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October 25th, 2010, 11:09 AM  #1 
Member Joined: Sep 2010 Posts: 32 Thanks: 0  My teacher says that...
...the reference angle to t= pi/4 (derived from arctan (1)) is 5pi/4. I don't get how it can be 5pi/4. I get that it should be 4pi/4?? Can someone please explain this? 
October 25th, 2010, 11:18 AM  #2 
Senior Member Joined: Jul 2010 From: St. Augustine, FL., U.S.A.'s oldest city Posts: 12,211 Thanks: 520 Math Focus: Calculus/ODEs  Re: My teacher says that...
What you got is what should be added to . Let's look at: or equivalently . Now, look at Let and where Since , we have This where we find that an integral multiple of is added to the original angle to get the reference angle. This makes sense too if we look at the unit circle and pick some arbitrary angle and the tangent of that angle. Now if we add some integral multiple of to that angle, the tangent value remains the same. Think of y = mx superimposed upon the unit circle. We know that for all x. Take a look at the two angles made by y and what their difference is. 
October 29th, 2010, 04:02 AM  #3 
Global Moderator Joined: Dec 2006 Posts: 20,262 Thanks: 1958 
For a given angle, its reference angle is defined as the absolute value of the angle in the interval [pi/2, pi/2] that differs from the given angle by an integer multiple of pi (i.e., the minimum unsigned difference between the given angle and an integer multiple of pi). For 5pi/4, the reference angle is pi/4, but not vice versa. The reference angle of an angle ? may equivalently be defined as arcsin(sin ?) or arccos(cos ?). 

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