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clockwise_rotation_diagram

Given a rotation $θ$ and a radius $r,$ how do I find the coordinate $(x,y)$?

I saw an answer like this:

From the picture, it seems that your circle has centre the origin, and radius $r.$ The rotation appears to be clockwise. And the question appears to be about where the point $(0,r)$ at the top of the circle ends up.

The point $(0,r)$ ends up at $x=r\sin θ, y=r\cos θ.$

In general, suppose that you are rotating about the origin clockwise through an angle $θ.$ Then the point $(s,t)$ ends up at $(u,v)$ where $u=s\cos θ+t\sin θ$ and $v=−s\sin θ+t\cos θ.$

can someone please explain to me what happened? I'm confused :(

Thanks a lot!

N. F. Taussig
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Kash Costello
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    "I saw an answer ... ." Where? Did you try asking the person who gave the answer? Which parts of the answer do you understand, and which do not make sense? Why don't they make sense? Do they seem to contradict other parts? What's the contradiction? If you edit your question so that people can see what you actually need, you might get better answers. – David K Oct 16 '18 at 20:30
  • Hi. The answer I saw was back in 2012. I don't think the person would reply in a few days or, ever, if I ask him/her about it which is why I decided to ask the same question. I just want confirmation on how it happened. I think my question was pretty clear? Like, how did they arrive at the answer? I am just confused so I need the explanation on this part. – Kash Costello Oct 16 '18 at 20:37
  • "How did they arrive at the answer?" is a much different question than "What happened?" What happened is simple, you started with $(s,t)$ and you applied a formula to get $(u,v).$ Hopefully the answer you got is satisfactory. – David K Oct 16 '18 at 21:40

1 Answers1

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We need simply to apply the definition of $\sin \theta$ and $\cos \theta$ and note that

  • $x=r\cos \left(\theta-\frac{\pi}2\right)=r\cos \left(\frac{\pi}2-\theta\right)=r\sin\theta$

  • $y=-r\sin\left(\theta-\frac{\pi}2\right)=r\sin \left(\frac{\pi}2-\theta\right)=r\cos\theta$

enter image description here

For the second part note that fro a clockwise rotation of $\theta$

  • $(0,1)$ goes to $(\sin \theta, \cos \theta)$

  • $(1,0)$ goes to $(\cos \theta,-\sin \theta)$

enter image description here

then a generic point

  • $(s,t)=s(1,0)+t(0,1)$

goes to

  • $(u,v)=s(\cos \theta,-\sin \theta)+t(\sin \theta, \cos \theta)=(s\cos \theta+t\sin \theta,-s\sin \theta+t\cos \theta)$
user
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  • Yes, I got it! Thank you so much! But what about the u and v part (last part of the question)???? How do you add them? – Kash Costello Oct 16 '18 at 20:35
  • @KashCostello We have that $(1,0)$ goes to $(\cos \theta,-\sin \theta)$ and $(0,1)$ goes to $(\sin \theta,\cos \theta)$ then note that $(s,t)=s(1,0)+t(0,1)$. – user Oct 16 '18 at 20:40
  • Hi, ok I'm being an idiot and stupid... can you explain further? – Kash Costello Oct 16 '18 at 20:43
  • @KashCostello No problem I want to clarify as I can. In the given answer we have determined that $(0,r)$ goes to $(r\sin \theta,r\cos \theta)$ then $(0,1)$ goes to $(\sin \theta,\cos \theta)$ and $(0,t)$ goes to $(t\sin \theta,t\cos \theta)$. Can you see that? – user Oct 16 '18 at 20:47
  • Yes I know that: (0,r) --> (rsinθ,rcosθ) AND (0,1) --> because r is 1 so it's just (sinθ,cosθ) AND (0,t) --> (tsinθ,tcosθ) because t became r. Did I understand correctly? I am curious about the (1,0) thing how it became (cosθ,−sinθ) – Kash Costello Oct 16 '18 at 20:53
  • @KashCostello For $(1,0)$ we can use the same method, just consider the same picture in the answer and take $\alpha=\theta$. – user Oct 16 '18 at 21:02
  • Hi, I still don't understand :(( I'm so sorry – Kash Costello Oct 16 '18 at 21:12
  • @KashCostello I revise the answer to clarify better what I mean. – user Oct 16 '18 at 21:13
  • Which answer are you talking about? I really appreciate your patience! I'm so sorry if I am still confused. Also, when you say (0,r) --> what does this mean? what is 0 to begin with? I know that the second one is r. What is the first one? – Kash Costello Oct 16 '18 at 21:17
  • like I want to know how this happened? --> u=scosθ+tsinθ and v=−ssinθ+tcosθ – Kash Costello Oct 16 '18 at 21:20
  • Last question and I promise I'm done. (scosθ+tsinθ,−ssinθ+tcosθ) --> when you say generic, this means that it's always like this? like a default? again, i want to thank you so much for all your help and patience!! T_T You rock! – Kash Costello Oct 16 '18 at 21:30
  • @KashCostello The key fact is that the rotation around the origin is a linear transformation, therefore if we know where $(1,0)$ and $(0,1)$ end up we can find where any generic point $(s,t)$ ends up as we have shown, Yes of course it is valid for any point of the plane (for a clockwise rotation of $\theta$). – user Oct 16 '18 at 21:34
  • got it! For a Counter clockwise rotation, will a generic point (s,t)=s(1,0)+t(0,1) go to --> (u,v) = (scosθ+tsinθ, ssinθ+tcosθ) ??? Is this right? – Kash Costello Oct 16 '18 at 21:37
  • @KashCostello Almost, it goes to $(s\cos \theta-t\sin \theta,s\sin \theta+t\cos \theta)$. Note that it suffices take $-\theta$ to obtain that from the previous one. – user Oct 16 '18 at 21:40
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    Ah yes yes, negative since it's in the 2nd quadrant. got it. omg I can't thank you enough for your patience and assistance. I owe you big time! Have a good day/evening!! <3 – Kash Costello Oct 16 '18 at 21:42
  • @KashCostello You are welcome! It's a plesure for me if I may help you in some way. Bye – user Oct 16 '18 at 21:46