Find the terminal point when the distance is not in terms of $pi$
$begingroup$
From Stewart Precalculus 5th edi, P407
I am not sure what to do here, in the textbook, Steward didn't provide any example as to finding the terminal point when the distance $t$ is an integer. I know how to find when the distance is in terms of $pi$.
Am I supposed to replace 1 with $frac{pi}{3}$, 2 with $frac{2pi}{3}$ and give an estimation?
algebra-precalculus trigonometry
edited Mar 9 '17 at 17:30
Community♦
1
asked May 23 '13 at 10:48
BenBen
170212
$endgroup$
add a comment |
$begingroup$
From Stewart Precalculus 5th edi, P407
I am not sure what to do here, in the textbook, Steward didn't provide any example as to finding the terminal point when the distance $t$ is an integer. I know how to find when the distance is in terms of $pi$.
Am I supposed to replace 1 with $frac{pi}{3}$, 2 with $frac{2pi}{3}$ and give an estimation?
algebra-precalculus trigonometry
edited Mar 9 '17 at 17:30
Community♦
1
asked May 23 '13 at 10:48
BenBen
170212
$endgroup$
add a comment |
$begingroup$
From Stewart Precalculus 5th edi, P407
I am not sure what to do here, in the textbook, Steward didn't provide any example as to finding the terminal point when the distance $t$ is an integer. I know how to find when the distance is in terms of $pi$.
Am I supposed to replace 1 with $frac{pi}{3}$, 2 with $frac{2pi}{3}$ and give an estimation?
algebra-precalculus trigonometry
edited Mar 9 '17 at 17:30
Community♦
1
asked May 23 '13 at 10:48
BenBen
170212
$endgroup$
From Stewart Precalculus 5th edi, P407
I am not sure what to do here, in the textbook, Steward didn't provide any example as to finding the terminal point when the distance $t$ is an integer. I know how to find when the distance is in terms of $pi$.
Am I supposed to replace 1 with $frac{pi}{3}$, 2 with $frac{2pi}{3}$ and give an estimation?
algebra-precalculus trigonometry
algebra-precalculus trigonometry
edited Mar 9 '17 at 17:30
Community♦
1
asked May 23 '13 at 10:48
BenBen
170212
edited Mar 9 '17 at 17:30
Community♦
1
asked May 23 '13 at 10:48
BenBen
170212
edited Mar 9 '17 at 17:30
Community♦
1
edited Mar 9 '17 at 17:30
Community♦
1
edited Mar 9 '17 at 17:30
Community♦
1
1
asked May 23 '13 at 10:48
BenBen
170212
asked May 23 '13 at 10:48
BenBen
170212
asked May 23 '13 at 10:48
BenBen
170212
170212
add a comment |
add a comment |
2 Answers
2
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oldest
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$begingroup$
Looking at the figure you can see that he has labels around the edges of the circle. The numbers $1,2,3,4,5,$ and $6$. The terminal point when $t=1$ would correspond to the location labelled with a $1$. The angle at that point is exactly $1$ radian which is about $57$ degrees. By counting the number of tics to get to $1$ you can determine how many radians each tick is worth and then use that to answer questions that involve a fractional number of radians such as $t=2.5$.
answered Jul 2 '13 at 20:41
SpencerSpencer
8,51012056
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add a comment |
$begingroup$
I looked at it a bit and was able to figure it out. First, draw a right triangle with the hypotenuse connecting the outer diameter (point 1) of the circle and the origin. The horizontal leg is the $x$ coordinate and the vertical is the $y$ coordinate of the terminal point. We know that $1$ radian is approximately $57$ degrees, and the radius of the circle is $1$. Knowing this, we can set up the trig ratio :
$sin(x) = dfrac{text{Opposite}}{text{Hypotenuse}}$
"$sin(57) = dfrac{y}{1}$" which can be simplified to "$sin(57) = y$"
If you plug this into a calculator, it will give you the length of the vertical leg, AKA the $y$ coordinate, which is approx $0.8$. Now that you know two of the side lengths of the triangle, you can use the Pythagorean theorem to figure out the third length, which is approx $0.5$. Since the horizontal leg is $0.5$ and the vertical leg is $0.8$, your coordinates will be $(0.5,0.8)$ when $t=1$. You can use this strategy for all the other problems. Hope this helps!
edited Dec 12 '18 at 6:10
Yadati Kiran
1,773619
answered Dec 12 '18 at 5:46
user625566user625566
111
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add a comment |
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2 Answers
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2 Answers
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$begingroup$
Looking at the figure you can see that he has labels around the edges of the circle. The numbers $1,2,3,4,5,$ and $6$. The terminal point when $t=1$ would correspond to the location labelled with a $1$. The angle at that point is exactly $1$ radian which is about $57$ degrees. By counting the number of tics to get to $1$ you can determine how many radians each tick is worth and then use that to answer questions that involve a fractional number of radians such as $t=2.5$.
answered Jul 2 '13 at 20:41
SpencerSpencer
8,51012056
$endgroup$
add a comment |
$begingroup$
Looking at the figure you can see that he has labels around the edges of the circle. The numbers $1,2,3,4,5,$ and $6$. The terminal point when $t=1$ would correspond to the location labelled with a $1$. The angle at that point is exactly $1$ radian which is about $57$ degrees. By counting the number of tics to get to $1$ you can determine how many radians each tick is worth and then use that to answer questions that involve a fractional number of radians such as $t=2.5$.
answered Jul 2 '13 at 20:41
SpencerSpencer
8,51012056
$endgroup$
add a comment |
$begingroup$
Looking at the figure you can see that he has labels around the edges of the circle. The numbers $1,2,3,4,5,$ and $6$. The terminal point when $t=1$ would correspond to the location labelled with a $1$. The angle at that point is exactly $1$ radian which is about $57$ degrees. By counting the number of tics to get to $1$ you can determine how many radians each tick is worth and then use that to answer questions that involve a fractional number of radians such as $t=2.5$.
answered Jul 2 '13 at 20:41
SpencerSpencer
8,51012056
$endgroup$
Looking at the figure you can see that he has labels around the edges of the circle. The numbers $1,2,3,4,5,$ and $6$. The terminal point when $t=1$ would correspond to the location labelled with a $1$. The angle at that point is exactly $1$ radian which is about $57$ degrees. By counting the number of tics to get to $1$ you can determine how many radians each tick is worth and then use that to answer questions that involve a fractional number of radians such as $t=2.5$.
answered Jul 2 '13 at 20:41
SpencerSpencer
8,51012056
answered Jul 2 '13 at 20:41
SpencerSpencer
8,51012056
answered Jul 2 '13 at 20:41
SpencerSpencer
8,51012056
answered Jul 2 '13 at 20:41
SpencerSpencer
8,51012056
8,51012056
add a comment |
add a comment |
$begingroup$
I looked at it a bit and was able to figure it out. First, draw a right triangle with the hypotenuse connecting the outer diameter (point 1) of the circle and the origin. The horizontal leg is the $x$ coordinate and the vertical is the $y$ coordinate of the terminal point. We know that $1$ radian is approximately $57$ degrees, and the radius of the circle is $1$. Knowing this, we can set up the trig ratio :
$sin(x) = dfrac{text{Opposite}}{text{Hypotenuse}}$
"$sin(57) = dfrac{y}{1}$" which can be simplified to "$sin(57) = y$"
If you plug this into a calculator, it will give you the length of the vertical leg, AKA the $y$ coordinate, which is approx $0.8$. Now that you know two of the side lengths of the triangle, you can use the Pythagorean theorem to figure out the third length, which is approx $0.5$. Since the horizontal leg is $0.5$ and the vertical leg is $0.8$, your coordinates will be $(0.5,0.8)$ when $t=1$. You can use this strategy for all the other problems. Hope this helps!
edited Dec 12 '18 at 6:10
Yadati Kiran
1,773619
answered Dec 12 '18 at 5:46
user625566user625566
111
$endgroup$
add a comment |
$begingroup$
I looked at it a bit and was able to figure it out. First, draw a right triangle with the hypotenuse connecting the outer diameter (point 1) of the circle and the origin. The horizontal leg is the $x$ coordinate and the vertical is the $y$ coordinate of the terminal point. We know that $1$ radian is approximately $57$ degrees, and the radius of the circle is $1$. Knowing this, we can set up the trig ratio :
$sin(x) = dfrac{text{Opposite}}{text{Hypotenuse}}$
"$sin(57) = dfrac{y}{1}$" which can be simplified to "$sin(57) = y$"
If you plug this into a calculator, it will give you the length of the vertical leg, AKA the $y$ coordinate, which is approx $0.8$. Now that you know two of the side lengths of the triangle, you can use the Pythagorean theorem to figure out the third length, which is approx $0.5$. Since the horizontal leg is $0.5$ and the vertical leg is $0.8$, your coordinates will be $(0.5,0.8)$ when $t=1$. You can use this strategy for all the other problems. Hope this helps!
edited Dec 12 '18 at 6:10
Yadati Kiran
1,773619
answered Dec 12 '18 at 5:46
user625566user625566
111
$endgroup$
add a comment |
$begingroup$
I looked at it a bit and was able to figure it out. First, draw a right triangle with the hypotenuse connecting the outer diameter (point 1) of the circle and the origin. The horizontal leg is the $x$ coordinate and the vertical is the $y$ coordinate of the terminal point. We know that $1$ radian is approximately $57$ degrees, and the radius of the circle is $1$. Knowing this, we can set up the trig ratio :
$sin(x) = dfrac{text{Opposite}}{text{Hypotenuse}}$
"$sin(57) = dfrac{y}{1}$" which can be simplified to "$sin(57) = y$"
If you plug this into a calculator, it will give you the length of the vertical leg, AKA the $y$ coordinate, which is approx $0.8$. Now that you know two of the side lengths of the triangle, you can use the Pythagorean theorem to figure out the third length, which is approx $0.5$. Since the horizontal leg is $0.5$ and the vertical leg is $0.8$, your coordinates will be $(0.5,0.8)$ when $t=1$. You can use this strategy for all the other problems. Hope this helps!
edited Dec 12 '18 at 6:10
Yadati Kiran
1,773619
answered Dec 12 '18 at 5:46
user625566user625566
111
$endgroup$
I looked at it a bit and was able to figure it out. First, draw a right triangle with the hypotenuse connecting the outer diameter (point 1) of the circle and the origin. The horizontal leg is the $x$ coordinate and the vertical is the $y$ coordinate of the terminal point. We know that $1$ radian is approximately $57$ degrees, and the radius of the circle is $1$. Knowing this, we can set up the trig ratio :
$sin(x) = dfrac{text{Opposite}}{text{Hypotenuse}}$
"$sin(57) = dfrac{y}{1}$" which can be simplified to "$sin(57) = y$"
If you plug this into a calculator, it will give you the length of the vertical leg, AKA the $y$ coordinate, which is approx $0.8$. Now that you know two of the side lengths of the triangle, you can use the Pythagorean theorem to figure out the third length, which is approx $0.5$. Since the horizontal leg is $0.5$ and the vertical leg is $0.8$, your coordinates will be $(0.5,0.8)$ when $t=1$. You can use this strategy for all the other problems. Hope this helps!
edited Dec 12 '18 at 6:10
Yadati Kiran
1,773619
answered Dec 12 '18 at 5:46
user625566user625566
111
edited Dec 12 '18 at 6:10
Yadati Kiran
1,773619
edited Dec 12 '18 at 6:10
Yadati Kiran
1,773619
edited Dec 12 '18 at 6:10
Yadati Kiran
1,773619
1,773619
answered Dec 12 '18 at 5:46
user625566user625566
111
answered Dec 12 '18 at 5:46
user625566user625566
111
answered Dec 12 '18 at 5:46
user625566user625566
111
111
add a comment |
add a comment |
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