Mixing
Sine function - getting coordinates from the graph.
Clash Royale CLAN TAG#URR8PPP
up vote
0
down vote
favorite
I have trouble with getting coordinates from the sine graph:
The format of the function is:
$f(x)=sin(x-k)+c$
I need to find the coordinates of A. Could you give me just a little hint where to start from? I understand that the period of the function is $2pi$, and that the horizontal coordinate is $frac2x3$, but I have no idea how to get the vertical coordinate which, according to the exercise, is $frac32$, all I can say is that it should be equal to $c+1$ as the maximum point is one because there is no vertical stretch, and the graph is translated up by $c$.
I am expected to solve the problem in this order.
functions trigonometry
asked Jul 30 at 8:07
Danielius
1135
add a comment |Â
up vote
0
down vote
favorite
I have trouble with getting coordinates from the sine graph:
The format of the function is:
$f(x)=sin(x-k)+c$
I need to find the coordinates of A. Could you give me just a little hint where to start from? I understand that the period of the function is $2pi$, and that the horizontal coordinate is $frac2x3$, but I have no idea how to get the vertical coordinate which, according to the exercise, is $frac32$, all I can say is that it should be equal to $c+1$ as the maximum point is one because there is no vertical stretch, and the graph is translated up by $c$.
I am expected to solve the problem in this order.
functions trigonometry
asked Jul 30 at 8:07
Danielius
1135
1
You need $c$ to write down the coordinates of $A$. As the question does not say that you have to give the numerical values, a valid answer could be $A(frac2pi3, 1+c)$
– trancelocation
Jul 30 at 9:22
@trancelocation Uhm, thats what I wrote, but they stated that the answer is 3/2. I came to a conclusion that this problem is not solvable with what they have taught us before, so thanks for all the help.
– Danielius
Jul 30 at 9:49
add a comment |Â
up vote
0
down vote
favorite
up vote
0
down vote
favorite
I have trouble with getting coordinates from the sine graph:
The format of the function is:
$f(x)=sin(x-k)+c$
I need to find the coordinates of A. Could you give me just a little hint where to start from? I understand that the period of the function is $2pi$, and that the horizontal coordinate is $frac2x3$, but I have no idea how to get the vertical coordinate which, according to the exercise, is $frac32$, all I can say is that it should be equal to $c+1$ as the maximum point is one because there is no vertical stretch, and the graph is translated up by $c$.
I am expected to solve the problem in this order.
functions trigonometry
asked Jul 30 at 8:07
Danielius
1135
I have trouble with getting coordinates from the sine graph:
The format of the function is:
$f(x)=sin(x-k)+c$
I need to find the coordinates of A. Could you give me just a little hint where to start from? I understand that the period of the function is $2pi$, and that the horizontal coordinate is $frac2x3$, but I have no idea how to get the vertical coordinate which, according to the exercise, is $frac32$, all I can say is that it should be equal to $c+1$ as the maximum point is one because there is no vertical stretch, and the graph is translated up by $c$.
I am expected to solve the problem in this order.
functions trigonometry
asked Jul 30 at 8:07
Danielius
1135
edited Jul 30 at 8:49
asked Jul 30 at 8:07
Danielius
1135
asked Jul 30 at 8:07
Danielius
1135
asked Jul 30 at 8:07
Danielius
1135
1135
1
You need $c$ to write down the coordinates of $A$. As the question does not say that you have to give the numerical values, a valid answer could be $A(frac2pi3, 1+c)$
– trancelocation
Jul 30 at 9:22
@trancelocation Uhm, thats what I wrote, but they stated that the answer is 3/2. I came to a conclusion that this problem is not solvable with what they have taught us before, so thanks for all the help.
– Danielius
Jul 30 at 9:49
add a comment |Â
1
You need $c$ to write down the coordinates of $A$. As the question does not say that you have to give the numerical values, a valid answer could be $A(frac2pi3, 1+c)$
– trancelocation
Jul 30 at 9:22
@trancelocation Uhm, thats what I wrote, but they stated that the answer is 3/2. I came to a conclusion that this problem is not solvable with what they have taught us before, so thanks for all the help.
– Danielius
Jul 30 at 9:49
1
1
You need $c$ to write down the coordinates of $A$. As the question does not say that you have to give the numerical values, a valid answer could be $A(frac2pi3, 1+c)$
– trancelocation
Jul 30 at 9:22
You need $c$ to write down the coordinates of $A$. As the question does not say that you have to give the numerical values, a valid answer could be $A(frac2pi3, 1+c)$
– trancelocation
Jul 30 at 9:22
@trancelocation Uhm, thats what I wrote, but they stated that the answer is 3/2. I came to a conclusion that this problem is not solvable with what they have taught us before, so thanks for all the help.
– Danielius
Jul 30 at 9:49
@trancelocation Uhm, thats what I wrote, but they stated that the answer is 3/2. I came to a conclusion that this problem is not solvable with what they have taught us before, so thanks for all the help.
– Danielius
Jul 30 at 9:49
add a comment |Â
4 Answers
4
active
oldest
votes
up vote
1
down vote
We have $0=f(2 pi)$ and $0=f(frac4 pi3)$. This gives two equations for $k$ and $c$.
Solve this system of equations. Then look for $x_0 in (0, frac4 pi3) $ such that $f'(x_0)=0$.
Then we have $A=f(x_0)$.
answered Jul 30 at 8:18
Fred
37k1237
Uhmm, I need to find $k$ and $c$ in next part, so I guess I cannot use this... What is $f'()$? A derivative? We haven't studied these yet...
– Danielius
Jul 30 at 8:21
add a comment |Â
up vote
1
down vote
I would start by figuring out $k$ and $c$. From the graph, you can see that
$f(0)= sin-k +c = - sink + c = 0 Rightarrow c = sink$
and also
$
fleft(frac4pi3 right) = sinleft(frac4pi3 - k right) + c = 0
$
Now we can use $sin(x-y) = sinx cosy - cosxsiny $
$$
Rightarrow sinleft(frac4pi3 - k right) = sinfrac4pi3 cosk - cosfrac4pi3sink = -fracsqrt32 cosk + frac12sink
$$
and therefore
$$
fleft(frac4pi3 right) = -fracsqrt32 cosk + frac12sink + c = 0
$$
We can also use $cosk = sqrt1- sin^2k = sqrt1-c^2$, giving an equation for $c$:
$$
-fracsqrt32 sqrt1-c^2 + frac12c + c = 0
$$
Can you continue from here?
answered Jul 30 at 8:22
Matti P.
1,306212
Yes I could have solved the problem like that, but the thing is that I first need to find the coordinates of $A$ and only then I need to find $k$ and $c$. Moreover, even though I know that $sin(x−y)=sin(x)cos(y)−cos(x)sin(y)$, this formula was not shown in the textbook yet :D So is there any way to solve this without using this formula and in correct order? Otherwise, this exercise is unsolvable using the ideas currently presented in the textbook.
– Danielius
Jul 30 at 8:30
And, it is not $frac4pi3$ but $frac4x3$ in the picture.
– Danielius
Jul 30 at 8:42
add a comment |Â
up vote
1
down vote
You have $$f(x)=sin(x-k)+c$$ Using the two points
$$fleft(frac4 pi 3right)=sinleft(frac4 pi 3-kright)+c= 0tag 1$$
$$f(2pi)=sin(2pi-k)+c=-sin(k)+c=0tag 2$$
Subract $(2)$ from $(1)$ to get a simple trigonometric equation in $k$, namely
$$sinleft(frac4 pi 3-kright)+sin(k)=0 $$ Use $sin(p)+sin(q)=??$, then solve it for $k$, and back to $(1)$ or $(2)$ get $c$ .
On the other hand, point $A$ seems to be a maximum. Since $$f'(x)=cos(x-k)$$ at this point $f'(x)=0$. Then $x= ?$; at this point compute $f(x)$ and you are done.
answered Jul 30 at 8:22
Claude Leibovici
111k1055126
Umm, is $f'()$ is a derivative? As we did not study these. Also, its not $frac4pi3$ but $frac4x3$ in the picture. Also, I cannot get c and k before writing the coordinates of A, is it really possible to do it in this order(first getting coordinates of A, then getting k and c)?
– Danielius
Jul 30 at 8:46
I edited my question and added the full problem.
– Danielius
Jul 30 at 8:49
@Danielius. As far as I can "see" it (I am almost blind), it is $frac4pi3$
– Claude Leibovici
Jul 30 at 9:59
OOPS, my bad. That makes a lot more sense.
– Danielius
Jul 30 at 10:31
add a comment |Â
up vote
1
down vote
- Symmetry gives the first negative minimum of $f(x)$ in the middle between $frac4pi3$ and $2pi$ at $x_m = frac2pi + frac4pi32= frac53pi$. Since $sin(frac32pi)= -1$ we get the shift
$$frac53pi- k = frac32pi Rightarrow boxedk = fracpi6$$ - $f(2pi) = sin(2pi - fracpi6) + c = 0$
$$Rightarrow boxedc = sin fracpi6 = frac12 $$
answered Jul 30 at 9:08
trancelocation
4,5701413
add a comment |Â
4 Answers
4
active
oldest
votes
4 Answers
4
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
We have $0=f(2 pi)$ and $0=f(frac4 pi3)$. This gives two equations for $k$ and $c$.
Solve this system of equations. Then look for $x_0 in (0, frac4 pi3) $ such that $f'(x_0)=0$.
Then we have $A=f(x_0)$.
answered Jul 30 at 8:18
Fred
37k1237
Uhmm, I need to find $k$ and $c$ in next part, so I guess I cannot use this... What is $f'()$? A derivative? We haven't studied these yet...
– Danielius
Jul 30 at 8:21
add a comment |Â
up vote
1
down vote
We have $0=f(2 pi)$ and $0=f(frac4 pi3)$. This gives two equations for $k$ and $c$.
Solve this system of equations. Then look for $x_0 in (0, frac4 pi3) $ such that $f'(x_0)=0$.
Then we have $A=f(x_0)$.
answered Jul 30 at 8:18
Fred
37k1237
Uhmm, I need to find $k$ and $c$ in next part, so I guess I cannot use this... What is $f'()$? A derivative? We haven't studied these yet...
– Danielius
Jul 30 at 8:21
add a comment |Â
up vote
1
down vote
up vote
1
down vote
We have $0=f(2 pi)$ and $0=f(frac4 pi3)$. This gives two equations for $k$ and $c$.
Solve this system of equations. Then look for $x_0 in (0, frac4 pi3) $ such that $f'(x_0)=0$.
Then we have $A=f(x_0)$.
answered Jul 30 at 8:18
Fred
37k1237
We have $0=f(2 pi)$ and $0=f(frac4 pi3)$. This gives two equations for $k$ and $c$.
Solve this system of equations. Then look for $x_0 in (0, frac4 pi3) $ such that $f'(x_0)=0$.
Then we have $A=f(x_0)$.
answered Jul 30 at 8:18
Fred
37k1237
answered Jul 30 at 8:18
Fred
37k1237
answered Jul 30 at 8:18
Fred
37k1237
answered Jul 30 at 8:18
Fred
37k1237
37k1237
Uhmm, I need to find $k$ and $c$ in next part, so I guess I cannot use this... What is $f'()$? A derivative? We haven't studied these yet...
– Danielius
Jul 30 at 8:21
add a comment |Â
Uhmm, I need to find $k$ and $c$ in next part, so I guess I cannot use this... What is $f'()$? A derivative? We haven't studied these yet...
– Danielius
Jul 30 at 8:21
Uhmm, I need to find $k$ and $c$ in next part, so I guess I cannot use this... What is $f'()$? A derivative? We haven't studied these yet...
– Danielius
Jul 30 at 8:21
Uhmm, I need to find $k$ and $c$ in next part, so I guess I cannot use this... What is $f'()$? A derivative? We haven't studied these yet...
– Danielius
Jul 30 at 8:21
add a comment |Â
up vote
1
down vote
I would start by figuring out $k$ and $c$. From the graph, you can see that
$f(0)= sin-k +c = - sink + c = 0 Rightarrow c = sink$
and also
$
fleft(frac4pi3 right) = sinleft(frac4pi3 - k right) + c = 0
$
Now we can use $sin(x-y) = sinx cosy - cosxsiny $
$$
Rightarrow sinleft(frac4pi3 - k right) = sinfrac4pi3 cosk - cosfrac4pi3sink = -fracsqrt32 cosk + frac12sink
$$
and therefore
$$
fleft(frac4pi3 right) = -fracsqrt32 cosk + frac12sink + c = 0
$$
We can also use $cosk = sqrt1- sin^2k = sqrt1-c^2$, giving an equation for $c$:
$$
-fracsqrt32 sqrt1-c^2 + frac12c + c = 0
$$
Can you continue from here?
answered Jul 30 at 8:22
Matti P.
1,306212
Yes I could have solved the problem like that, but the thing is that I first need to find the coordinates of $A$ and only then I need to find $k$ and $c$. Moreover, even though I know that $sin(x−y)=sin(x)cos(y)−cos(x)sin(y)$, this formula was not shown in the textbook yet :D So is there any way to solve this without using this formula and in correct order? Otherwise, this exercise is unsolvable using the ideas currently presented in the textbook.
– Danielius
Jul 30 at 8:30
And, it is not $frac4pi3$ but $frac4x3$ in the picture.
– Danielius
Jul 30 at 8:42
add a comment |Â
up vote
1
down vote
I would start by figuring out $k$ and $c$. From the graph, you can see that
$f(0)= sin-k +c = - sink + c = 0 Rightarrow c = sink$
and also
$
fleft(frac4pi3 right) = sinleft(frac4pi3 - k right) + c = 0
$
Now we can use $sin(x-y) = sinx cosy - cosxsiny $
$$
Rightarrow sinleft(frac4pi3 - k right) = sinfrac4pi3 cosk - cosfrac4pi3sink = -fracsqrt32 cosk + frac12sink
$$
and therefore
$$
fleft(frac4pi3 right) = -fracsqrt32 cosk + frac12sink + c = 0
$$
We can also use $cosk = sqrt1- sin^2k = sqrt1-c^2$, giving an equation for $c$:
$$
-fracsqrt32 sqrt1-c^2 + frac12c + c = 0
$$
Can you continue from here?
answered Jul 30 at 8:22
Matti P.
1,306212
Yes I could have solved the problem like that, but the thing is that I first need to find the coordinates of $A$ and only then I need to find $k$ and $c$. Moreover, even though I know that $sin(x−y)=sin(x)cos(y)−cos(x)sin(y)$, this formula was not shown in the textbook yet :D So is there any way to solve this without using this formula and in correct order? Otherwise, this exercise is unsolvable using the ideas currently presented in the textbook.
– Danielius
Jul 30 at 8:30
And, it is not $frac4pi3$ but $frac4x3$ in the picture.
– Danielius
Jul 30 at 8:42
add a comment |Â
up vote
1
down vote
up vote
1
down vote
I would start by figuring out $k$ and $c$. From the graph, you can see that
$f(0)= sin-k +c = - sink + c = 0 Rightarrow c = sink$
and also
$
fleft(frac4pi3 right) = sinleft(frac4pi3 - k right) + c = 0
$
Now we can use $sin(x-y) = sinx cosy - cosxsiny $
$$
Rightarrow sinleft(frac4pi3 - k right) = sinfrac4pi3 cosk - cosfrac4pi3sink = -fracsqrt32 cosk + frac12sink
$$
and therefore
$$
fleft(frac4pi3 right) = -fracsqrt32 cosk + frac12sink + c = 0
$$
We can also use $cosk = sqrt1- sin^2k = sqrt1-c^2$, giving an equation for $c$:
$$
-fracsqrt32 sqrt1-c^2 + frac12c + c = 0
$$
Can you continue from here?
answered Jul 30 at 8:22
Matti P.
1,306212
I would start by figuring out $k$ and $c$. From the graph, you can see that
$f(0)= sin-k +c = - sink + c = 0 Rightarrow c = sink$
and also
$
fleft(frac4pi3 right) = sinleft(frac4pi3 - k right) + c = 0
$
Now we can use $sin(x-y) = sinx cosy - cosxsiny $
$$
Rightarrow sinleft(frac4pi3 - k right) = sinfrac4pi3 cosk - cosfrac4pi3sink = -fracsqrt32 cosk + frac12sink
$$
and therefore
$$
fleft(frac4pi3 right) = -fracsqrt32 cosk + frac12sink + c = 0
$$
We can also use $cosk = sqrt1- sin^2k = sqrt1-c^2$, giving an equation for $c$:
$$
-fracsqrt32 sqrt1-c^2 + frac12c + c = 0
$$
Can you continue from here?
answered Jul 30 at 8:22
Matti P.
1,306212
answered Jul 30 at 8:22
Matti P.
1,306212
answered Jul 30 at 8:22
Matti P.
1,306212
answered Jul 30 at 8:22
Matti P.
1,306212
1,306212
Yes I could have solved the problem like that, but the thing is that I first need to find the coordinates of $A$ and only then I need to find $k$ and $c$. Moreover, even though I know that $sin(x−y)=sin(x)cos(y)−cos(x)sin(y)$, this formula was not shown in the textbook yet :D So is there any way to solve this without using this formula and in correct order? Otherwise, this exercise is unsolvable using the ideas currently presented in the textbook.
– Danielius
Jul 30 at 8:30
And, it is not $frac4pi3$ but $frac4x3$ in the picture.
– Danielius
Jul 30 at 8:42
add a comment |Â
Yes I could have solved the problem like that, but the thing is that I first need to find the coordinates of $A$ and only then I need to find $k$ and $c$. Moreover, even though I know that $sin(x−y)=sin(x)cos(y)−cos(x)sin(y)$, this formula was not shown in the textbook yet :D So is there any way to solve this without using this formula and in correct order? Otherwise, this exercise is unsolvable using the ideas currently presented in the textbook.
– Danielius
Jul 30 at 8:30
And, it is not $frac4pi3$ but $frac4x3$ in the picture.
– Danielius
Jul 30 at 8:42
Yes I could have solved the problem like that, but the thing is that I first need to find the coordinates of $A$ and only then I need to find $k$ and $c$. Moreover, even though I know that $sin(x−y)=sin(x)cos(y)−cos(x)sin(y)$, this formula was not shown in the textbook yet :D So is there any way to solve this without using this formula and in correct order? Otherwise, this exercise is unsolvable using the ideas currently presented in the textbook.
– Danielius
Jul 30 at 8:30
Yes I could have solved the problem like that, but the thing is that I first need to find the coordinates of $A$ and only then I need to find $k$ and $c$. Moreover, even though I know that $sin(x−y)=sin(x)cos(y)−cos(x)sin(y)$, this formula was not shown in the textbook yet :D So is there any way to solve this without using this formula and in correct order? Otherwise, this exercise is unsolvable using the ideas currently presented in the textbook.
– Danielius
Jul 30 at 8:30
And, it is not $frac4pi3$ but $frac4x3$ in the picture.
– Danielius
Jul 30 at 8:42
And, it is not $frac4pi3$ but $frac4x3$ in the picture.
– Danielius
Jul 30 at 8:42
add a comment |Â
up vote
1
down vote
You have $$f(x)=sin(x-k)+c$$ Using the two points
$$fleft(frac4 pi 3right)=sinleft(frac4 pi 3-kright)+c= 0tag 1$$
$$f(2pi)=sin(2pi-k)+c=-sin(k)+c=0tag 2$$
Subract $(2)$ from $(1)$ to get a simple trigonometric equation in $k$, namely
$$sinleft(frac4 pi 3-kright)+sin(k)=0 $$ Use $sin(p)+sin(q)=??$, then solve it for $k$, and back to $(1)$ or $(2)$ get $c$ .
On the other hand, point $A$ seems to be a maximum. Since $$f'(x)=cos(x-k)$$ at this point $f'(x)=0$. Then $x= ?$; at this point compute $f(x)$ and you are done.
answered Jul 30 at 8:22
Claude Leibovici
111k1055126
Umm, is $f'()$ is a derivative? As we did not study these. Also, its not $frac4pi3$ but $frac4x3$ in the picture. Also, I cannot get c and k before writing the coordinates of A, is it really possible to do it in this order(first getting coordinates of A, then getting k and c)?
– Danielius
Jul 30 at 8:46
I edited my question and added the full problem.
– Danielius
Jul 30 at 8:49
@Danielius. As far as I can "see" it (I am almost blind), it is $frac4pi3$
– Claude Leibovici
Jul 30 at 9:59
OOPS, my bad. That makes a lot more sense.
– Danielius
Jul 30 at 10:31
add a comment |Â
up vote
1
down vote
You have $$f(x)=sin(x-k)+c$$ Using the two points
$$fleft(frac4 pi 3right)=sinleft(frac4 pi 3-kright)+c= 0tag 1$$
$$f(2pi)=sin(2pi-k)+c=-sin(k)+c=0tag 2$$
Subract $(2)$ from $(1)$ to get a simple trigonometric equation in $k$, namely
$$sinleft(frac4 pi 3-kright)+sin(k)=0 $$ Use $sin(p)+sin(q)=??$, then solve it for $k$, and back to $(1)$ or $(2)$ get $c$ .
On the other hand, point $A$ seems to be a maximum. Since $$f'(x)=cos(x-k)$$ at this point $f'(x)=0$. Then $x= ?$; at this point compute $f(x)$ and you are done.
answered Jul 30 at 8:22
Claude Leibovici
111k1055126
Umm, is $f'()$ is a derivative? As we did not study these. Also, its not $frac4pi3$ but $frac4x3$ in the picture. Also, I cannot get c and k before writing the coordinates of A, is it really possible to do it in this order(first getting coordinates of A, then getting k and c)?
– Danielius
Jul 30 at 8:46
I edited my question and added the full problem.
– Danielius
Jul 30 at 8:49
@Danielius. As far as I can "see" it (I am almost blind), it is $frac4pi3$
– Claude Leibovici
Jul 30 at 9:59
OOPS, my bad. That makes a lot more sense.
– Danielius
Jul 30 at 10:31
add a comment |Â
up vote
1
down vote
up vote
1
down vote
You have $$f(x)=sin(x-k)+c$$ Using the two points
$$fleft(frac4 pi 3right)=sinleft(frac4 pi 3-kright)+c= 0tag 1$$
$$f(2pi)=sin(2pi-k)+c=-sin(k)+c=0tag 2$$
Subract $(2)$ from $(1)$ to get a simple trigonometric equation in $k$, namely
$$sinleft(frac4 pi 3-kright)+sin(k)=0 $$ Use $sin(p)+sin(q)=??$, then solve it for $k$, and back to $(1)$ or $(2)$ get $c$ .
On the other hand, point $A$ seems to be a maximum. Since $$f'(x)=cos(x-k)$$ at this point $f'(x)=0$. Then $x= ?$; at this point compute $f(x)$ and you are done.
answered Jul 30 at 8:22
Claude Leibovici
111k1055126
You have $$f(x)=sin(x-k)+c$$ Using the two points
$$fleft(frac4 pi 3right)=sinleft(frac4 pi 3-kright)+c= 0tag 1$$
$$f(2pi)=sin(2pi-k)+c=-sin(k)+c=0tag 2$$
Subract $(2)$ from $(1)$ to get a simple trigonometric equation in $k$, namely
$$sinleft(frac4 pi 3-kright)+sin(k)=0 $$ Use $sin(p)+sin(q)=??$, then solve it for $k$, and back to $(1)$ or $(2)$ get $c$ .
On the other hand, point $A$ seems to be a maximum. Since $$f'(x)=cos(x-k)$$ at this point $f'(x)=0$. Then $x= ?$; at this point compute $f(x)$ and you are done.
answered Jul 30 at 8:22
Claude Leibovici
111k1055126
edited Jul 30 at 8:41
answered Jul 30 at 8:22
Claude Leibovici
111k1055126
answered Jul 30 at 8:22
Claude Leibovici
111k1055126
answered Jul 30 at 8:22
Claude Leibovici
111k1055126
111k1055126
Umm, is $f'()$ is a derivative? As we did not study these. Also, its not $frac4pi3$ but $frac4x3$ in the picture. Also, I cannot get c and k before writing the coordinates of A, is it really possible to do it in this order(first getting coordinates of A, then getting k and c)?
– Danielius
Jul 30 at 8:46
I edited my question and added the full problem.
– Danielius
Jul 30 at 8:49
@Danielius. As far as I can "see" it (I am almost blind), it is $frac4pi3$
– Claude Leibovici
Jul 30 at 9:59
OOPS, my bad. That makes a lot more sense.
– Danielius
Jul 30 at 10:31
add a comment |Â
Umm, is $f'()$ is a derivative? As we did not study these. Also, its not $frac4pi3$ but $frac4x3$ in the picture. Also, I cannot get c and k before writing the coordinates of A, is it really possible to do it in this order(first getting coordinates of A, then getting k and c)?
– Danielius
Jul 30 at 8:46
I edited my question and added the full problem.
– Danielius
Jul 30 at 8:49
@Danielius. As far as I can "see" it (I am almost blind), it is $frac4pi3$
– Claude Leibovici
Jul 30 at 9:59
OOPS, my bad. That makes a lot more sense.
– Danielius
Jul 30 at 10:31
Umm, is $f'()$ is a derivative? As we did not study these. Also, its not $frac4pi3$ but $frac4x3$ in the picture. Also, I cannot get c and k before writing the coordinates of A, is it really possible to do it in this order(first getting coordinates of A, then getting k and c)?
– Danielius
Jul 30 at 8:46
Umm, is $f'()$ is a derivative? As we did not study these. Also, its not $frac4pi3$ but $frac4x3$ in the picture. Also, I cannot get c and k before writing the coordinates of A, is it really possible to do it in this order(first getting coordinates of A, then getting k and c)?
– Danielius
Jul 30 at 8:46
I edited my question and added the full problem.
– Danielius
Jul 30 at 8:49
I edited my question and added the full problem.
– Danielius
Jul 30 at 8:49
@Danielius. As far as I can "see" it (I am almost blind), it is $frac4pi3$
– Claude Leibovici
Jul 30 at 9:59
@Danielius. As far as I can "see" it (I am almost blind), it is $frac4pi3$
– Claude Leibovici
Jul 30 at 9:59
OOPS, my bad. That makes a lot more sense.
– Danielius
Jul 30 at 10:31
OOPS, my bad. That makes a lot more sense.
– Danielius
Jul 30 at 10:31
add a comment |Â
up vote
1
down vote
- Symmetry gives the first negative minimum of $f(x)$ in the middle between $frac4pi3$ and $2pi$ at $x_m = frac2pi + frac4pi32= frac53pi$. Since $sin(frac32pi)= -1$ we get the shift
$$frac53pi- k = frac32pi Rightarrow boxedk = fracpi6$$ - $f(2pi) = sin(2pi - fracpi6) + c = 0$
$$Rightarrow boxedc = sin fracpi6 = frac12 $$
answered Jul 30 at 9:08
trancelocation
4,5701413
add a comment |Â
up vote
1
down vote
- Symmetry gives the first negative minimum of $f(x)$ in the middle between $frac4pi3$ and $2pi$ at $x_m = frac2pi + frac4pi32= frac53pi$. Since $sin(frac32pi)= -1$ we get the shift
$$frac53pi- k = frac32pi Rightarrow boxedk = fracpi6$$ - $f(2pi) = sin(2pi - fracpi6) + c = 0$
$$Rightarrow boxedc = sin fracpi6 = frac12 $$
answered Jul 30 at 9:08
trancelocation
4,5701413
add a comment |Â
up vote
1
down vote
up vote
1
down vote
- Symmetry gives the first negative minimum of $f(x)$ in the middle between $frac4pi3$ and $2pi$ at $x_m = frac2pi + frac4pi32= frac53pi$. Since $sin(frac32pi)= -1$ we get the shift
$$frac53pi- k = frac32pi Rightarrow boxedk = fracpi6$$ - $f(2pi) = sin(2pi - fracpi6) + c = 0$
$$Rightarrow boxedc = sin fracpi6 = frac12 $$
answered Jul 30 at 9:08
trancelocation
4,5701413
- Symmetry gives the first negative minimum of $f(x)$ in the middle between $frac4pi3$ and $2pi$ at $x_m = frac2pi + frac4pi32= frac53pi$. Since $sin(frac32pi)= -1$ we get the shift
$$frac53pi- k = frac32pi Rightarrow boxedk = fracpi6$$ - $f(2pi) = sin(2pi - fracpi6) + c = 0$
$$Rightarrow boxedc = sin fracpi6 = frac12 $$
answered Jul 30 at 9:08
trancelocation
4,5701413
answered Jul 30 at 9:08
trancelocation
4,5701413
answered Jul 30 at 9:08
trancelocation
4,5701413
answered Jul 30 at 9:08
trancelocation
4,5701413
4,5701413
add a comment |Â
add a comment |Â
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f2866758%2fsine-function-getting-coordinates-from-the-graph%23new-answer', 'question_page');
);
Post as a guest
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
1
You need $c$ to write down the coordinates of $A$. As the question does not say that you have to give the numerical values, a valid answer could be $A(frac2pi3, 1+c)$
– trancelocation
Jul 30 at 9:22
@trancelocation Uhm, thats what I wrote, but they stated that the answer is 3/2. I came to a conclusion that this problem is not solvable with what they have taught us before, so thanks for all the help.
– Danielius
Jul 30 at 9:49