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

\left(\begin{array}{c}1\\0\end{array}\right)\mbox{ or }\left(\begin{array}{c}0\\1\end{array}\right)

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

You'll need the sum of angles formulas for cos and sin, \cos(\theta_{0}+\theta) and \sin(\theta_{0}+\theta).
(\theta_{0} 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:

v = a\cdot \left(\begin{array}{c}1\\0\end{array}\right)<br /> + b\cdot \left(\begin{array}{c}0\\1\end{array}\right)

so the rotated vector will be:

rot_\theta(v) = a \cdot rot_\theta (\left(\begin{array}{c}1\\0\end{array}\right) )<br /> + b\cdot rot_\theta ( \left(\begin{array}{c}0\\1\end{array}\right) )

You can first determine c and d in:

rot_\theta (\left(\begin{array}{c}1\\0\end{array}\right) ) = <br /> c\cdot \left(\begin{array}{c}1\\0\end{array}\right)<br /> +<br /> d\cdot \left(\begin{array}{c}0\\1\end{array}\right)

and e and f in:

rot_\theta (\left(\begin{array}{c}0\\1\end{array}\right) ) = <br /> e\cdot \left(\begin{array}{c}1\\0\end{array}\right)<br /> +<br /> f\cdot \left(\begin{array}{c}0\\1\end{array}\right)

and then substitute those in your expression for rot_\theta(v).