# New coordinates from rotation of axis

1. May 3, 2015

### rajeshmarndi

1. The problem statement, all variables and given/known data
There is a point P(x,y) and now I rotate the x-y axis, say by θ degree. What will be the coordinates of P from this new axis.

I have google but found formula for new coordinates when the points is rotated by θ degree. So I tried my own. So is there other simplified formula for the above situation.

2. Relevant equations

3. The attempt at a solution
Plz see the attached figure.
AC-AD is the new axis and (x',y') are the new coordinates of point P.

y'=PC
cos θ= PC/PB = y'/PB
y'= PB cos θ

PB= y- BE
tan θ= BE/x , BE = x tanθ
PB = y - xtanθ
y'= (y- xtanθ) cosθ
= ycosθ - xsinθ --------------eq(1)

x'= AB + BC
sinθ=BE/AB
AB=BE/sinθ, tanθ=BE/x
BE=xtanθ
AB=xtanθ/sinθ = x/cosθ

tanθ = BC/y'
BC= y'tanθ
= (ycosθ-xsinθ)tanθ , ( y' from eq(1) )
= ysinθ - x sinθtanθ
x' = AB + BC
= x/cosθ + ysinθ - x sinθtanθ

So the new coordinates are
x'= x/cosθ + ysinθ - x sinθtanθ
y'= ycosθ - xsinθ

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2. May 3, 2015

### Orodruin

Staff Emeritus
Can you think of a way of rewriting x(1/cosθ - sinθ tanθ) to a form which is slightly more pleasant to look at? Otherwise this formula is correct.

3. May 3, 2015

### rajeshmarndi

Thanks.
x'= x/cosθ + ysinθ - x sinθtanθ
x'= x/cosθ - xsin2θ/cosθ + ysinθ
x'= x(1-sin2θ)/cosθ) + ysinθ
x'= x(cos2θ/cosθ) + ysinθ
x'= xcosθ + ysinθ

Now it is pleasant to look at.

This looks very much similar to, when a points is rotated by θ. The new coordinates are,

x' = x cos θ - y sin θ
y' = y cos θ + x sin θ

Only difference is the -/+ sign. Seems like the difference is because as we increase the θ(when a points is rotated), x' gets shorter. Thats just a guess with a first look.

4. May 3, 2015

### Orodruin

Staff Emeritus
Yes, the sign depends on the active-vs-passive transformation, i.e., if you rotate the points or the coordinate system.

5. May 3, 2015

### Ray Vickson

If you fix the point and rotate the coordinate system, the new coordinates $(x',y')$ are given by
$$x' = \cos(\theta) x + \sin(\theta) y \\ y' = -\sin(\theta) x + \cos(\theta) y$$
If you fix the coordinate system and rotate the point, the new coordinates $(x',y')$ are given by
$$x' = \cos(\theta) x - \sin(\theta) y\\ y' = \sin(\theta) x + \cos(\theta) y$$