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  1. How does $ {\sqrt 2 \over 2} = \cos (45 \text{ degrees })$?

  2. Is my graph (the one underneath the original) accurate with how I've depicted the representation of the triangle that the trig function represent? What I mean is, the blue triangle is the pre-rotated block, the green is the post-rotated block, and the purple is the rotated change (45 degrees) between them.

  3. How do these trig functions in this matrix represent a clockwise rotation? (Like, why does "$-\sin \theta $ " in the bottom left mean clockwise rotation... and "$- \sin \theta $ " in the upper right mean counter clockwise? Why not "$-\cos \theta $ "? $$\begin{bmatrix}\cos \theta &\sin \theta \\-\sin \theta & \cos \theta \end{bmatrix}$$

enter image description here

Any help in understanding the trig representations of a rotation would be extremely helpful! Thanks

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For the third question: If you believe that the matrix for counter clockwise rotation is correct, then to obtain the clockwise matrix, just replace $\phi$ by $-\phi$. Rules of trigonometry will then tell you that $\cos -\phi = \cos \phi$ and $\sin -\phi = -\sin \phi$, leading to the clockwise matrix you have. –  TMM Apr 16 '13 at 19:50
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All of the trigonometry will be clear if you examine what happens to the points $(1,0)$ and $(0,1)$ under these transformations. After they have moved, drop a perpendicular vertically and a line through the origin and consider the triangle formed. They will be (sometimes degenerate) triangles with hypotenuese 1 and then you will see why each of their legs has measure $\sin(\phi)$ or $\cos(\phi)$ etc.

Here's what I mean: after a $\pi/6$ rotation counterclockwise, the point $(1,0)$ has moved to $(\sqrt{3}/2,1/2)$. This point, in addition to $(0,0)$ and the point directly below it on the $x$ axis, $(\sqrt{3}/2,0$ form a right triangle with hypoteneuse $1$. Look at lengths of the short sides of the triangle. Try to do the same thing with an angle $\phi$ between 0 and $\pi/2$, and analyze what the sides of the triangle have to be in terms of $\sin(\phi)$ and $\cos(\phi)$.

Because a rotation in the plane is totally determined by how it moves points on the unit circle, this is all you have to understand.

You don't actually need a representation for both clockwise and counterclockwise. You can use the counterclockwise one all the time, if you agree that a clockwise rotation would be a negative counterclockwise rotation. That is, if you want to perform a clockwise rotation of $\pi/4$ radians, then you should use $\phi=-\pi/4$ in the counterclockwise rotation representation.

The fact that $\sin(-\phi)=-\sin(\phi)$ accounts for the change in the sign of sine between the two representations, and the fact that the $\cos(\phi)$ doesn't change is because $\cos(-\phi)=\cos(\phi)$. You may as well just pick the counterclockwise representation scheme, and perform both clockwise and counterclockwise rotations with it.


To provide some extra evidence that it makes sense these are rotation matrices, you can check to see that the columns of these matrices always have Euclidean length 1 (easy application of the $\sin^2(x)+\cos^2(x)=1$ identity.) Moreover, they are orthogonal to each other. That means they are orthogonal matrices, and consequently represent rotations. They satisfy $UU^T=U^TU=I_2$. This demonstrates that $U^T=U^{-1}$, and now you'll notice that the transpose of the counterclockwise representation gives you the clockwise representation! Of course, rotating clockwise and rotating counterclockwise by $\phi$ radians are inverse operations.

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drop a perpendicular vertically and consider the triangle formed - what does this mean? and Because a rotation in the plane is totally determined by how it moves points on the unit circle, this is all you have to understand. do you think you can diagram this point out? –  Growler Apr 16 '13 at 19:38
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@Growler I wish I knew how to do pictures well on here! I'll clarify my explanation and include an example asap. –  rschwieb Apr 16 '13 at 19:40
    
I guess you could do it in paint or some other tool then append the picture to your answer? –  Growler Apr 16 '13 at 19:41
    
@Growler OK, I finished adding a bit. Let me know if I should say more. I think you'll be able to draw it based on that. –  rschwieb Apr 16 '13 at 19:45
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