View Full Version : Trig identites
lkjs
Nov 13, 2009, 06:24 AM
ay I came this question:
expand and simplify:
tan( pi/2 - y)
I used the compound angle formula for tan(x-y), but don't know what do when I have tan( pi/2), since I take it to as being non-defined since its an asymptote, but the answer for the question comes as cot (y), can someone please explain and show the necessary steps taken to solve the equation. Thanks a lot.
P.S: I realise there is an exact question in a previous thread, but the reply wasn't very forthcoming. Thank you.
Unknown008
Nov 13, 2009, 06:55 AM
I don't know if that is allowed, but I came up with this:
tan(\frac{\pi}{2} - y) = \frac{tan\frac{\pi}{2} - tany}{1+tan\frac{\pi}{2}tany}
= \frac{\frac{sin \pi/2} {cos \pi/2 } - tany}{1+\frac{sin\pi/2}{cos\pi/2}{tany}}
= \frac{\frac{1} {0 } - tany}{1+\frac{1}{0}{tany}}
= \frac{\frac{1} {0 }}{1+\frac{1}{0}{tany}} - \frac{tany}{1+\frac{1}{0}{tany}}
= \frac{\frac{1} {0 }}{1+\frac{1}{0}{tany}} - \cancel{\frac{tany}{1+\frac{1}{0}{tany}}}
= \frac{\frac{1} {0 }}{\frac{0+tany}{0}}
= \frac{1}{ tany}
I say again, that I don't know if that's the way to do it, but I came up with this... :o
s_cianci
Nov 13, 2009, 07:12 AM
Take it one step further ; 1/(tan y) = cot y.
ArcSine
Nov 13, 2009, 07:15 AM
Here's an approach relying on two of the 'co-function' identities...
tan \left( \frac{\pi}{2} \ - \ y \right) \ = \ \frac{sin\left(\frac{\pi}{2} \ - \ y ) }{cos\left(\frac{\pi}{2} \ - \ y) } \ = \ \frac{\text{cos}(y)}{\text{sin}(y)} \ = \ \text{cot}(y)
That second step is where the two co-function IDs kick in.
lkjs
Nov 13, 2009, 11:57 PM
Hey thanks for your replies, so the first solution is actually legitimate?
galactus
Nov 14, 2009, 05:30 AM
I don't know if that is allowed, but I came up with this:
tan(\frac{\pi}{2} - y) = \frac{tan\frac{\pi}{2} - tany}{1+tan\frac{\pi}{2}tany}
= \frac{\frac{sin \pi/2} {cos \pi/2 } - tany}{1+\frac{sin\pi/2}{cos\pi/2}{tany}}
= \frac{\frac{1} {0 } - tany}{1+\frac{1}{0}{tany}}
= \frac{\frac{1} {0 }}{1+\frac{1}{0}{tany}} - \frac{tany}{1+\frac{1}{0}{tany}}
= \frac{\frac{1} {0 }}{1+\frac{1}{0}{tany}} - \cancel{\frac{tany}{1+\frac{1}{0}{tany}}}
= \frac{\frac{1} {0 }}{\frac{0+tany}{0}}
= \frac{1}{ tany}
I say again, that I don't know if that's the way to do it, but I came up with this... :o
Good use of LaTex. I like the cancel you used. You learned me something. I always wondered how to do that on this site. I tried a greenie, but it would not let me.
Perito
Nov 14, 2009, 05:30 AM
Yes. Both solutions are legitimate.
Unknown008
Nov 14, 2009, 07:18 AM
Phew, I wasn't sure f we could 'play' with the zero like an ordinary number :o