# Thread: Trigonometry: Using Double Angle Formula for Tangent to Prove an Equality

1. ## Trigonometry: Using Double Angle Formula for Tangent to Prove an Equality

1. The problem statement, all variables and given/known data
Show that $tan(4arctan(1/5))$ is equivalent to $120/119$

2. Relevant equations
Double Angle Formula for tan:
$tan(2A) =\frac{2tanA}{1-\tan^2A}$

3. The attempt at a solution
From my understanding you have to apply the double angle formula twice.
Set x = arctan(1/5) and plug that into the double angle formula to solve for tan(2arctan1/5) (which is basically 2x). Now I set 2x = y, and plug y into the double angle formula to solve for 2y, which is 4arctan1/5.

sub x = arctan(1/5) into the double angle formula:
$tan(2A) =\frac{2tanA}{1-\tan^2A}$
$tan(2arctan(1/5)) =\frac{2tan(arctan1/5)}{1-\tan^2(arctan1/5)}$
Apply tan(arctanx) = x into the formula, resulting with:
$tan(2arctan(1/5)) =\frac{2(1/5)}{1-1/5arctan1/5}$
Simplify the numerator:
$tan(2arctan(1/5)) =\frac{2/5}{1-1/5arctan1/5}$

At this point I'm stuck, did I mess up in simplifying the denominator because I would like to be able to simplify it further so I don't have to use the final result above when I plug it back into a double angle formula once more. If that is correct, how would I work solve for
$tan(2(\frac{2/5}{1-1/5arctan1/5})) =\frac{2tan(\frac{2/5}{1-1/5arctan1/5})}{1-\tan^2(\frac{2/5}{1-1/5arctan1/5})}$?

Thanks for any assistance.

2. Hi Ali812, welcome to MHF.
2arctan(1/5) = arctan[2/5/(1-1/25) = arctan(10/24) = arctan(5/12)
Similarly
2arctan(5/12) = ...........?

3. Originally Posted by sa-ri-ga-ma
Hi Ali812, welcome to MHF.
2arctan(1/5) = arctan[2/5/(1-1/25) = arctan(10/24) = arctan(5/12)
Similarly
2arctan(5/12) = ...........?
2arctan(5/12) = arctan[2(5/12)/(1-25/144)] = arctan[(10/12)/(119/144)] = arctan[(120/144)/(119/144)] = arctan[120/119]

And since I'm plugging these into a double formula for tangent, the tan and arctan cancel out and I'm left with 120/119! Wow thank you very much!