numbers, it follows that the range of tan θ is the set of all real numbers.










share|cite|improve this question











$endgroup$





This question already has an answer here:




  • On The Range Of The Tangent

    7 answers




I need someone to review the following proof to find the range of tan $theta$, whether it is acceptable or not:

Starting with |cos$theta|le$ 1

Square both sides: |cos$theta|^2le 1^2$

Simplify: $cos^2thetale 1$

Use reciprocal property for inequality: $frac{1}{cos^2theta}gefrac{1}{1}$


Use reciprocal identity: $sec^2thetage1$


Use Pythagorean identity: $1+tan^2thetage1$

Simplify: $tan^2thetage 0$

Since the square of any real number is always greater or equal to 0, it follows that the range of tan $theta$ is the set of all real numbers.
***************************************************************************

I need someone to review the following proof whether it is acceptable or not.

Using the rectangular coordinates definition of $tan theta=frac{y}{x}$ where $theta$ is in standard position and (x,y) is a point on the terminal side,

tan $theta=frac{y}{x}$

=$frac{y-0}{x-0}$

= Slope of the terminal side (The terminal side starts from the origin)

Since the slope of a straight line is the set of all real numbers, it follows that the range of tan θ is the set of all real numbers.





This question already has an answer here:




  • On The Range Of The Tangent

    7 answers








trigonometry proof-verification






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edited Dec 5 '18 at 3:38







user622701

















asked Dec 4 '18 at 9:03









user622701user622701

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marked as duplicate by N. F. Taussig, Cesareo, Vidyanshu Mishra, Leila, user159517 Dec 4 '18 at 14:53


This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.






marked as duplicate by N. F. Taussig, Cesareo, Vidyanshu Mishra, Leila, user159517 Dec 4 '18 at 14:53


This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.














  • $begingroup$
    $(x^2)^2=x^4ge0$ but this does not imply that the range of $x^2$ is the whole real line. Morover, $cos^2thetage 0$ but the range of $costheta$ is $[-1,1]$.
    $endgroup$
    – Tito Eliatron
    Dec 4 '18 at 9:05












  • $begingroup$
    In fact, the range of tangent is R. This is the basis of a textbook example of a symmetric distribution with undefined mean called the Cauchy distribution, whose cumulative distribution function is the arctangent function with domain R.
    $endgroup$
    – GNUSupporter 8964民主女神 地下教會
    Dec 4 '18 at 9:19


















  • $begingroup$
    $(x^2)^2=x^4ge0$ but this does not imply that the range of $x^2$ is the whole real line. Morover, $cos^2thetage 0$ but the range of $costheta$ is $[-1,1]$.
    $endgroup$
    – Tito Eliatron
    Dec 4 '18 at 9:05












  • $begingroup$
    In fact, the range of tangent is R. This is the basis of a textbook example of a symmetric distribution with undefined mean called the Cauchy distribution, whose cumulative distribution function is the arctangent function with domain R.
    $endgroup$
    – GNUSupporter 8964民主女神 地下教會
    Dec 4 '18 at 9:19
















$begingroup$
$(x^2)^2=x^4ge0$ but this does not imply that the range of $x^2$ is the whole real line. Morover, $cos^2thetage 0$ but the range of $costheta$ is $[-1,1]$.
$endgroup$
– Tito Eliatron
Dec 4 '18 at 9:05






$begingroup$
$(x^2)^2=x^4ge0$ but this does not imply that the range of $x^2$ is the whole real line. Morover, $cos^2thetage 0$ but the range of $costheta$ is $[-1,1]$.
$endgroup$
– Tito Eliatron
Dec 4 '18 at 9:05














$begingroup$
In fact, the range of tangent is R. This is the basis of a textbook example of a symmetric distribution with undefined mean called the Cauchy distribution, whose cumulative distribution function is the arctangent function with domain R.
$endgroup$
– GNUSupporter 8964民主女神 地下教會
Dec 4 '18 at 9:19




$begingroup$
In fact, the range of tangent is R. This is the basis of a textbook example of a symmetric distribution with undefined mean called the Cauchy distribution, whose cumulative distribution function is the arctangent function with domain R.
$endgroup$
– GNUSupporter 8964民主女神 地下教會
Dec 4 '18 at 9:19










1 Answer
1






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oldest

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2












$begingroup$

This is not correct. You deduce, after several steps that $tan^2thetageqslant0$. But this holds for every function. Are you going to say that the range of every function from $mathbb R$ into $mathbb R$ is $mathbb R$?






share|cite|improve this answer











$endgroup$













  • $begingroup$
    I agree. when $[f(x)]^2geq 0$ it is not necessary that $f(x)$ has negative values.
    $endgroup$
    – Hussain-Alqatari
    Dec 4 '18 at 9:08




















1 Answer
1






active

oldest

votes








1 Answer
1






active

oldest

votes









active

oldest

votes






active

oldest

votes









2












$begingroup$

This is not correct. You deduce, after several steps that $tan^2thetageqslant0$. But this holds for every function. Are you going to say that the range of every function from $mathbb R$ into $mathbb R$ is $mathbb R$?






share|cite|improve this answer











$endgroup$













  • $begingroup$
    I agree. when $[f(x)]^2geq 0$ it is not necessary that $f(x)$ has negative values.
    $endgroup$
    – Hussain-Alqatari
    Dec 4 '18 at 9:08


















2












$begingroup$

This is not correct. You deduce, after several steps that $tan^2thetageqslant0$. But this holds for every function. Are you going to say that the range of every function from $mathbb R$ into $mathbb R$ is $mathbb R$?






share|cite|improve this answer











$endgroup$













  • $begingroup$
    I agree. when $[f(x)]^2geq 0$ it is not necessary that $f(x)$ has negative values.
    $endgroup$
    – Hussain-Alqatari
    Dec 4 '18 at 9:08
















2












2








2





$begingroup$

This is not correct. You deduce, after several steps that $tan^2thetageqslant0$. But this holds for every function. Are you going to say that the range of every function from $mathbb R$ into $mathbb R$ is $mathbb R$?






share|cite|improve this answer











$endgroup$



This is not correct. You deduce, after several steps that $tan^2thetageqslant0$. But this holds for every function. Are you going to say that the range of every function from $mathbb R$ into $mathbb R$ is $mathbb R$?







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Dec 8 '18 at 16:12

























answered Dec 4 '18 at 9:06









José Carlos SantosJosé Carlos Santos

155k22124227




155k22124227












  • $begingroup$
    I agree. when $[f(x)]^2geq 0$ it is not necessary that $f(x)$ has negative values.
    $endgroup$
    – Hussain-Alqatari
    Dec 4 '18 at 9:08




















  • $begingroup$
    I agree. when $[f(x)]^2geq 0$ it is not necessary that $f(x)$ has negative values.
    $endgroup$
    – Hussain-Alqatari
    Dec 4 '18 at 9:08


















$begingroup$
I agree. when $[f(x)]^2geq 0$ it is not necessary that $f(x)$ has negative values.
$endgroup$
– Hussain-Alqatari
Dec 4 '18 at 9:08






$begingroup$
I agree. when $[f(x)]^2geq 0$ it is not necessary that $f(x)$ has negative values.
$endgroup$
– Hussain-Alqatari
Dec 4 '18 at 9:08





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