For any positive integer n and any angle θ , show that in the group S L ( 2 , R ) , [ cos θ − sin θ sin θ cos θ ] = [ cos n θ − sin n θ sin n θ cos n θ ] .Use this formula to find the order of [ cos 60 ° − sin 60 ° sin 60 ° cos 60 ° ] = [ cos 2 ° − sin 2 ° sin 2 ° cos 2 ° ] .(Geometrically, [ cos θ − sin θ sin θ cos θ ] represents a rotation of the plane θ degrees.)
For any positive integer n and any angle θ , show that in the group S L ( 2 , R ) , [ cos θ − sin θ sin θ cos θ ] = [ cos n θ − sin n θ sin n θ cos n θ ] .Use this formula to find the order of [ cos 60 ° − sin 60 ° sin 60 ° cos 60 ° ] = [ cos 2 ° − sin 2 ° sin 2 ° cos 2 ° ] .(Geometrically, [ cos θ − sin θ sin θ cos θ ] represents a rotation of the plane θ degrees.)
Solution Summary: The author explains that G is the symmetry group of a circle. As n increases to infinity, the polygon becomes symmetric.
For any positive integer n and any angle
θ
, show that in the group
S
L
(
2
,
R
)
,
[
cos
θ
−
sin
θ
sin
θ
cos
θ
]
=
[
cos
n
θ
−
sin
n
θ
sin
n
θ
cos
n
θ
]
.Use this formula to find the order of
[
cos
60
°
−
sin
60
°
sin
60
°
cos
60
°
]
=
[
cos
2
°
−
sin
2
°
sin
2
°
cos
2
°
]
.(Geometrically,
[
cos
θ
−
sin
θ
sin
θ
cos
θ
]
represents a rotation of the plane
θ
degrees.)
Solutions of inequalitie
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Mic
Is (-3, 2) a solution of 7x+9y > -3?
Choose 1 answer:
A
Yes
B
No
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