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 referring to cannot be left blank. I just tried doing that.

 

[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 note that in my proof, i did not have the $4S$ and instead worked with $S$ ... that was wrong. Your working clearly shows that error on my part. Cheers.

 

[Gavran]

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

There are four formulas of this kind.

Euler's formula
$$ \arctan\frac12+\arctan\frac13=\frac\pi4 $$
Hermann's formula
$$ 2\arctan\frac12-\arctan\frac17=\frac\pi4 $$
Hutton's or Vega's formula
$$ 2\arctan\frac13+\arctan\frac17=\frac\pi4 $$
Machin's formula
$$ 4\arctan\frac15-\arctan\frac1{239}=\frac\pi4 $$
And all of them can be proved by using the trigonometric identity $$ \tan(\alpha\pm\beta)=\frac{\tan\alpha\pm\tan\beta}{1\mp\tan\alpha\tan\beta} $$ or $$ \arctan x\pm\arctan y=\arctan\frac{x\pm y}{1\mp xy} $$.
Also, they can be proved by using complex numbers. Let us take Herman’s formula as an example.

$\begin{align} 2\arctan\frac12-\arctan\frac17&=\arctan\frac12+\arctan\frac12-\arctan\frac17\nonumber\\ &=\arg(\frac{(2+i)(2+i)}{(7+i)})\nonumber\\ &=\arg((2+i)(2+i)(7-i))\nonumber\\ &=\arg((4+4i-1)(7-i))\nonumber\\ &=\arg(28+28i-7-4i+4+i)\nonumber\\ &=\arg(25+25i)\nonumber\\ &=\frac\pi4\nonumber \end{align}$

 

[neilparker62]

$$\tan x=\frac{1}{5} \implies \tan2x=\frac{5}{12} \implies \tan4x=\frac{120}{119}$$
$$\tan \left( \tan^{-1} \left( \frac{1}{239} \right)+\frac{\pi}{4}\right)=\frac{1+\frac{1}{239}}{1-\frac{1}{239}}=\frac{120}{119}$$

 

[Steve4Physics]

For an essentially geometric proof (no standard trig' identities used) see this 3 minute (appox.) video.

 

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

a different line for $\tan 4S$

$\tan (4S)= \tan (2S +2S) = \dfrac{\tan 2S + \tan 2S}{1 - \tan^2 (2S)}$

$\tan (2S) = \dfrac{5}{12}$

therefore,
$\tan 4S= \left[ \dfrac{\dfrac{5}{12} + \dfrac{5}{12}}{1-\left[\dfrac{5}{12}\right]^2}\right]$
$\tan 4S= \left[\dfrac{\dfrac{10}{12}}{\dfrac{119}{144}}\right] = \dfrac{10}{12} ×\dfrac{144}{119}= \dfrac{120}{119}$
...

$\tan (4S-T) =\left[\dfrac{\dfrac{120}{119}-\dfrac{1}{239}}{1+\dfrac{120}{119}×\dfrac{1}{239}}\right] = \dfrac{28561}{28441} × \dfrac{28441}{28561}= 1$

$\tan^{-1} \tan(4S-T) = \tan^{-1} (1)$
$4S-T= \dfrac {π}{4}$

 

[neilparker62]

For an essentially geometric proof (no standard trig' identities used) see this 3 minute (appox.) video.

Neat geometric derivation of $\tan(A+B)=\frac{\tan A + \tan B}{1-\tan A \tan B}$ if we divide all terms in $\frac{ay+bx}{by-ax}$ by $by$.

 

[neilparker62]

"All in one" derivation of compound angle formulae - based on the video construction.

 

[chwala]

There are four formulas of this kind.

Euler's formula
$$ \arctan\frac12+\arctan\frac13=\frac\pi4 $$
Hermann's formula
$$ 2\arctan\frac12-\arctan\frac17=\frac\pi4 $$
Hutton's or Vega's formula
$$ 2\arctan\frac13+\arctan\frac17=\frac\pi4 $$
Machin's formula
$$ 4\arctan\frac15-\arctan\frac1{239}=\frac\pi4 $$
And all of them can be proved by using the trigonometric identity $$ \tan(\alpha\pm\beta)=\frac{\tan\alpha\pm\tan\beta}{1\mp\tan\alpha\tan\beta} $$ or $$ \arctan x\pm\arctan y=\arctan\frac{x\pm y}{1\mp xy} $$.
Also, they can be proved by using complex numbers. Let us take Herman’s formula as an example.

$\begin{align} 2\arctan\frac12-\arctan\frac17&=\arctan\frac12+\arctan\frac12-\arctan\frac17\nonumber\\ &=\arg(\frac{(2+i)(2+i)}{(7+i)})\nonumber\\ &=\arg((2+i)(2+i)(7-i))\nonumber\\ &=\arg((4+4i-1)(7-i))\nonumber\\ &=\arg(28+28i-7-4i+4+i)\nonumber\\ &=\arg(25+25i)\nonumber\\ &=\frac\pi4\nonumber \end{align}$

With complex numbers, it makes it even easier,...

 

[chwala]

There are four formulas of this kind.

Euler's formula
$$ \arctan\frac12+\arctan\frac13=\frac\pi4 $$
Hermann's formula
$$ 2\arctan\frac12-\arctan\frac17=\frac\pi4 $$
Hutton's or Vega's formula
$$ 2\arctan\frac13+\arctan\frac17=\frac\pi4 $$
Machin's formula
$$ 4\arctan\frac15-\arctan\frac1{239}=\frac\pi4 $$
And all of them can be proved by using the trigonometric identity $$ \tan(\alpha\pm\beta)=\frac{\tan\alpha\pm\tan\beta}{1\mp\tan\alpha\tan\beta} $$ or $$ \arctan x\pm\arctan y=\arctan\frac{x\pm y}{1\mp xy} $$.
Also, they can be proved by using complex numbers. Let us take Herman’s formula as an example.

$\begin{align} 2\arctan\frac12-\arctan\frac17&=\arctan\frac12+\arctan\frac12-\arctan\frac17\nonumber\\ &=\arg(\frac{(2+i)(2+i)}{(7+i)})\nonumber\\ &=\arg((2+i)(2+i)(7-i))\nonumber\\ &=\arg((4+4i-1)(7-i))\nonumber\\ &=\arg(28+28i-7-4i+4+i)\nonumber\\ &=\arg(25+25i)\nonumber\\ &=\frac\pi4\nonumber \end{align}$

@Gavran ...you left out $50$ i.e in the denominator... can you see that?

ought to be
...$$=\arg(0.5+0.5i)=\dfrac{π}{4}$$

Cheers.

 

[anuttarasammyak]

@Gavran ...you left out 50 i.e in the denominator... can you see that?

Arg function depends on direction not magnitude of complex number as you got the same result with him. He made use of it for easy calculation

[EDIT]
in

Let a,b,c be complex numbers
$$arg(abc)=arg(a)+arg(b)+arg(c)$$
$$arg(a^*)=arg(a^{-1})=-arg(a)$$

 

[mathwonk]

Courant, Differential and Integral Calculus, vol. 1, pp. 352-3.