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 October 25th, 2010, 10:09 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, 10:18 AM #2 Senior Member     Joined: Jul 2010 From: St. Augustine, FL., U.S.A.'s oldest city Posts: 12,211 Thanks: 521 Math Focus: Calculus/ODEs Re: My teacher says that... What you got is what should be added to $\dfrac{\pi}{4}$. $\dfrac{\pi}{4}+\pi=\dfrac{5\pi}{4}$ Let's look at: $\theta=\arctan(x)$ or equivalently $\tan(\theta)=x$. Now, look at $\tan(\alpha \pm \beta)= \frac{\tan \alpha \pm \tan \beta}{1 \mp \tan \alpha \tan \beta}$ Let $\alpha=\theta$ and $\beta=k\pi$ where $k\in \mathbb Z$ $\tan(\theta + k\pi)= \dfrac{\tan \theta + \tan k\pi}{1 - \tan \theta \tan k\pi}$ Since $\tan k\pi=0$, we have $\tan(\theta + k\pi)=\tan\theta$ This where we find that an integral multiple of $\pi$ 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 $\pi$ to that angle, the tangent value remains the same. Think of y = mx superimposed upon the unit circle. We know that $m=\tan\theta$ for all x. Take a look at the two angles made by y and what their difference is.
 October 29th, 2010, 03:02 AM #3 Global Moderator   Joined: Dec 2006 Posts: 21,020 Thanks: 2256 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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