Given
#arctan({6x-8x^2}/{1-12x^2}) - arctan({4x}/{1-4x^2}) = arctan 2x #
show #|2x| < 1/sqrt{3}#
Wow. I don't know if this refers solely to the principal values of the inverse tangent. Let's see if it matters.
The tangent difference angle formula is
# tan (a-b) = {tan a - tan b}/{1 + tan a tan b }#
Taking tangents of the given,
#tan( arctan({6x-8x^2}/{1-12x^2}) - arctan({4x}/{1-4x^2}) ) = tan arctan 2x #
# { {6x-8x^2}/{1-12x^2} - {4x}/{1-4x^2} } / { 1 + {6x-8x^2}/{1-12x^2} cdot {4x}/{1-4x^2} } = 2x #
# {6x-8x^2}/{1-12x^2} - {4x}/{1-4x^2} = 2x ( 1 + {6x-8x^2}/{1-12x^2} cdot {4x}/{1-4x^2} ) #
# (6x-8x^2)(1-4x^2) -4x ( 1-12x^2) = 2x ( (1-12x^2) (1-4x^2 ) + 4x (6x-8x^2) ) #
#96 x^5 - 96 x^4 - 8 x^3 + 8 x^2 = 0#
#8x^2 ( 12 x^3 - 12 x^2 - x + 1) = 0#
We can set #x=1# is a solution so #x-1# is a factor.
#8x^2 (x-1) ( 12 x^2 - 1) = 0#
#x = 0 or x = 1 or x = pm 1/{2sqrt{3}}#
We didn't prove what we set out to prove. We found a counterexample, #x=1.# Let's check it.
# t = arctan({6-8}/{1-12}) - arctan({4}/{1-4}) #
# = arctan({6-8}/{1-12}) - arctan({4}/{1-4}) #
#tan t = ( 2/11 + 4/3 ) / ( 1 + (2/11)(-4/3) ) =2 = 2(1) quad sqrt #
In terms of principal value we're subtracting a fourth quadrant angle from a first quadrant and get a positive result, so we're still in the first quadrant, so #x=1# is a solution even restricted to the principal values.
