Proof: Coordinate Rotation Around (0,0)

[Maybe_Memorie]

Homework Statement

Prove that the coordinates of the point (x',y') where the counter-clockwise rotation through the angle @ around (0,0) brings the given point (x,y) are

x' = xcos@ - ysin@
y' = xsin@ + ycos@

Hint: show that for the points (x,y) = (1,0) and (x,y) = (0,1) directly,
and use the fact that the vector (x,y) is equal to the combination
x.(1,0) + y.(0,1)

Homework Equations

For vectors u and v, angle @ between them
u.v = |u||v|Cos@

The Attempt at a Solution

I don't want to be told how to do it, I would prefer if someone would kind of tease the solution out of me, if you know what i mean..

I've included a diagram, showing my interpretation of the question.

I've tried a few different approaches for the question.
I used the fact that tan@ = (m1 - m2)/(1 +m1m2).
I got the slopes of the lines being y/x and y'/x'. When I plugged everything in and rewrote tan as sin/cos, I got the required formulae, but they were both being divided by each other.

I also used the dot product, put this just resulted with a lot of squares which doesn't help.

I don't entirely understand the hint also.

 

[gerben]

Use the hint that was given.

First draw a rectangle with corners (0,0) and (x,y) (with sides parallel to x- and y-axis) and then draw the rectangle that you get when the whole plane is rotated by an angle @.

Then you can see in your drawing that the new x' and y' are just sums of the height and width of the rectangle multiplied by sines and/or cosines of @

 

[Maybe_Memorie]

I don't really understand what the hint means.

Okay, I'll try that.

 

[JThompson]

The hint is basically saying that if your starting vectors are either

[tex]\left(\begin{array}{c}1\\0\end{array}\right)\mbox{ or }\left(\begin{array}{c}0\\1\end{array}\right)[/tex]

then it's pretty easy to figure out what the new coordinates will be if you rotate it through an angle [tex]\theta[/tex].

You'll need the sum of angles formulas for cos and sin, [tex]\cos(\theta_{0}+\theta)[/tex] and [tex]\sin(\theta_{0}+\theta)[/tex].
([tex]\theta_{0}[/tex] is the initial trig angle for the vector).

Does this make sense?

 

[Maybe_Memorie]

It makes sense, but I have no idea where to start.

 

[gerben]

You original vector is:

[tex]v = a\cdot \left(\begin{array}{c}1\\0\end{array}\right)
+ b\cdot \left(\begin{array}{c}0\\1\end{array}\right)[/tex]

so the rotated vector will be:

[tex]rot_\theta(v) = a \cdot rot_\theta (\left(\begin{array}{c}1\\0\end{array}\right) )
+ b\cdot rot_\theta ( \left(\begin{array}{c}0\\1\end{array}\right) )[/tex]

You can first determine c and d in:

[tex] rot_\theta (\left(\begin{array}{c}1\\0\end{array}\right) ) =
c\cdot \left(\begin{array}{c}1\\0\end{array}\right)
+
d\cdot \left(\begin{array}{c}0\\1\end{array}\right)[/tex]

and e and f in:

[tex]rot_\theta (\left(\begin{array}{c}0\\1\end{array}\right) ) =
e\cdot \left(\begin{array}{c}1\\0\end{array}\right)
+
f\cdot \left(\begin{array}{c}0\\1\end{array}\right)[/tex]

and then substitute those in your expression for [tex]rot_\theta(v) [/tex].