Solutions to $(x-x_0)cos(x_0)+sin(x_0)-sin(x)=0$ Announcing the arrival of Valued Associate #679: Cesar Manara Planned maintenance scheduled April 23, 2019 at 23:30 UTC (7:30pm US/Eastern)Solution set of $cos(cos(cos(cos(x)))) = sin(sin(sin(sin(x))))$Solutions of a nonlinear trigonometric equationSolution to $bsin(theta)cos(phi)+acos(theta)sin(phi)=0$ for $phi$Value of $x$ in $sin^-1(x)+sin^-1(1-x)=cos^-1(x)$solution of the equation $2sintheta/2=cos(theta(n-1/2))?$Find point $P$ so that the tangent to the function plot passes through originExplain the concept behind solving $sin(x)cos(x) + cos(x) = 0$, from Paul's Online Math NotesHow can I solve for the area under sine rotated 45 degrees?A line tangent to the graph of sinRelationship between 2 definitions of sin. One using differential equations and one using the unit circle.
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Solutions to $(x-x_0)cos(x_0)+sin(x_0)-sin(x)=0$
Announcing the arrival of Valued Associate #679: Cesar Manara
Planned maintenance scheduled April 23, 2019 at 23:30 UTC (7:30pm US/Eastern)Solution set of $cos(cos(cos(cos(x)))) = sin(sin(sin(sin(x))))$Solutions of a nonlinear trigonometric equationSolution to $bsin(theta)cos(phi)+acos(theta)sin(phi)=0$ for $phi$Value of $x$ in $sin^-1(x)+sin^-1(1-x)=cos^-1(x)$solution of the equation $2sintheta/2=cos(theta(n-1/2))?$Find point $P$ so that the tangent to the function plot passes through originExplain the concept behind solving $sin(x)cos(x) + cos(x) = 0$, from Paul's Online Math NotesHow can I solve for the area under sine rotated 45 degrees?A line tangent to the graph of sinRelationship between 2 definitions of sin. One using differential equations and one using the unit circle.
$begingroup$
A tangent line to the sine function at the point $x_0, sin(x_0)$, will intersect the sine at the points where
$$(x-x_0)cos(x_0)+sin(x_0)-sin(x)=0$$
The solutions to that equation looks like this:
There's the obvious $x_0=x$ solution, and all the other solutions are the same function, only shifted along the $x_0=x$ line. I'm interested in these solutions, but since they're all the same function but shifted, I'm focusing on the middle one.
If I plot just the middle one:
I've tried finding it analytically, to no avail, and functions I thought looked similar (for example $mathrmSi(x)$) are not solutions to the equation.
Can anyone help me to be able to plot this function without just doing Newton steps to solve it?
I'm interested in the solution for all values of $x_0$ in the range $left[-frac12pi; frac12piright]$ and all $xin mathbbR$ (the same as my second plot)
trigonometry tangent-line
asked Apr 2 at 13:59
AtnasAtnas
21115
$endgroup$
add a comment |
$begingroup$
A tangent line to the sine function at the point $x_0, sin(x_0)$, will intersect the sine at the points where
$$(x-x_0)cos(x_0)+sin(x_0)-sin(x)=0$$
The solutions to that equation looks like this:
There's the obvious $x_0=x$ solution, and all the other solutions are the same function, only shifted along the $x_0=x$ line. I'm interested in these solutions, but since they're all the same function but shifted, I'm focusing on the middle one.
If I plot just the middle one:
I've tried finding it analytically, to no avail, and functions I thought looked similar (for example $mathrmSi(x)$) are not solutions to the equation.
Can anyone help me to be able to plot this function without just doing Newton steps to solve it?
I'm interested in the solution for all values of $x_0$ in the range $left[-frac12pi; frac12piright]$ and all $xin mathbbR$ (the same as my second plot)
trigonometry tangent-line
asked Apr 2 at 13:59
AtnasAtnas
21115
$endgroup$
$begingroup$
Please calarify whether you want all valid values of $a$ or whether you want to draw the equation (you need a specific $a$ value for this).
$endgroup$
– NoChance
Apr 2 at 14:49
$begingroup$
@NoChance I updated my question ($a$ is now $x_0$), and in order to plot the function, I need many values for $x_0$ (it is on the second axis)
$endgroup$
– Atnas
Apr 2 at 15:11
add a comment |
$begingroup$
A tangent line to the sine function at the point $x_0, sin(x_0)$, will intersect the sine at the points where
$$(x-x_0)cos(x_0)+sin(x_0)-sin(x)=0$$
The solutions to that equation looks like this:
There's the obvious $x_0=x$ solution, and all the other solutions are the same function, only shifted along the $x_0=x$ line. I'm interested in these solutions, but since they're all the same function but shifted, I'm focusing on the middle one.
If I plot just the middle one:
I've tried finding it analytically, to no avail, and functions I thought looked similar (for example $mathrmSi(x)$) are not solutions to the equation.
Can anyone help me to be able to plot this function without just doing Newton steps to solve it?
I'm interested in the solution for all values of $x_0$ in the range $left[-frac12pi; frac12piright]$ and all $xin mathbbR$ (the same as my second plot)
trigonometry tangent-line
asked Apr 2 at 13:59
AtnasAtnas
21115
$endgroup$
A tangent line to the sine function at the point $x_0, sin(x_0)$, will intersect the sine at the points where
$$(x-x_0)cos(x_0)+sin(x_0)-sin(x)=0$$
The solutions to that equation looks like this:
There's the obvious $x_0=x$ solution, and all the other solutions are the same function, only shifted along the $x_0=x$ line. I'm interested in these solutions, but since they're all the same function but shifted, I'm focusing on the middle one.
If I plot just the middle one:
I've tried finding it analytically, to no avail, and functions I thought looked similar (for example $mathrmSi(x)$) are not solutions to the equation.
Can anyone help me to be able to plot this function without just doing Newton steps to solve it?
I'm interested in the solution for all values of $x_0$ in the range $left[-frac12pi; frac12piright]$ and all $xin mathbbR$ (the same as my second plot)
trigonometry tangent-line
trigonometry tangent-line
asked Apr 2 at 13:59
AtnasAtnas
21115
asked Apr 2 at 13:59
AtnasAtnas
21115
edited Apr 2 at 15:17
Atnas
asked Apr 2 at 13:59
AtnasAtnas
21115
asked Apr 2 at 13:59
AtnasAtnas
21115
asked Apr 2 at 13:59
AtnasAtnas
21115
21115
$begingroup$
Please calarify whether you want all valid values of $a$ or whether you want to draw the equation (you need a specific $a$ value for this).
$endgroup$
– NoChance
Apr 2 at 14:49
$begingroup$
@NoChance I updated my question ($a$ is now $x_0$), and in order to plot the function, I need many values for $x_0$ (it is on the second axis)
$endgroup$
– Atnas
Apr 2 at 15:11
add a comment |
$begingroup$
Please calarify whether you want all valid values of $a$ or whether you want to draw the equation (you need a specific $a$ value for this).
$endgroup$
– NoChance
Apr 2 at 14:49
$begingroup$
@NoChance I updated my question ($a$ is now $x_0$), and in order to plot the function, I need many values for $x_0$ (it is on the second axis)
$endgroup$
– Atnas
Apr 2 at 15:11
$begingroup$
Please calarify whether you want all valid values of $a$ or whether you want to draw the equation (you need a specific $a$ value for this).
$endgroup$
– NoChance
Apr 2 at 14:49
$begingroup$
Please calarify whether you want all valid values of $a$ or whether you want to draw the equation (you need a specific $a$ value for this).
$endgroup$
– NoChance
Apr 2 at 14:49
$begingroup$
@NoChance I updated my question ($a$ is now $x_0$), and in order to plot the function, I need many values for $x_0$ (it is on the second axis)
$endgroup$
– Atnas
Apr 2 at 15:11
$begingroup$
@NoChance I updated my question ($a$ is now $x_0$), and in order to plot the function, I need many values for $x_0$ (it is on the second axis)
$endgroup$
– Atnas
Apr 2 at 15:11
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
Let $$P(x_0,sin(x_0))$$ then the equation of the tangent line is given by $$y=cos(x_0)(x-x_0)+sin(x_0)$$ so we get
$$sin(x)=cos(x_0)(x-x_0)+sin(x_0)$$ Substituting $x=a$ we get
$$sin(a)=cos(x_0)(a-x_0)+sin(x_0)$$ there is a difference to your equation, you have
$$(a-x)cos(a)+sin(x)-sin(a)=0$$?
answered Apr 2 at 14:39
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
79.6k42867
$endgroup$
$begingroup$
I had swapped $x$ and $a$ in my definition of the equation, sorry about that. I have updated my question (and adopted your $x_0$ as the name as I think it makes it clearer)
$endgroup$
– Atnas
Apr 2 at 15:15
add a comment |
$begingroup$
It was too long for a comment, so I post it as an answer.:
Let try this question as a simple case: from point (0,0) which is between $-pi/2$ and $pi/2$ a line tangent to the curve $y=sin x$ is drawn. find the coordinates of point the line touches the curve. We can estimate the coordinates as follows: draw a line from (0, 0) to minimum point (3pi/2, -1). the gradient of line is:
$tan (alpha)=frac-13pi/2 ≈-0.2123$ and $alpha ≈-12^o$.
the point of tangent is near $(x=3pi/2, y=-1)$. Let $x_0=180+[(90-12)=78^o]$.
$cos (180+78)=-cos 78=-0.2079$ is the gradient of tangent line due to the formula which is near the value of [$tan (-12^o)$]. So the coordinates of point of touch a is almost $(x_a=180 +78=258 ≈44pi/30, y_a=-0.978)$.If the point of intersection is $(x_0, y_0)$ and point of touch is $(x_a, y_a)$ we can write:
$m=fracy_a-y_0x_a-x_0=cos(x_a)$
This is a sort of Diophantine equation.I took minimum point $(3pi/2, -1)$ for rough estimation of tangent coordinates.
answered Apr 6 at 20:23
siroussirous
1,8131514
$endgroup$
$begingroup$
I'm confused about what you're doing. At (0,0) $x_0=0$, and the derivative of $sin$ is $cos$ so the slope at that point is $cos(0)=1$. And for exactly $x_0=0$, it only touches the curve once, and that's at the tangent point $x=0$.
$endgroup$
– Atnas
Apr 7 at 14:04
$begingroup$
@Atnas, sorry about it I edited my answer.
$endgroup$
– sirous
Apr 7 at 18:52
$begingroup$
I'm still confused because the only solution for $x_0=0$ is $x=0$ as I wrote before. Here's an illustration: i.imgur.com/t4hHc5e.png You can see that the red line only intersects the blue at $x=0$
$endgroup$
– Atnas
Apr 8 at 8:11
$begingroup$
@Atnas, I can not open your illustrations. If you want to understand what I mean, plot $y=sin (x)$, draw a line from origin tangent to curve near point $(3pi/2, -1)$.Another example: draw a line tangent to curve at point $(5pi/4, -sqrt 2/2)$, it crosses the curve at point $(x_0, y_0)$. to find $x_0$ and $y_0$ you have to solve equation:$sin(x) +(sqrt 2/2)x +(sqrt 2/2)[1-5pi/4]=0$. I think numerical method works.
$endgroup$
– sirous
Apr 8 at 12:05
add a comment |
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2 Answers
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active
oldest
votes
2 Answers
2
active
oldest
votes
active
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active
oldest
votes
$begingroup$
Let $$P(x_0,sin(x_0))$$ then the equation of the tangent line is given by $$y=cos(x_0)(x-x_0)+sin(x_0)$$ so we get
$$sin(x)=cos(x_0)(x-x_0)+sin(x_0)$$ Substituting $x=a$ we get
$$sin(a)=cos(x_0)(a-x_0)+sin(x_0)$$ there is a difference to your equation, you have
$$(a-x)cos(a)+sin(x)-sin(a)=0$$?
answered Apr 2 at 14:39
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
79.6k42867
$endgroup$
$begingroup$
I had swapped $x$ and $a$ in my definition of the equation, sorry about that. I have updated my question (and adopted your $x_0$ as the name as I think it makes it clearer)
$endgroup$
– Atnas
Apr 2 at 15:15
add a comment |
$begingroup$
Let $$P(x_0,sin(x_0))$$ then the equation of the tangent line is given by $$y=cos(x_0)(x-x_0)+sin(x_0)$$ so we get
$$sin(x)=cos(x_0)(x-x_0)+sin(x_0)$$ Substituting $x=a$ we get
$$sin(a)=cos(x_0)(a-x_0)+sin(x_0)$$ there is a difference to your equation, you have
$$(a-x)cos(a)+sin(x)-sin(a)=0$$?
answered Apr 2 at 14:39
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
79.6k42867
$endgroup$
$begingroup$
I had swapped $x$ and $a$ in my definition of the equation, sorry about that. I have updated my question (and adopted your $x_0$ as the name as I think it makes it clearer)
$endgroup$
– Atnas
Apr 2 at 15:15
add a comment |
$begingroup$
Let $$P(x_0,sin(x_0))$$ then the equation of the tangent line is given by $$y=cos(x_0)(x-x_0)+sin(x_0)$$ so we get
$$sin(x)=cos(x_0)(x-x_0)+sin(x_0)$$ Substituting $x=a$ we get
$$sin(a)=cos(x_0)(a-x_0)+sin(x_0)$$ there is a difference to your equation, you have
$$(a-x)cos(a)+sin(x)-sin(a)=0$$?
answered Apr 2 at 14:39
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
79.6k42867
$endgroup$
Let $$P(x_0,sin(x_0))$$ then the equation of the tangent line is given by $$y=cos(x_0)(x-x_0)+sin(x_0)$$ so we get
$$sin(x)=cos(x_0)(x-x_0)+sin(x_0)$$ Substituting $x=a$ we get
$$sin(a)=cos(x_0)(a-x_0)+sin(x_0)$$ there is a difference to your equation, you have
$$(a-x)cos(a)+sin(x)-sin(a)=0$$?
answered Apr 2 at 14:39
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
79.6k42867
answered Apr 2 at 14:39
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
79.6k42867
answered Apr 2 at 14:39
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
79.6k42867
answered Apr 2 at 14:39
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
79.6k42867
79.6k42867
$begingroup$
I had swapped $x$ and $a$ in my definition of the equation, sorry about that. I have updated my question (and adopted your $x_0$ as the name as I think it makes it clearer)
$endgroup$
– Atnas
Apr 2 at 15:15
add a comment |
$begingroup$
I had swapped $x$ and $a$ in my definition of the equation, sorry about that. I have updated my question (and adopted your $x_0$ as the name as I think it makes it clearer)
$endgroup$
– Atnas
Apr 2 at 15:15
$begingroup$
I had swapped $x$ and $a$ in my definition of the equation, sorry about that. I have updated my question (and adopted your $x_0$ as the name as I think it makes it clearer)
$endgroup$
– Atnas
Apr 2 at 15:15
$begingroup$
I had swapped $x$ and $a$ in my definition of the equation, sorry about that. I have updated my question (and adopted your $x_0$ as the name as I think it makes it clearer)
$endgroup$
– Atnas
Apr 2 at 15:15
add a comment |
$begingroup$
It was too long for a comment, so I post it as an answer.:
Let try this question as a simple case: from point (0,0) which is between $-pi/2$ and $pi/2$ a line tangent to the curve $y=sin x$ is drawn. find the coordinates of point the line touches the curve. We can estimate the coordinates as follows: draw a line from (0, 0) to minimum point (3pi/2, -1). the gradient of line is:
$tan (alpha)=frac-13pi/2 ≈-0.2123$ and $alpha ≈-12^o$.
the point of tangent is near $(x=3pi/2, y=-1)$. Let $x_0=180+[(90-12)=78^o]$.
$cos (180+78)=-cos 78=-0.2079$ is the gradient of tangent line due to the formula which is near the value of [$tan (-12^o)$]. So the coordinates of point of touch a is almost $(x_a=180 +78=258 ≈44pi/30, y_a=-0.978)$.If the point of intersection is $(x_0, y_0)$ and point of touch is $(x_a, y_a)$ we can write:
$m=fracy_a-y_0x_a-x_0=cos(x_a)$
This is a sort of Diophantine equation.I took minimum point $(3pi/2, -1)$ for rough estimation of tangent coordinates.
answered Apr 6 at 20:23
siroussirous
1,8131514
$endgroup$
$begingroup$
I'm confused about what you're doing. At (0,0) $x_0=0$, and the derivative of $sin$ is $cos$ so the slope at that point is $cos(0)=1$. And for exactly $x_0=0$, it only touches the curve once, and that's at the tangent point $x=0$.
$endgroup$
– Atnas
Apr 7 at 14:04
$begingroup$
@Atnas, sorry about it I edited my answer.
$endgroup$
– sirous
Apr 7 at 18:52
$begingroup$
I'm still confused because the only solution for $x_0=0$ is $x=0$ as I wrote before. Here's an illustration: i.imgur.com/t4hHc5e.png You can see that the red line only intersects the blue at $x=0$
$endgroup$
– Atnas
Apr 8 at 8:11
$begingroup$
@Atnas, I can not open your illustrations. If you want to understand what I mean, plot $y=sin (x)$, draw a line from origin tangent to curve near point $(3pi/2, -1)$.Another example: draw a line tangent to curve at point $(5pi/4, -sqrt 2/2)$, it crosses the curve at point $(x_0, y_0)$. to find $x_0$ and $y_0$ you have to solve equation:$sin(x) +(sqrt 2/2)x +(sqrt 2/2)[1-5pi/4]=0$. I think numerical method works.
$endgroup$
– sirous
Apr 8 at 12:05
add a comment |
$begingroup$
It was too long for a comment, so I post it as an answer.:
Let try this question as a simple case: from point (0,0) which is between $-pi/2$ and $pi/2$ a line tangent to the curve $y=sin x$ is drawn. find the coordinates of point the line touches the curve. We can estimate the coordinates as follows: draw a line from (0, 0) to minimum point (3pi/2, -1). the gradient of line is:
$tan (alpha)=frac-13pi/2 ≈-0.2123$ and $alpha ≈-12^o$.
the point of tangent is near $(x=3pi/2, y=-1)$. Let $x_0=180+[(90-12)=78^o]$.
$cos (180+78)=-cos 78=-0.2079$ is the gradient of tangent line due to the formula which is near the value of [$tan (-12^o)$]. So the coordinates of point of touch a is almost $(x_a=180 +78=258 ≈44pi/30, y_a=-0.978)$.If the point of intersection is $(x_0, y_0)$ and point of touch is $(x_a, y_a)$ we can write:
$m=fracy_a-y_0x_a-x_0=cos(x_a)$
This is a sort of Diophantine equation.I took minimum point $(3pi/2, -1)$ for rough estimation of tangent coordinates.
answered Apr 6 at 20:23
siroussirous
1,8131514
$endgroup$
$begingroup$
I'm confused about what you're doing. At (0,0) $x_0=0$, and the derivative of $sin$ is $cos$ so the slope at that point is $cos(0)=1$. And for exactly $x_0=0$, it only touches the curve once, and that's at the tangent point $x=0$.
$endgroup$
– Atnas
Apr 7 at 14:04
$begingroup$
@Atnas, sorry about it I edited my answer.
$endgroup$
– sirous
Apr 7 at 18:52
$begingroup$
I'm still confused because the only solution for $x_0=0$ is $x=0$ as I wrote before. Here's an illustration: i.imgur.com/t4hHc5e.png You can see that the red line only intersects the blue at $x=0$
$endgroup$
– Atnas
Apr 8 at 8:11
$begingroup$
@Atnas, I can not open your illustrations. If you want to understand what I mean, plot $y=sin (x)$, draw a line from origin tangent to curve near point $(3pi/2, -1)$.Another example: draw a line tangent to curve at point $(5pi/4, -sqrt 2/2)$, it crosses the curve at point $(x_0, y_0)$. to find $x_0$ and $y_0$ you have to solve equation:$sin(x) +(sqrt 2/2)x +(sqrt 2/2)[1-5pi/4]=0$. I think numerical method works.
$endgroup$
– sirous
Apr 8 at 12:05
add a comment |
$begingroup$
It was too long for a comment, so I post it as an answer.:
Let try this question as a simple case: from point (0,0) which is between $-pi/2$ and $pi/2$ a line tangent to the curve $y=sin x$ is drawn. find the coordinates of point the line touches the curve. We can estimate the coordinates as follows: draw a line from (0, 0) to minimum point (3pi/2, -1). the gradient of line is:
$tan (alpha)=frac-13pi/2 ≈-0.2123$ and $alpha ≈-12^o$.
the point of tangent is near $(x=3pi/2, y=-1)$. Let $x_0=180+[(90-12)=78^o]$.
$cos (180+78)=-cos 78=-0.2079$ is the gradient of tangent line due to the formula which is near the value of [$tan (-12^o)$]. So the coordinates of point of touch a is almost $(x_a=180 +78=258 ≈44pi/30, y_a=-0.978)$.If the point of intersection is $(x_0, y_0)$ and point of touch is $(x_a, y_a)$ we can write:
$m=fracy_a-y_0x_a-x_0=cos(x_a)$
This is a sort of Diophantine equation.I took minimum point $(3pi/2, -1)$ for rough estimation of tangent coordinates.
answered Apr 6 at 20:23
siroussirous
1,8131514
$endgroup$
It was too long for a comment, so I post it as an answer.:
Let try this question as a simple case: from point (0,0) which is between $-pi/2$ and $pi/2$ a line tangent to the curve $y=sin x$ is drawn. find the coordinates of point the line touches the curve. We can estimate the coordinates as follows: draw a line from (0, 0) to minimum point (3pi/2, -1). the gradient of line is:
$tan (alpha)=frac-13pi/2 ≈-0.2123$ and $alpha ≈-12^o$.
the point of tangent is near $(x=3pi/2, y=-1)$. Let $x_0=180+[(90-12)=78^o]$.
$cos (180+78)=-cos 78=-0.2079$ is the gradient of tangent line due to the formula which is near the value of [$tan (-12^o)$]. So the coordinates of point of touch a is almost $(x_a=180 +78=258 ≈44pi/30, y_a=-0.978)$.If the point of intersection is $(x_0, y_0)$ and point of touch is $(x_a, y_a)$ we can write:
$m=fracy_a-y_0x_a-x_0=cos(x_a)$
This is a sort of Diophantine equation.I took minimum point $(3pi/2, -1)$ for rough estimation of tangent coordinates.
answered Apr 6 at 20:23
siroussirous
1,8131514
edited Apr 7 at 19:05
answered Apr 6 at 20:23
siroussirous
1,8131514
answered Apr 6 at 20:23
siroussirous
1,8131514
answered Apr 6 at 20:23
siroussirous
1,8131514
1,8131514
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I'm confused about what you're doing. At (0,0) $x_0=0$, and the derivative of $sin$ is $cos$ so the slope at that point is $cos(0)=1$. And for exactly $x_0=0$, it only touches the curve once, and that's at the tangent point $x=0$.
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– Atnas
Apr 7 at 14:04
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@Atnas, sorry about it I edited my answer.
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– sirous
Apr 7 at 18:52
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I'm still confused because the only solution for $x_0=0$ is $x=0$ as I wrote before. Here's an illustration: i.imgur.com/t4hHc5e.png You can see that the red line only intersects the blue at $x=0$
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– Atnas
Apr 8 at 8:11
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@Atnas, I can not open your illustrations. If you want to understand what I mean, plot $y=sin (x)$, draw a line from origin tangent to curve near point $(3pi/2, -1)$.Another example: draw a line tangent to curve at point $(5pi/4, -sqrt 2/2)$, it crosses the curve at point $(x_0, y_0)$. to find $x_0$ and $y_0$ you have to solve equation:$sin(x) +(sqrt 2/2)x +(sqrt 2/2)[1-5pi/4]=0$. I think numerical method works.
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– sirous
Apr 8 at 12:05
add a comment |
$begingroup$
I'm confused about what you're doing. At (0,0) $x_0=0$, and the derivative of $sin$ is $cos$ so the slope at that point is $cos(0)=1$. And for exactly $x_0=0$, it only touches the curve once, and that's at the tangent point $x=0$.
$endgroup$
– Atnas
Apr 7 at 14:04
$begingroup$
@Atnas, sorry about it I edited my answer.
$endgroup$
– sirous
Apr 7 at 18:52
$begingroup$
I'm still confused because the only solution for $x_0=0$ is $x=0$ as I wrote before. Here's an illustration: i.imgur.com/t4hHc5e.png You can see that the red line only intersects the blue at $x=0$
$endgroup$
– Atnas
Apr 8 at 8:11
$begingroup$
@Atnas, I can not open your illustrations. If you want to understand what I mean, plot $y=sin (x)$, draw a line from origin tangent to curve near point $(3pi/2, -1)$.Another example: draw a line tangent to curve at point $(5pi/4, -sqrt 2/2)$, it crosses the curve at point $(x_0, y_0)$. to find $x_0$ and $y_0$ you have to solve equation:$sin(x) +(sqrt 2/2)x +(sqrt 2/2)[1-5pi/4]=0$. I think numerical method works.
$endgroup$
– sirous
Apr 8 at 12:05
$begingroup$
I'm confused about what you're doing. At (0,0) $x_0=0$, and the derivative of $sin$ is $cos$ so the slope at that point is $cos(0)=1$. And for exactly $x_0=0$, it only touches the curve once, and that's at the tangent point $x=0$.
$endgroup$
– Atnas
Apr 7 at 14:04
$begingroup$
I'm confused about what you're doing. At (0,0) $x_0=0$, and the derivative of $sin$ is $cos$ so the slope at that point is $cos(0)=1$. And for exactly $x_0=0$, it only touches the curve once, and that's at the tangent point $x=0$.
$endgroup$
– Atnas
Apr 7 at 14:04
$begingroup$
@Atnas, sorry about it I edited my answer.
$endgroup$
– sirous
Apr 7 at 18:52
$begingroup$
@Atnas, sorry about it I edited my answer.
$endgroup$
– sirous
Apr 7 at 18:52
$begingroup$
I'm still confused because the only solution for $x_0=0$ is $x=0$ as I wrote before. Here's an illustration: i.imgur.com/t4hHc5e.png You can see that the red line only intersects the blue at $x=0$
$endgroup$
– Atnas
Apr 8 at 8:11
$begingroup$
I'm still confused because the only solution for $x_0=0$ is $x=0$ as I wrote before. Here's an illustration: i.imgur.com/t4hHc5e.png You can see that the red line only intersects the blue at $x=0$
$endgroup$
– Atnas
Apr 8 at 8:11
$begingroup$
@Atnas, I can not open your illustrations. If you want to understand what I mean, plot $y=sin (x)$, draw a line from origin tangent to curve near point $(3pi/2, -1)$.Another example: draw a line tangent to curve at point $(5pi/4, -sqrt 2/2)$, it crosses the curve at point $(x_0, y_0)$. to find $x_0$ and $y_0$ you have to solve equation:$sin(x) +(sqrt 2/2)x +(sqrt 2/2)[1-5pi/4]=0$. I think numerical method works.
$endgroup$
– sirous
Apr 8 at 12:05
$begingroup$
@Atnas, I can not open your illustrations. If you want to understand what I mean, plot $y=sin (x)$, draw a line from origin tangent to curve near point $(3pi/2, -1)$.Another example: draw a line tangent to curve at point $(5pi/4, -sqrt 2/2)$, it crosses the curve at point $(x_0, y_0)$. to find $x_0$ and $y_0$ you have to solve equation:$sin(x) +(sqrt 2/2)x +(sqrt 2/2)[1-5pi/4]=0$. I think numerical method works.
$endgroup$
– sirous
Apr 8 at 12:05
add a comment |
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$begingroup$
Please calarify whether you want all valid values of $a$ or whether you want to draw the equation (you need a specific $a$ value for this).
$endgroup$
– NoChance
Apr 2 at 14:49
$begingroup$
@NoChance I updated my question ($a$ is now $x_0$), and in order to plot the function, I need many values for $x_0$ (it is on the second axis)
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– Atnas
Apr 2 at 15:11