Proof of range of tan $theta$ [duplicate]
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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.
trigonometry proof-verification
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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.
add a comment |
$begingroup$
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.
trigonometry proof-verification
$endgroup$
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.
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$(x^2)^2=x^4ge0$ 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 $[-1,1]$.
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– Tito Eliatron
Dec 4 '18 at 9:05
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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.
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– GNUSupporter 8964民主女神 地下教會
Dec 4 '18 at 9:19
add a comment |
$begingroup$
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.
trigonometry proof-verification
$endgroup$
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
trigonometry proof-verification
edited Dec 5 '18 at 3:38
user622701
asked Dec 4 '18 at 9:03
user622701user622701
63
63
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
add a comment |
$begingroup$
$(x^2)^2=x^4ge0$ 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
add a comment |
1 Answer
1
active
oldest
votes
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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 $mathbb R$?
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I agree. when $[f(x)]^2geq 0$ it is not necessary that $f(x)$ has negative values.
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– Hussain-Alqatari
Dec 4 '18 at 9:08
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
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 $mathbb R$?
$endgroup$
$begingroup$
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
add a comment |
$begingroup$
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 $mathbb R$?
$endgroup$
$begingroup$
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
add a comment |
$begingroup$
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 $mathbb R$?
$endgroup$
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 $mathbb R$?
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
add a comment |
$begingroup$
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
add a comment |
$begingroup$
$(x^2)^2=x^4ge0$ 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