Equivalent Trig Expressions

Aug 2011
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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!
 

greg1313

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Oct 2008
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London, Ontario, Canada - The Forest City
\(\displaystyle \tan\left(x + \frac{\pi}{2}\right) = -\cot(x)\)

\(\displaystyle \frac{13\pi}{14} + \frac{\pi}{2} = \frac{10\pi}{7}\)

\(\displaystyle -\tan(x) = \tan(-x) \Rightarrow \tan\left(-\frac{10\pi}{7}\right) = \tan\left(-\frac{3\pi}{7}\right),\,z = - \frac{3\pi}{7}\)
 
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Jan 2012
9
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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.
 
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greg1313

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Oct 2008
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More accurately, \(\displaystyle z = -\frac{3\pi}{7} + k\pi,\,k \in \mathbb{Z}\)
 
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skipjack

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Dec 2006
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I suspect that the book's answer was $4\pi$/7.