"Provearccos(x−1)(x+1)=-2arctanx4+π2." solved and explained - Plainmath

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Calculus and Analysis

Answered question

2022-04-19

Prove $arc \cos \frac{(\sqrt{x} - 1)}{(\sqrt{x} + 1)} = - 2 \arctan \sqrt[4]{x} + \frac{π}{2}$ .

Answer & Explanation

RizerMix

Expert2023-04-28Added 656 answers

To prove that

\arccos (\frac{\sqrt{x} - 1}{\sqrt{x} + 1}) = - 2 \arctan (\sqrt{4 x} + \frac{π}{2})

, we'll start by finding the value of

\cos (- 2 \arctan (\sqrt{4 x} + \frac{π}{2}))

.
Using the identity

\cos (2 θ) = \frac{1}{\tan^{2} (θ) + 1}

, we have:

\cos (- 2 \arctan (\sqrt{4 x} + \frac{π}{2})) & = \cos (\arctan (\frac{- 2 \sqrt{x}}{\sqrt{4 x} + π}))

= \frac{1}{\sqrt{1 + {(\frac{- 2 \sqrt{x}}{\sqrt{4 x} + π})}^{2}}}

= \frac{1}{\frac{2 x + 2 π \sqrt{x} + π^{2}}{(\sqrt{4 x} + π)^{2}}}

= \frac{(\sqrt{4 x} + π)^{2}}{2 x + 2 π \sqrt{x} + π^{2}}

= \frac{4 x + 2 π \sqrt{x} + π^{2}}{2 x + 2 π \sqrt{x} + π^{2}}

= 1 - \frac{2 x}{2 x + 2 π \sqrt{x} + π^{2}}

Next, we'll simplify

\frac{\sqrt{x} - 1}{\sqrt{x} + 1}

:

\frac{\sqrt{x} - 1}{\sqrt{x} + 1} & = \frac{(\sqrt{x} - 1) (\sqrt{x} - 1)}{(\sqrt{x} + 1) (\sqrt{x} - 1)}

= \frac{x - 2 \sqrt{x} + 1}{x - 1}

= 1 - \frac{2 \sqrt{x}}{x - 1}

Now, we can take the

\arccos

of this expression:

\arccos (\frac{\sqrt{x} - 1}{\sqrt{x} + 1}) = \arccos (1 - \frac{2 \sqrt{x}}{x - 1})

= \arcsin (\frac{2 \sqrt{x}}{x - 1})

= \arcsin (\frac{2 \sqrt{x}}{x - 1} \cdot \frac{x + 1}{x + 1})

= \arcsin (\frac{2 \sqrt{x} (x + 1)}{x^{2} - 1})

Comparing this result to our expression for

\cos (- 2 \arctan (\sqrt{4 x} + \frac{π}{2}))

, we can see that:

\arccos (\frac{\sqrt{x} - 1}{\sqrt{x} + 1}) = - 2 \arctan (\sqrt{4 x} + \frac{π}{2})

as required.

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