Prove that ## 4\tan^{-1}\left[\dfrac{1}{5}\right]- \tan^{-1}\left[\dfrac{1}{239}\right]= \dfrac{π}{4}##

[chwala]

Homework Statement

Prove that,
$4\tan^{-1}\left[\dfrac{1}{5}\right]- \tan^{-1}\left[\dfrac{1}{239}\right]= \dfrac{π}{4}$

Relevant Equations

Trig. identities

I let,

$4\tan^{-1}\left[\dfrac{1}{5}\right]- \tan^{-1}\left[\dfrac{1}{239}\right]= \dfrac{π}{4}$
$\tan^{-1}\left[\dfrac{1}{5}\right]- \dfrac{1}{4}\tan^{-1}\left[\dfrac{1}{239}\right]= \dfrac{π}{16}$

Then i let, $\tan^{-1}\left[\dfrac{1}{5}\right] = α , \tan^{-1}\left[\dfrac{1}{239}\right]=β$

$⇒\tan α=\left[\dfrac{1}{5}\right],\tan β =\left[\dfrac{1}{239}\right],$

$\tan (α-\dfrac{β}{4})= \left[\dfrac{\dfrac{1}{5}- \dfrac{1}{239×4}}{1+ \dfrac{1}{5}⋅\dfrac{1}{239×4}}\right]$

$\tan (α-\dfrac{β}{4})= \left[\dfrac{951}{4780} × \dfrac{4780}{4781}\right]$

$\tan (α-\dfrac{β}{4})=\left[\dfrac{951}{4781}\right]$

$\tan^{-1}(\tan (α-\dfrac{β}{4})≅11.25^0 = \dfrac{π}{16}$

$4[\tan^{-1}(\tan (α-\dfrac {β}{4})]≅45^0 = \dfrac{π}{4}$

I had a problem dealing with the $4$ in $4\tan^{-1}\dfrac{1}{5}$... there may be a better approach...

 

[anuttarasammyak]

I think of the calculation procedure
$$\tan^{-1}\frac{1}{5}-\tan^{-1}\frac{1}{239}=\tan^{-1}A, A=\frac{117}{598}$$
if my math is good. Then
$$\tan^{-1}\frac{1}{5}+\tan^{-1}A=\tan^{-1}B$$
$$\tan^{-1}\frac{1}{5}+\tan^{-1}B=\tan^{-1}C$$
$$\tan^{-1}\frac{1}{5}+\tan^{-1}C=\tan^{-1}D$$
We expect D=1.

[EDIT]
$$\tan(2\tan^{-1}\frac{1}{5})=\frac{5}{12}$$
$$\tan(4\tan^{-1}\frac{1}{5})=\frac{120}{119}$$
$$\tan(4\tan^{-1}\frac{1}{5}-\tan^{-1}\frac{1}{239})=1$$

 

[MatinSAR]

I'm not sure what you did ...

What I've done:

Prove that: $4\tan^{-1}\left[\dfrac{1}{5}\right]- \tan^{-1}\left[\dfrac{1}{239}\right]= \dfrac{π}{4}$

Let:
$\tan^{-1} \frac{1}{5} = S$...................so................... $tanS=\dfrac{1}{5}$
$\tan^{-1} \frac{1}{239} = T$..............so................... $tanT=\dfrac{1}{239}$
So
$4S - T = X$ (We want to prove that $X=\dfrac{π}{4}$)
$tan(4S - T)=tanX$

We know that $tan(4S-T)=\dfrac{tan(4S)-tanT}{1+tan(4S)tanT}$
So we need to find tan(4S):

$tan(4S)=\dfrac{4tanS(1-tan^{2}S)}{1-6tan^{2}S+tan^{4}S}=\dfrac{(4/5)(24/25)}{1-(6/25)+1/625}=\dfrac{120}{119}$

Now we calculate $tan(4S-T)$ :

$tan(4S-T)=\dfrac{(120/119)-(1/239)}{1+(120/119)(1/239)}=\dfrac{ \dfrac{239*120-119}{119*239} }{ \dfrac{119*239+120}{119*239} }$

$=\dfrac{239*120-119}{239*119+120}=\dfrac{239*119+239-119}{239*119+120}=\dfrac{239*119+120}{239*119+120}=1$
We proved that $tan(4S-T)=1$ so we can say that $4S-T=\dfrac{π}{4}$.

Remember that $4S-T =4\tan^{-1}\left[\dfrac{1}{5}\right]- \tan^{-1}\left[\dfrac{1}{239}\right]$.

 

[chwala]

I'm not sure what you did ...

What I've done:

Prove that: $4\tan^{-1}\left[\dfrac{1}{5}\right]- \tan^{-1}\left[\dfrac{1}{239}\right]= \dfrac{π}{4}$

Let:
$\tan^{-1} \frac{1}{5} = S$...................so................... $tanS=\dfrac{1}{5}$
$\tan^{-1} \frac{1}{239} = T$..............so................... $tanT=\dfrac{1}{239}$
So
$4S - T = X$ (We want to prove that $X=\dfrac{π}{4}$)
$tan(4S - T)=tanX$

We know that $tan(4S-T)=\dfrac{tan(4S)-tanT}{1+tan(4S)tanT}$
So we need to find tan(4S):

$tan(4S)=\dfrac{4tanS(1-tan^{2}S)}{1-6tan^{2}S+tan^{4}S}=\dfrac{(4/5)(24/25)}{1-(6/25)+1/625}=\dfrac{120}{119}$

Now we calculate $tan(4S-T)$ :

$tan(4S-T)=\dfrac{(120/119)-(1/239)}{1+(120/119)(1/239)}=\dfrac{ \dfrac{239*120-119}{119*239} }{ \dfrac{119*239+120}{119*239} }$

$=\dfrac{239*120-119}{239*119+120}=\dfrac{239*119+239-119}{239*119+120}=\dfrac{239*119+120}{239*119+120}=1$
We proved that $tan(4S-T)=1$ so we can say that $4S-T=\dfrac{π}{4}$.

Remember that $4S-T =4\tan^{-1}\left[\dfrac{1}{5}\right]- \tan^{-1}\left[\dfrac{1}{239}\right]$.

I am not sure on my working. I could be wrong.

 

[MatinSAR]

I am not sure on my working. I could be wrong.

How did you find out that $tan(\dfrac{\beta}{4})=\dfrac{1}{239*4}$?
It's wrong.

 

[chwala]

How did you find out that $tan(\dfrac{\beta}{4})=\dfrac{1}{239*4}$?
It's wrong.

I divided each term by $4$.

 

[MatinSAR]

I divided each term by $4$.

According to you $\dfrac{tan\beta} {4}=tan\dfrac{\beta} {4}$ but it's not correct.

Fore example :
$\dfrac{1} {4} tan \pi =0$
But $tan(\dfrac{\pi} {4} )=1$

 

[chwala]

I am not sure on my working. I could be wrong.

How did you simplify your $\tan (4S)$? I had tried $\tan (2S + 2S)$ expansion and I noted that it was quite long with the substitutions... Am assuming you used the same approach to realize your rhs.

 

[anuttarasammyak]

Try to get tan 2S first. Then you can proceed to tan4S.

 

[chwala]

Try to get tan 2S first. Then you can proceed to tan4S.

That should be easy... I think I had different equations from start... thanks though...

 

[MatinSAR]

How did you simplify your $\tan (4S)$? I had tried $\tan (2S + 2S)$ expansion and I noted that it was quite long with the substitutions... Am assuming you used the same approach to realize your rhs.

I haven't proved it in the post.
You can google tan4x formula there are plenty of sites which proved the formula. It's not hard.
Did you understand your mistake in post #1?

 

[chwala]

I haven't proved it in the post.
You can google tan4x formula there are plenty of sites which proved the formula. It's not hard.
Did you understand your mistake in post #1?

Yes I did...and I replied in post $10$ that the simplification that I was asking is as easy as abc. I had assumed that it was the same equation that I had in my hard copy book but I just counter checked and realized that my equations were different. Cheers man!

 

[Mark44]

Some comments:

I let,

$4\tan^{-1}\left[\dfrac{1}{5}\right]- \tan^{-1}\left[\dfrac{1}{239}\right]= \dfrac{π}{4}$
$\tan^{-1}\left[\dfrac{1}{5}\right]- \dfrac{1}{4}\tan^{-1}\left[\dfrac{1}{239}\right]= \dfrac{π}{16}$

The first equation above is what you're supposed to prove, so it is generally invalid to assume ("let") a statement you're trying to prove. Only under very specific conditions (*) is it valid to make this sort of assumption.

* Each step is reversible; i.e., by performing a one-to-one operation on each side of the equation or inequality.

Then i let, $\tan^{-1}\left[\dfrac{1}{5}\right] = α , \tan^{-1}\left[\dfrac{1}{239}\right]=β$

$⇒\tan α=\left[\dfrac{1}{5}\right],\tan β =\left[\dfrac{1}{239}\right],$

This is what you should do first, not assume that the equation you're trying to prove is true.

Relevant Equations:
Trig. identities

Unless you list specific identities you should leave this section blank.

Also, as already mentioned by @MatinSAR $\dfrac{tan\beta} {4} \ne tan\dfrac{\beta} {4}$

 

[chwala]

Some comments:

The first equation above is what you're supposed to prove, so it is generally invalid to assume ("let") a statement you're trying to prove. Only under very specific conditions (*) is it valid to make this sort of assumption.

* Each step is reversible; i.e., by performing a one-to-one operation on each side of the equation or inequality.
This is what you should do first, not assume that the equation you're trying to prove is true.

Unless you list specific identities you should leave this section blank.

Also, as already mentioned by @MatinSAR $\dfrac{tan\beta} {4} \ne tan\dfrac{\beta} {4}$

The section you are reffering to cannot be left blank. I just tried doing that.