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 January 25th, 2012, 12:59 PM #1 Newbie   Joined: Aug 2011 Posts: 24 Thanks: 0 Equivalent Trig Expressions Given that cot (13pi/14)= tan z, find angle z. So I used cot x=tan(pi/2-x), and I got the answer -3pi/7, but the answer says it's positive, so I am confused. Thanks in advance!
 January 25th, 2012, 01:20 PM #2 Global Moderator     Joined: Oct 2008 From: London, Ontario, Canada - The Forest City Posts: 7,963 Thanks: 1148 Math Focus: Elementary mathematics and beyond Re: Equivalent Trig Expressions $\displaystyle \tan\left(x + \frac{\pi}{2}\right) = -\cot(x)$ $\displaystyle \frac{13\pi}{14} + \frac{\pi}{2} = \frac{10\pi}{7}$ $\displaystyle -\tan(x) = \tan(-x) \Rightarrow \tan\left(-\frac{10\pi}{7}\right) = \tan\left(-\frac{3\pi}{7}\right),\,z = - \frac{3\pi}{7}$ Last edited by skipjack; August 24th, 2019 at 04:24 PM.
 January 25th, 2012, 01:36 PM #3 Newbie   Joined: Jan 2012 Posts: 9 Thanks: 0 Re: Equivalent Trig Expressions Yes, it is correct and tg must be negative because cotangent is in II. Quadrant. Maybe in your book there is positive solution because if you add 2pi to -3pi/7 you get the same angle 2pi+(-3pi/7)=11pi/7 but in positive direction. Last edited by skipjack; August 24th, 2019 at 04:25 PM.
 January 25th, 2012, 02:52 PM #4 Newbie   Joined: Aug 2011 Posts: 24 Thanks: 0 Re: Equivalent Trig Expressions thank you!
 January 25th, 2012, 03:34 PM #5 Global Moderator     Joined: Oct 2008 From: London, Ontario, Canada - The Forest City Posts: 7,963 Thanks: 1148 Math Focus: Elementary mathematics and beyond Re: Equivalent Trig Expressions More accurately, $\displaystyle z = -\frac{3\pi}{7} + k\pi,\,k \in \mathbb{Z}$ Last edited by skipjack; August 24th, 2019 at 04:26 PM.
 August 24th, 2019, 04:30 PM #6 Global Moderator   Joined: Dec 2006 Posts: 20,969 Thanks: 2219 I suspect that the book's answer was $4\pi$/7.

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