# nasty trig induction

• August 28th 2006, 03:45 AM
jacs
nasty 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
• August 28th 2006, 01:53 PM
galactus
Add $tan^{-1}(\frac{1}{2(n+1)^{2}})$ to both sides

That would be next in your series.

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

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

It does, but can you show how?.
• August 28th 2006, 03:32 PM
jacs
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
• August 28th 2006, 04:57 PM
ThePerfectHacker
For $n=1$
The left hand side is,
$\tan^{-1} \frac{1}{2\cdot 1^2}=\tan^{-1} \frac{1}{2}=\frac{\pi}{4}$
The right hand side is,
$\frac{\pi}{4}-\tan^{-1} \frac{1}{2(1)+1}=\frac{\pi}{4}-\mbox{ some non-zero number}$
No wonder you cannot prove it!
• August 28th 2006, 06:14 PM
galactus
Quote:

Originally Posted by ThePerfectHacker
For $n=1$
The left hand side is,
$\tan^{-1} \frac{1}{2\cdot 1^2}=\tan^{-1} \frac{1}{2}=\frac{\pi}{4}$
The right hand side is,
$\frac{\pi}{4}-\tan^{-1} \frac{1}{2(1)+1}=\frac{\pi}{4}-\mbox{ some non-zero number}$
No wonder you cannot prove it!

Hey PH, I believe you made a slight error. ;)

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

Left side:

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

It equals 0.463647609001

The right side:

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

It is true for n=1
• August 28th 2006, 06:25 PM
ThePerfectHacker
Quote:

Originally Posted by galactus
Hey PH, I believe you made a slight error...

I agree. But I bet you cannot prove me wrong :D

Remember mathematicians do not accept numerical, graphical, logical, obvious results ;)
• August 28th 2006, 08:30 PM
CaptainBlack
Quote:

Originally Posted by ThePerfectHacker
I agree. But I bet you cannot prove me wrong :D

Remember mathematicians do not accept numerical, graphical, logical, obvious results ;)

He has proven you wrong - he just has not proven that:

$
\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
• August 29th 2006, 05:29 AM
galactus
Yes, Cap'N. That method proves the general result, also.

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

Skipping ahead a lot of algebra:

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

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