October 15th, 2006, 12:16 AM
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| Senior Member | | Join Date: Apr 2006
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Quote:
Originally Posted by Glaysher  Then is answer simply:
[-sin(sin(sin(sin(sin(1))))), sin(sin(sin(sin(sin(1)))))?
Because that seems too simple to be right. I was trying to think of a way of writing it exactly in terms of pi | Or [sin(sin(sin(sin(1)))), sin(sin(sin(1)))]? sin[0,2pi] = [-1,1],
sin[-1,1] = [sin(1),1],
sin[sin(1),1] = [sin(sin(1)),sin(1)],
sin[sin(sin(1)),sin(1)] = [sin(sin(sin(1))),sin(sin(1))],
sin[sin(sin(sin(1))),sin(sin(1))] = [sin(sin(sin(sin(1)))),sin(sin(sin(1)))].
Last edited by JakeD; October 15th, 2006 at 12:32 AM.
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