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

  • #1
chwala
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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...
 
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  • #2
I think of the calculation procedure
[tex]\tan^{-1}\frac{1}{5}-\tan^{-1}\frac{1}{239}=\tan^{-1}A, A=\frac{117}{598}[/tex]
if my math is good. Then
[tex]\tan^{-1}\frac{1}{5}+\tan^{-1}A=\tan^{-1}B[/tex]
[tex]\tan^{-1}\frac{1}{5}+\tan^{-1}B=\tan^{-1}C[/tex]
[tex]\tan^{-1}\frac{1}{5}+\tan^{-1}C=\tan^{-1}D[/tex]
We expect D=1.

[EDIT]
[tex]\tan(2\tan^{-1}\frac{1}{5})=\frac{5}{12}[/tex]
[tex]\tan(4\tan^{-1}\frac{1}{5})=\frac{120}{119}[/tex]
[tex]\tan(4\tan^{-1}\frac{1}{5}-\tan^{-1}\frac{1}{239})=1[/tex]
 
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  • #3
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]##.
 
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  • #4
MatinSAR said:
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.
 
  • #5
chwala said:
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.
 
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  • #6
MatinSAR said:
How did you find out that ##tan(\dfrac{\beta}{4})=\dfrac{1}{239*4}##?
It's wrong.
I divided each term by ##4##.
 
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  • #7
chwala said:
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 ##
 
  • #8
chwala said:
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.
 
  • #9
Try to get tan 2S first. Then you can proceed to tan4S.
 
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  • #10
anuttarasammyak said:
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...
 
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  • #11
chwala said:
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?
 
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  • #12
MatinSAR said:
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!
 
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  • #13
Some comments:
chwala said:
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.
chwala said:
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.

chwala said:
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} ##
 
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  • #14
Mark44 said:
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.
 

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

What is the significance of the equation \( 4\tan^{-1}\left[\dfrac{1}{5}\right]- \tan^{-1}\left[\dfrac{1}{239}\right]= \dfrac{π}{4} \) in mathematics?

This equation is a famous identity used in the proof of the value of π (pi). It is known as Machin's formula, which was discovered by John Machin in 1706. This identity allows for the computation of π to a high degree of accuracy using the arctangent function.

How do you prove the equation \( 4\tan^{-1}\left[\dfrac{1}{5}\right]- \tan^{-1}\left[\dfrac{1}{239}\right]= \dfrac{π}{4} \)?

The proof involves using the addition and subtraction formulas for the arctangent function. Specifically, you can use the identity for the arctangent of a sum and difference of two angles and verify that both sides of the equation simplify to the same value. Detailed steps include expressing each arctangent in terms of complex numbers and then simplifying.

Why is Machin's formula useful for calculating π?

Machin's formula is useful because it converges very quickly. When calculating π, the arctangent series converges much faster for smaller arguments. By breaking down the calculation of π into a series involving smaller arguments, Machin's formula allows for efficient and precise computation of π.

What are the addition and subtraction formulas for the arctangent function?

The addition formula for arctangent is \( \tan^{-1}(x) + \tan^{-1}(y) = \tan^{-1}\left(\frac{x+y}{1-xy}\right) \) when \( xy < 1 \). The subtraction formula is \( \tan^{-1}(x) - \tan^{-1}(y) = \tan^{-1}\left(\frac{x-y}{1+xy}\right) \). These formulas are essential in proving identities involving arctangent functions.

Can Machin's formula be generalized or extended?

Yes, Machin's formula can be generalized. Variations and extensions of Machin-like formulas exist where the arctangent terms can be different, but they all aim to exploit the rapid convergence properties of the arctangent series. These generalizations are used in various algorithms for computing π to many decimal places.

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