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What are the polar coordinates of $(2sqrt3, 2)$?
The Next CEO of Stack OverflowWhat are the polar coordinates of the origin?seemingly simple question about polar coordinatesLimit using polar coordinates?Setup region of integration for polar coordinatesHow can $r$ be negative when dealing with polar coordinates?Triple Integration: from Cartesian to Polar CoordinatesWhat are the characteristics of functions that look the same in both polar and rectangular graph?Solve differential equation by using polar coordinatesCalculating a nasty integral with polar coordinatesUnderstanding polar coordinates
$begingroup$
My answer to this is $(4,fracπ6)$. But a calculator said that $(-4,frac7π6)$ is also an answer, and there are infinitely many solutions. Is that correct?
algebra-precalculus trigonometry polar-coordinates
asked Mar 19 at 22:23
harpey1111harpey1111
707
$endgroup$
add a comment |
$begingroup$
My answer to this is $(4,fracπ6)$. But a calculator said that $(-4,frac7π6)$ is also an answer, and there are infinitely many solutions. Is that correct?
algebra-precalculus trigonometry polar-coordinates
asked Mar 19 at 22:23
harpey1111harpey1111
707
$endgroup$
add a comment |
$begingroup$
My answer to this is $(4,fracπ6)$. But a calculator said that $(-4,frac7π6)$ is also an answer, and there are infinitely many solutions. Is that correct?
algebra-precalculus trigonometry polar-coordinates
asked Mar 19 at 22:23
harpey1111harpey1111
707
$endgroup$
My answer to this is $(4,fracπ6)$. But a calculator said that $(-4,frac7π6)$ is also an answer, and there are infinitely many solutions. Is that correct?
algebra-precalculus trigonometry polar-coordinates
algebra-precalculus trigonometry polar-coordinates
asked Mar 19 at 22:23
harpey1111harpey1111
707
asked Mar 19 at 22:23
harpey1111harpey1111
707
asked Mar 19 at 22:23
harpey1111harpey1111
707
asked Mar 19 at 22:23
harpey1111harpey1111
707
asked Mar 19 at 22:23
harpey1111harpey1111
707
707
add a comment |
add a comment |
1 Answer
1
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$begingroup$
Your solution is correct. Now, concerning $left(-4,frac7pi6right)$, it is indeed true that$$-4left(cosleft(frac7pi6right),sinleft(frac7pi6right)right)=left(2sqrt3,2right),$$but $-4<0$, and therefore that answer is wrong. And your answer is the only one which uses an angle from $[0,2pi)$.
answered Mar 19 at 22:30
José Carlos SantosJosé Carlos Santos
171k23132240
$endgroup$
$begingroup$
$7pi/6 in [0, 2pi)$
$endgroup$
– ErotemeObelus
Mar 19 at 22:49
$begingroup$
But $-4$ is negative.
$endgroup$
– José Carlos Santos
Mar 19 at 22:51
$begingroup$
@harpey By convention the first number in the polar form is a radius $rge 0$, it is positive since it is a distance from the origin. Even if $-4exp(i7pi/6)=4exp(ipi/6)$ the radius $r$ is still $4$ and only the rhs is the "true" polar form. The same for angles, they need to be in $[0,2pi)$. Even though adding multiple of $2pi$ doesn't changes the result, the polar form is the one with the angle in the conventional interval.
$endgroup$
– zwim
Mar 19 at 23:02
add a comment |
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1 Answer
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1 Answer
1
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oldest
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$begingroup$
Your solution is correct. Now, concerning $left(-4,frac7pi6right)$, it is indeed true that$$-4left(cosleft(frac7pi6right),sinleft(frac7pi6right)right)=left(2sqrt3,2right),$$but $-4<0$, and therefore that answer is wrong. And your answer is the only one which uses an angle from $[0,2pi)$.
answered Mar 19 at 22:30
José Carlos SantosJosé Carlos Santos
171k23132240
$endgroup$
$begingroup$
$7pi/6 in [0, 2pi)$
$endgroup$
– ErotemeObelus
Mar 19 at 22:49
$begingroup$
But $-4$ is negative.
$endgroup$
– José Carlos Santos
Mar 19 at 22:51
$begingroup$
@harpey By convention the first number in the polar form is a radius $rge 0$, it is positive since it is a distance from the origin. Even if $-4exp(i7pi/6)=4exp(ipi/6)$ the radius $r$ is still $4$ and only the rhs is the "true" polar form. The same for angles, they need to be in $[0,2pi)$. Even though adding multiple of $2pi$ doesn't changes the result, the polar form is the one with the angle in the conventional interval.
$endgroup$
– zwim
Mar 19 at 23:02
add a comment |
$begingroup$
Your solution is correct. Now, concerning $left(-4,frac7pi6right)$, it is indeed true that$$-4left(cosleft(frac7pi6right),sinleft(frac7pi6right)right)=left(2sqrt3,2right),$$but $-4<0$, and therefore that answer is wrong. And your answer is the only one which uses an angle from $[0,2pi)$.
answered Mar 19 at 22:30
José Carlos SantosJosé Carlos Santos
171k23132240
$endgroup$
$begingroup$
$7pi/6 in [0, 2pi)$
$endgroup$
– ErotemeObelus
Mar 19 at 22:49
$begingroup$
But $-4$ is negative.
$endgroup$
– José Carlos Santos
Mar 19 at 22:51
$begingroup$
@harpey By convention the first number in the polar form is a radius $rge 0$, it is positive since it is a distance from the origin. Even if $-4exp(i7pi/6)=4exp(ipi/6)$ the radius $r$ is still $4$ and only the rhs is the "true" polar form. The same for angles, they need to be in $[0,2pi)$. Even though adding multiple of $2pi$ doesn't changes the result, the polar form is the one with the angle in the conventional interval.
$endgroup$
– zwim
Mar 19 at 23:02
add a comment |
$begingroup$
Your solution is correct. Now, concerning $left(-4,frac7pi6right)$, it is indeed true that$$-4left(cosleft(frac7pi6right),sinleft(frac7pi6right)right)=left(2sqrt3,2right),$$but $-4<0$, and therefore that answer is wrong. And your answer is the only one which uses an angle from $[0,2pi)$.
answered Mar 19 at 22:30
José Carlos SantosJosé Carlos Santos
171k23132240
$endgroup$
Your solution is correct. Now, concerning $left(-4,frac7pi6right)$, it is indeed true that$$-4left(cosleft(frac7pi6right),sinleft(frac7pi6right)right)=left(2sqrt3,2right),$$but $-4<0$, and therefore that answer is wrong. And your answer is the only one which uses an angle from $[0,2pi)$.
answered Mar 19 at 22:30
José Carlos SantosJosé Carlos Santos
171k23132240
answered Mar 19 at 22:30
José Carlos SantosJosé Carlos Santos
171k23132240
answered Mar 19 at 22:30
José Carlos SantosJosé Carlos Santos
171k23132240
answered Mar 19 at 22:30
José Carlos SantosJosé Carlos Santos
171k23132240
171k23132240
$begingroup$
$7pi/6 in [0, 2pi)$
$endgroup$
– ErotemeObelus
Mar 19 at 22:49
$begingroup$
But $-4$ is negative.
$endgroup$
– José Carlos Santos
Mar 19 at 22:51
$begingroup$
@harpey By convention the first number in the polar form is a radius $rge 0$, it is positive since it is a distance from the origin. Even if $-4exp(i7pi/6)=4exp(ipi/6)$ the radius $r$ is still $4$ and only the rhs is the "true" polar form. The same for angles, they need to be in $[0,2pi)$. Even though adding multiple of $2pi$ doesn't changes the result, the polar form is the one with the angle in the conventional interval.
$endgroup$
– zwim
Mar 19 at 23:02
add a comment |
$begingroup$
$7pi/6 in [0, 2pi)$
$endgroup$
– ErotemeObelus
Mar 19 at 22:49
$begingroup$
But $-4$ is negative.
$endgroup$
– José Carlos Santos
Mar 19 at 22:51
$begingroup$
@harpey By convention the first number in the polar form is a radius $rge 0$, it is positive since it is a distance from the origin. Even if $-4exp(i7pi/6)=4exp(ipi/6)$ the radius $r$ is still $4$ and only the rhs is the "true" polar form. The same for angles, they need to be in $[0,2pi)$. Even though adding multiple of $2pi$ doesn't changes the result, the polar form is the one with the angle in the conventional interval.
$endgroup$
– zwim
Mar 19 at 23:02
$begingroup$
$7pi/6 in [0, 2pi)$
$endgroup$
– ErotemeObelus
Mar 19 at 22:49
$begingroup$
$7pi/6 in [0, 2pi)$
$endgroup$
– ErotemeObelus
Mar 19 at 22:49
$begingroup$
But $-4$ is negative.
$endgroup$
– José Carlos Santos
Mar 19 at 22:51
$begingroup$
But $-4$ is negative.
$endgroup$
– José Carlos Santos
Mar 19 at 22:51
$begingroup$
@harpey By convention the first number in the polar form is a radius $rge 0$, it is positive since it is a distance from the origin. Even if $-4exp(i7pi/6)=4exp(ipi/6)$ the radius $r$ is still $4$ and only the rhs is the "true" polar form. The same for angles, they need to be in $[0,2pi)$. Even though adding multiple of $2pi$ doesn't changes the result, the polar form is the one with the angle in the conventional interval.
$endgroup$
– zwim
Mar 19 at 23:02
$begingroup$
@harpey By convention the first number in the polar form is a radius $rge 0$, it is positive since it is a distance from the origin. Even if $-4exp(i7pi/6)=4exp(ipi/6)$ the radius $r$ is still $4$ and only the rhs is the "true" polar form. The same for angles, they need to be in $[0,2pi)$. Even though adding multiple of $2pi$ doesn't changes the result, the polar form is the one with the angle in the conventional interval.
$endgroup$
– zwim
Mar 19 at 23:02
add a comment |
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