Question #d14f6

1 Answer
Mar 1, 2018

\qquad \qquad \quad "the solutions are:" \qquad \ \ \ x \ = \ \pm sqrt{ -2 + 2sqrt{ 5 } }/{ 2 } \quad.

Explanation:

"We want to solve:"

\qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad tan^{-1} (x) \ = \ cos^{-1} (x).

"We can proceed as follows:"

\qquad \qquad \qquad \qquad \qquad \quad tan[ tan^{-1} (x) ] \ = \ tan[ cos^{-1} (x) ]. \qquad \qquad \qquad \qquad \qquad \ (I)

\qquad \quad \ rArr \quad "now using:" \qquad tan^{-1}( tan ( x ) ) = x, \quad "in" \ \ (I) \quad rArr

\qquad \qquad \qquad \qquad \qquad \qquad \qquad \quad x \ = \ tan[ cos^{-1} (x) ].

"Emphasizing:"

\qquad \qquad \qquad \qquad \qquad \qquad \qquad \ \ x \ = \ tan[ \underbrace{ cos^{-1} (x) }_{\theta} ] \qquad \qquad \qquad \qquad \qquad \ (II)

\qquad \quad \ rArr \quad "now using:" \qquad 1 + tan^2 \theta = sec^2 \theta \quad rArr

\qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \ \ \ tan \theta = sqrt{ sec^2 \theta - 1 } \quad rArr

\qquad \qquad \qquad \quad \ \ \ tan \theta = sqrt{ 1/{ cos^2 \theta } - 1 }, \quad "in" \ \ (II) \quad rArr

\qquad \qquad \qquad \qquad \qquad \qquad \qquad x \ = \ sqrt{ 1/{ cos^2 ( cos^{-1} (x) ) } - 1 }

\qquad \qquad \qquad \qquad \qquad \qquad \qquad x \ = \ sqrt{ 1/{ [ cos ( cos^{-1} (x) ) ]^2 - 1 } }. \qquad \qquad \qquad \qquad \quad (III)

\qquad \quad \ rArr \quad "now using:" \qquad cos^{-1}( cos( x ) ) = x, \quad "in" \ \ (III) \quad rArr

\qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \quad x \ = \ sqrt{ 1/{ x^2 ) - 1 }.

\qquad :. \qquad \qquad \qquad \qquad \qquad \qquad \quad x^2 \ = \ 1/{ x^2 ) - 1.

\qquad :. \qquad \qquad \qquad \qquad \ \ x^2 [ x^2 ] \ = \ x^2 [ 1/{ x^2 ) - 1 ].

\qquad :. \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \ x^4 \ = \ 1 - x^2.

\qquad :. \qquad \qquad \qquad \qquad \qquad \qquad \qquad x^4 + x^2 - 1 \ = \ 0.

:. \qquad \qquad \qquad \qquad \qquad \qquad \quad ( x^2 )^2 + ( x^2 ) - 1 \ = \ 0. \qquad \qquad \qquad \qquad \qquad \qquad \quad \ (IV)

"Now let (for convenience): " \qquad u \ = \ x^2. \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \quad (V)

"So (IV) becomes: "

\qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \quad u^2 + u - 1 \ = \ 0.

"So by the Quadratic Formula: "

\qquad \qquad \qquad \qquad \qquad \qquad \quad u \ = \ { -1 \pm sqrt{ (1)^2 - 4 (1) (-1) } }/ (2 cdot 1 ).

\qquad :. \qquad \qquad \qquad \qquad \qquad \quad \ u \ = \ { -1 \pm sqrt{ 1 + 4 } }/2.

\qquad :. \qquad \qquad \qquad \qquad \qquad \quad \quad \ u \ = \ { -1 \pm sqrt{ 5 } }/2.

\qquad :. \qquad u \ = \ { -1 - sqrt{ 5 } }/2 \qquad \qquad "or \qquad \qquad u \ = \ { -1 + sqrt{ 5 } }/2.

"So by (V): "

\qquad \qquad \qquad x^2 \ = \ - { 1 + sqrt{ 5 } }/2 \qquad \qquad "and" \qquad \qquad x^2 \ = \ { -1 + sqrt{ 5 } }/2.

"Looking at the original equation, at the top, we see" \ x \ "must be"
"a real number because the domain of" \ \ tan(x) \ \ "is, in particular,"
"a subset of the real numbers. The first pair of equations in the"
"previous yields no real solutions for" \ x, \ "as" \ - { 1 + sqrt{ 5 } }/2 \ "is"
"negative. And the second pair yields two real solutions, as"
\ { -1 + sqrt{ 5 } }/2 \ "is positive [ certainly" \ sqrt{ 5 } > 1 \ "]."

\qquad :. \qquad \qquad \qquad \qquad \qquad \qquad x^2 \ = \ { -1 + sqrt{ 5 } }/2, \qquad "only".

\qquad :. \qquad \qquad \qquad \qquad \qquad \quad x \ = \ \pm sqrt{ { -1 + sqrt{ 5 } }/2 }.

\qquad :. \qquad \qquad \qquad \qquad \quad x \ = \ \pm sqrt{ { 2 ( -1 + sqrt{ 5 } ) }/{ 2 cdot 2 } }.

\qquad :. \qquad \qquad \qquad \qquad \qquad x \ = \ \pm sqrt{ -2 + 2sqrt{ 5 } }/{ 2 }.

"These are our solutions !!"

"So, summarizing:"

\qquad \quad o. \ "the solutions of:" \qquad tan^{-1} (x) = cos^{-1} (x)

\qquad \qquad \qquad \qquad \qquad \qquad \qquad \quad \quad \ "are:" \qquad \qquad \quad \ \ x \ = \ \pm sqrt{ -2 + 2sqrt{ 5 } }/{ 2 } \quad. \qquad \qquad square