Here is a nasty trig induction question i dont even pretend to know how to begin. I think it is a little out of the scope of my course, but would like to try it, just in case.

thanks jacs

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- Aug 28th 2006, 03:45 AMjacsnasty trig induction
Here is a nasty trig induction question i dont even pretend to know how to begin. I think it is a little out of the scope of my course, but would like to try it, just in case.

thanks jacs - Aug 28th 2006, 01:53 PMgalactus
Add $\displaystyle tan^{-1}(\frac{1}{2(n+1)^{2}})$ to both sides

That would be next in your series.

$\displaystyle tan^{-1}(\frac{1}{2\cdot{1^{2}}})+tan^{-1}(\frac{1}{2\cdot{2^{2}}})+$$\displaystyle ...........+tan^{-1}(\frac{1}{2n^{2}})+tan^{-1}(\frac{1}{2(n+1)^{2}})$$\displaystyle =\frac{\pi}{4}-tan^{-1}(\frac{1}{2n+1})+tan^{-1}(\frac{1}{2(n+1)^{2}})$

Now, your mission is to show that $\displaystyle \frac{\pi}{4}-tan^{-1}(\frac{1}{2n+1})+tan^{-1}(\frac{1}{2(n+1)^{2}})$$\displaystyle =\frac{\pi}{4}-tan^{-1}(\frac{1}{2(n+1)+1})=\frac{\pi}{4}-tan^{1}(\frac{1}{2n+3})$

It does, but can you show how?. - Aug 28th 2006, 03:32 PMjacs
I am afraid i have no idea how to proceed at all. This question i think is a little beyond what were are supposed to be able to do at our course level. It was presented as a challenge question, but even with your start, i don't know how to proceed.

jacs - Aug 28th 2006, 04:57 PMThePerfectHacker
For $\displaystyle n=1$

The left hand side is,

$\displaystyle \tan^{-1} \frac{1}{2\cdot 1^2}=\tan^{-1} \frac{1}{2}=\frac{\pi}{4}$

The right hand side is,

$\displaystyle \frac{\pi}{4}-\tan^{-1} \frac{1}{2(1)+1}=\frac{\pi}{4}-\mbox{ some non-zero number}$

No wonder you cannot prove it! - Aug 28th 2006, 06:14 PMgalactusQuote:

Originally Posted by**ThePerfectHacker**

$\displaystyle tan^{-1}(1)=\frac{\pi}{4}$

Left side:

$\displaystyle tan^{-1}(\frac{1}{2})\neq\frac{\pi}{4}$

It equals**0.463647609001**

The right side:

$\displaystyle \frac{\pi}{4}-tan^{-1}(\frac{1}{3})=0.46347609001$

It is true for n=1 - Aug 28th 2006, 06:25 PMThePerfectHackerQuote:

Originally Posted by**galactus**

Remember mathematicians do not accept numerical, graphical, logical, obvious results ;) - Aug 28th 2006, 08:30 PMCaptainBlackQuote:

Originally Posted by**ThePerfectHacker**

$\displaystyle

\arctan(1/2)=\frac{\pi}{4}-\arctan(1/3)

$

which follows immediately from taking tan of both sides and the trig identity

for the tan of a difference of two angles.

RonL - Aug 29th 2006, 05:29 AMgalactus
Yes, Cap'N. That method proves the general result, also.

$\displaystyle tan(u-v)=\frac{tan(u)-tan(v)}{1+tan(u)tan(v)}$

Skipping ahead a lot of algebra:

$\displaystyle \frac{\frac{1}{2(n+1)^{2}}-\frac{1}{2n+1}}{1+\frac{1}{2(n+1)^{2}}\cdot\frac{1 }{2n+1}}=\frac{-1}{2n+3}$

$\displaystyle tan(\frac{-1}{2n+3})=-tan(\frac{1}{2n+3})$