Rotation Matrix: Finding Two Expressions & Verifying Equivalence

yoyo
Messages
21
Reaction score
0

Homework Statement


A vector x in R^2 is rotate twice through an angle theta (it is rotated through theta and again through theta). Find two expressions for the matrix representing this rotation. Verify that these two expressions are equivalent


Homework Equations


rotation matrix R=[cos, -sin; sin, cos]


The Attempt at a Solution



I can only think of one expression:

(R)(R)(x).

Could (R^2)(x) be the other one? How would i prove that this is equivalent?
 
Last edited:
Physics news on Phys.org
How about just rotating once by 2theta?
 
follow up question: a vector x in R^2 is rotated n times through an angle theta. Find two expressions for the matrix representing this rotation. what identity is implied.

if what Ziox said is true then it should be [cos (theta)n, -sin(theta)n; sin (theta)n, cos(theta)n]

but i don't see what identity this implies?
 
yoyo said:
follow up question: a vector x in R^2 is rotated n times through an angle theta. Find two expressions for the matrix representing this rotation. [/color]what identity is implied.

if what Ziox said is true then it should be [cos (theta)n, -sin(theta)n; sin (theta)n, cos(theta)n]

but i don't see what identity this implies?

You should probably find a second expression for the matrix first.
 
First, what is
\left(\begin{array}{cc}cos(\theta) & -sin(\theta)\\ sin(\theta) & cos(\theta)\end{array}\right)^2?

Second, can you use trig identities to write that in terms of cos(2\theta) and sin(2\theta)?
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
Back
Top