Hello

Thank you Henrik for your answer. Now it is perfectly working. I am wondering why the resolution of the tilted coordinate system is much lower than the other one. Should't Mathematica be able to rotate an image without decreasing the resolution? Maybe this is the advantage of Ottos solution.

Regards,

Peter

Hi Peter,

sorry for this issue! I still do not fully understand how this could have happened - obviously there was a change from version 9.0 (where I developed my first "solution") to version 10.2 (where I can now verify the problem). Anyway, here is the code which should run for version 10.2; it is even somewhat simpler:

img0 = Plot[0, {x, -10, 10}, PlotRange -> {{-10, 10}, {-5, 5}}, 
   ImageSize -> 500];
img1 = ImageRotate[
   Plot[3 + Sin[x], {x, -10, 10}, PlotRange -> {-5, 5}, 
    PlotStyle -> Green, ImageSize -> 500], -10 \[Degree]];
ImageCompose[img0, img1, {251, 150}]

Regards -- Henrik

Hi Peter,

here comes a crude approach, but maybe this goes in the right direction:

img0 = Image@
   RotateLeft[ (* `RotateLeft` is for fine tuning ... *)
    ImageData@
     ColorConvert[
      Plot[0, {x, -10, 10}, PlotRange -> {{-10, 10}, {-4, 4}}, 
       ImageSize -> 500], "Grayscale"], {0, 2}];

img1 = ImageRotate[
   Plot[3 + Sin[x], {x, -10, 10}, PlotRange -> {-4, 4}, 
    PlotStyle -> Green, ImageSize -> 500], -10 \[Degree], 
   Background -> White];

ImageMultiply[img1, img0]

which gives:

Regards -- Henrik

Hello Peter,

Not sure whether this is what you are looking for. At least I hope it will provide a starting point:

ParametricPlot[
 Evaluate[Dot[RotationMatrix[\[Pi]/12], {x, Cos[x]}]], {x, -2*\[Pi], 
  2*\[Pi]}]

Not sure how to paste the image but it looks like a rotated cosine graph. Notice that the default AspectRatio for ParametricPlot is Automatic, meaning the horizontal and vertical scales are equal. This is different than the default value in Plot. This may become important if you want to display them together.

OL.