Could someone give an idea on how to complete this? Suppose I have a 2x2 matrix, where all entries but the bottom left are sqrt(2)/2. The bottom left however, is -sqrt(2)/2. With that being said, how I started it was by plugging in the values; for instance, the top left entry of the matrix is cos(sqrt(2)/2), bottom right is the same thing, etc. But I have a feeling this is wrong; any help here?
Describing Rotations Geometrically
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linear-algebra
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0Yes, that is exactly what I'm using in the book! I'm just confused as to how to draw it; I asked in another question about projection, so the problem with me is that I don't quite understand the actual drawing itself...if that makes any sense. Perhaps you could get me started; the theta respectively for the top left entry would be pi/4, the bottom right would be pi/4 too, the top right entry would be pi/4, and the bottom left entry would be 5pi/4. Thanks! – 2012-03-20
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0Your $\theta$ cannot change between the entries (Some can't be $\pi \over 4$ with others being $5\pi \over 4$). $\theta = \frac{\pi}{4}$ means you are rotating by 45 degrees counterclockwise. – 2012-03-20
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0Oh, you're right. Oops, sorry about that. So this leads me to my next question, how would you draw this geometrically? Would I draw a line to let's say, pi/4, then rotate it somewhere? To me, the main question is, where is my starting line? Thanks! – 2012-03-20