Co-ordinate plane rotation

If I have a point (x,y) and I rotate the axises by some amount. Why is x' = xcosθ+ysinθ and y'=-xsinθ+ycosθ?
 Look at it this way. Suppose (x,y) has polar coords R, phi. Then the rotated point has polar coords R, phi + theta. So the new rectangular coords should be x' = Rcos(phi+theta) y'= Rsin(phi+theta) Now use the angle addition formulas for cosine and sine and use the fact that x=Rcos(phi) y=Rsin(phi)
 Angle addition formulas for what? I don't understand you fully :(

Co-ordinate plane rotation

I understand that x' = rcos(theta+phi) and y'= rsin(theta+phi) and that x=rcos theta and y= rsin theta.
How do I use this to get x'= xcos theta + ysin theta
And y= -xsin theta + ycos theta
 The last sentence is actually y'. Sorry
 cos (α + β) = cos α cos β − sin α sin β sin (α + β) = sin α cos β + cos α sin β Using this, I can write, x'=rcos theta cos phi - sin theta sin phi and y'= rsin theta cos phi + cos theta sin phi and x= rcos phi y= rsin phi Now what do I do?
 Now substitute x=rcos(phi) and y=rsin(phi) into those expressions. You will get formulas that are almost the same as what you started with. The difference is due to the following: The formula you derived answers the following. Given a counter clockwise rotation of the point, what are its new coordinates. Your original question was this: If we rotate the coordinate axes, what are the new coordinates with respect to the rotated axes. To answer this we have to realize that rotating the coordinate axes counterclockwise is equivalent to rotating the points clockwise. So your new coordinates will be the coordinates you get after rotating your point clockwise. But the formula you derived is valid for counterclockwise rotations. To convert it, you must substitute -theta in place of theta. Then you will get the formula that you first asked about.