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Converting Polar Coordinates to Regular Coordinates

0

1

$begingroup$

If you guys didn't know, I have my quiz tomorrow and I have one last thing to ask to this Community! I am completely confused on how to convert polar coordinates to regular coordinates. The teacher gave us this example:

Convert to Polar Coordinates:
$(3 , -45^circ)$

$x = rcos(theta) = 2cos(-45^circ) = 3(frac{sqrt{2}}{2})$

$y = rcos(theta) = 3sin(-45 ^circ) = 3(frac{-sqrt{2}}{2})$

$(frac{3sqrt{2}}{2} , frac{-3sqrt{2}}{2})$

Ok, she did that and gave us this (one of of the two) for a review:

$(6 , frac{-2pi}{3})$

Well then I had completely no idea..(I know the equation though)

I did this:

$x = rcos(theta)$
$x = rsin(theta)$

I basically didn't know what to put at the coefficient of $cos$ and $sin$. please help. Thanks a lot for reading!

trigonometry polar-coordinates

share|cite|improve this question

edited May 13 '16 at 0:18

N. F. Taussig

45.4k103358

asked May 13 '16 at 0:11

amanuel2amanuel2

271111

$endgroup$

• 
  
  
  

  
  3
  

  
  
  
  
  
  

  
  $begingroup$
  You've copied the example down incorrectly (either here or in your notes). You should have: $$x = rcos(theta) = color{red}3cos(-45^circ) = 3(frac{sqrt{2}}{2})$$
  $endgroup$
  – Ian Miller
  May 13 '16 at 0:21
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Guess i wrote it incorrectly in my notes... wow @IanMiller
  $endgroup$
  – amanuel2
  May 13 '16 at 0:22
  

  
  
  
  


• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  $begingroup$
  The example you gave is a conversion from polar coordinates to rectangular coordinates. The polar coordinates are $(r, theta) = (3, -45^circ)$. The rectangular coordinates are $(x, y) = (frac{3sqrt{2}}{2}, -frac{3sqrt{2}}{2})$. Your formula for converting from polar coordinates to rectangular coordinates should read $x = rcostheta$, $color{red}{y} = rsintheta$. The coefficient for $costheta$ and $sintheta$ is the value of $r$. Does this help?
  $endgroup$
  – N. F. Taussig
  May 13 '16 at 0:23
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Yes @N.F.Taussig It does thanks a lot, for you guys helping me through this! I got this concept!
  $endgroup$
  – amanuel2
  May 13 '16 at 0:26
  

  
  
  
  

add a comment | 

0

1

$begingroup$

If you guys didn't know, I have my quiz tomorrow and I have one last thing to ask to this Community! I am completely confused on how to convert polar coordinates to regular coordinates. The teacher gave us this example:

Convert to Polar Coordinates:
$(3 , -45^circ)$

$x = rcos(theta) = 2cos(-45^circ) = 3(frac{sqrt{2}}{2})$

$y = rcos(theta) = 3sin(-45 ^circ) = 3(frac{-sqrt{2}}{2})$

$(frac{3sqrt{2}}{2} , frac{-3sqrt{2}}{2})$

Ok, she did that and gave us this (one of of the two) for a review:

$(6 , frac{-2pi}{3})$

Well then I had completely no idea..(I know the equation though)

I did this:

$x = rcos(theta)$
$x = rsin(theta)$

I basically didn't know what to put at the coefficient of $cos$ and $sin$. please help. Thanks a lot for reading!

trigonometry polar-coordinates

share|cite|improve this question

edited May 13 '16 at 0:18

N. F. Taussig

45.4k103358

asked May 13 '16 at 0:11

amanuel2amanuel2

271111

$endgroup$

• 
  
  
  

  
  3
  

  
  
  
  
  
  

  
  $begingroup$
  You've copied the example down incorrectly (either here or in your notes). You should have: $$x = rcos(theta) = color{red}3cos(-45^circ) = 3(frac{sqrt{2}}{2})$$
  $endgroup$
  – Ian Miller
  May 13 '16 at 0:21
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Guess i wrote it incorrectly in my notes... wow @IanMiller
  $endgroup$
  – amanuel2
  May 13 '16 at 0:22
  

  
  
  
  


• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  $begingroup$
  The example you gave is a conversion from polar coordinates to rectangular coordinates. The polar coordinates are $(r, theta) = (3, -45^circ)$. The rectangular coordinates are $(x, y) = (frac{3sqrt{2}}{2}, -frac{3sqrt{2}}{2})$. Your formula for converting from polar coordinates to rectangular coordinates should read $x = rcostheta$, $color{red}{y} = rsintheta$. The coefficient for $costheta$ and $sintheta$ is the value of $r$. Does this help?
  $endgroup$
  – N. F. Taussig
  May 13 '16 at 0:23
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Yes @N.F.Taussig It does thanks a lot, for you guys helping me through this! I got this concept!
  $endgroup$
  – amanuel2
  May 13 '16 at 0:26
  

  
  
  
  

add a comment | 

$begingroup$

If you guys didn't know, I have my quiz tomorrow and I have one last thing to ask to this Community! I am completely confused on how to convert polar coordinates to regular coordinates. The teacher gave us this example:

Convert to Polar Coordinates:
$(3 , -45^circ)$

$x = rcos(theta) = 2cos(-45^circ) = 3(frac{sqrt{2}}{2})$

$y = rcos(theta) = 3sin(-45 ^circ) = 3(frac{-sqrt{2}}{2})$

$(frac{3sqrt{2}}{2} , frac{-3sqrt{2}}{2})$

Ok, she did that and gave us this (one of of the two) for a review:

$(6 , frac{-2pi}{3})$

Well then I had completely no idea..(I know the equation though)

I did this:

$x = rcos(theta)$
$x = rsin(theta)$

I basically didn't know what to put at the coefficient of $cos$ and $sin$. please help. Thanks a lot for reading!

trigonometry polar-coordinates

share|cite|improve this question

edited May 13 '16 at 0:18

N. F. Taussig

45.4k103358

asked May 13 '16 at 0:11

amanuel2amanuel2

271111

$endgroup$

share|cite|improve this question

edited May 13 '16 at 0:18

N. F. Taussig

45.4k103358

asked May 13 '16 at 0:11

amanuel2amanuel2

271111

share|cite|improve this question

edited May 13 '16 at 0:18

N. F. Taussig

45.4k103358

asked May 13 '16 at 0:11

amanuel2amanuel2

271111

edited May 13 '16 at 0:18

N. F. Taussig

45.4k103358

edited May 13 '16 at 0:18

N. F. Taussig

45.4k103358

asked May 13 '16 at 0:11

amanuel2amanuel2

271111

asked May 13 '16 at 0:11

amanuel2amanuel2

271111

1 Answer
1

active

oldest

votes

3

$begingroup$

When writing coordinates in polar notation you've written them as $(r,theta)$. So these are the values you should stick into your formulae for $x$ and $y$.

The transformation from polar to rectangular can be seen as:

$$(r,theta)rightarrow(rcostheta,rsintheta)$$

So for your example you get:

$$left(6,-frac{2pi}{3}right)rightarrowleft(6cosleft(-frac{2pi}{3}right),6sinleft(-frac{2pi}{3}right)right)$$

You can then use your knowledge of exact values to get:

$$(-3,-3sqrt{3})$$

share|cite|improve this answer

answered May 13 '16 at 0:18

Ian MillerIan Miller

10.5k11439

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Wow that way is much easier than my teacher's way... Ian thanks a lot sir! Thanks a lot for helping me through guys @N.F.Taussig , Ian
  $endgroup$
  – amanuel2
  May 13 '16 at 0:25
  
  
  

  
  
  
  

add a comment | 

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3

$begingroup$

When writing coordinates in polar notation you've written them as $(r,theta)$. So these are the values you should stick into your formulae for $x$ and $y$.

The transformation from polar to rectangular can be seen as:

$$(r,theta)rightarrow(rcostheta,rsintheta)$$

So for your example you get:

$$left(6,-frac{2pi}{3}right)rightarrowleft(6cosleft(-frac{2pi}{3}right),6sinleft(-frac{2pi}{3}right)right)$$

You can then use your knowledge of exact values to get:

$$(-3,-3sqrt{3})$$

share|cite|improve this answer

answered May 13 '16 at 0:18

Ian MillerIan Miller

10.5k11439

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Wow that way is much easier than my teacher's way... Ian thanks a lot sir! Thanks a lot for helping me through guys @N.F.Taussig , Ian
  $endgroup$
  – amanuel2
  May 13 '16 at 0:25
  
  
  

  
  
  
  

add a comment | 

3

$begingroup$

When writing coordinates in polar notation you've written them as $(r,theta)$. So these are the values you should stick into your formulae for $x$ and $y$.

The transformation from polar to rectangular can be seen as:

$$(r,theta)rightarrow(rcostheta,rsintheta)$$

So for your example you get:

$$left(6,-frac{2pi}{3}right)rightarrowleft(6cosleft(-frac{2pi}{3}right),6sinleft(-frac{2pi}{3}right)right)$$

You can then use your knowledge of exact values to get:

$$(-3,-3sqrt{3})$$

share|cite|improve this answer

answered May 13 '16 at 0:18

Ian MillerIan Miller

10.5k11439

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Wow that way is much easier than my teacher's way... Ian thanks a lot sir! Thanks a lot for helping me through guys @N.F.Taussig , Ian
  $endgroup$
  – amanuel2
  May 13 '16 at 0:25
  
  
  

  
  
  
  

add a comment | 

$begingroup$

When writing coordinates in polar notation you've written them as $(r,theta)$. So these are the values you should stick into your formulae for $x$ and $y$.

The transformation from polar to rectangular can be seen as:

$$(r,theta)rightarrow(rcostheta,rsintheta)$$

So for your example you get:

$$left(6,-frac{2pi}{3}right)rightarrowleft(6cosleft(-frac{2pi}{3}right),6sinleft(-frac{2pi}{3}right)right)$$

You can then use your knowledge of exact values to get:

$$(-3,-3sqrt{3})$$

share|cite|improve this answer

answered May 13 '16 at 0:18

Ian MillerIan Miller

10.5k11439

$endgroup$

share|cite|improve this answer

answered May 13 '16 at 0:18

Ian MillerIan Miller

10.5k11439

answered May 13 '16 at 0:18

Ian MillerIan Miller

10.5k11439

answered May 13 '16 at 0:18

Ian MillerIan Miller

10.5k11439