Here is the question:
(a) Find the inverse of the rotation matrix R_theta=(row1 cos(theta), -sin(theta) row2 sin(theta), cos(theta)) Where theta is a fixed angle.
(b) Use your result to solve the system
x=acos(theta)-bsin(theta)
y=asin(theta)+bcos(theta)
for a and b in terms of x and y.
(c) Prove that, for any real number a, and 0
I got part a: Rinverse_theta=( row1(cos(theta), sin(theta) row2(-sin(theta), cos(theta))
Now i'm unsure of what it is asking me to do in part b.... any help interpreting the question would be greatly appreciated.
(a) Find the inverse of the rotation matrix R_theta=(row1 cos(theta), -sin(theta) row2 sin(theta), cos(theta)) Where theta is a fixed angle.
(b) Use your result to solve the system
x=acos(theta)-bsin(theta)
y=asin(theta)+bcos(theta)
for a and b in terms of x and y.
(c) Prove that, for any real number a, and 0
I got part a: Rinverse_theta=( row1(cos(theta), sin(theta) row2(-sin(theta), cos(theta))
Now i'm unsure of what it is asking me to do in part b.... any help interpreting the question would be greatly appreciated.
-
Part b, as requested:
Consider the matrix equivalent to the equations in (b). This equation can be written as:
[x,y] = R_theta * [a,b]
Rinverse_theta [x,y] = Rinverse_theta * R_theta * [a,b]
Rinverse_theta [x,y] = 1 * [a,b]
[a,b] = Rinverse_theta [x,y]
Now turn this back into the two eqns for a and b in terms of x and y using the Rinverse_theta matrix you calculated.
Consider the matrix equivalent to the equations in (b). This equation can be written as:
[x,y] = R_theta * [a,b]
Rinverse_theta [x,y] = Rinverse_theta * R_theta * [a,b]
Rinverse_theta [x,y] = 1 * [a,b]
[a,b] = Rinverse_theta [x,y]
Now turn this back into the two eqns for a and b in terms of x and y using the Rinverse_theta matrix you calculated.