Expert Answer to "Derivative of cos−11+x2 using substitution"

pierdoodsu

Answered question

2021-12-30

Derivative of

\cos^{- 1} \sqrt{\frac{1 + x}{2}}

using substitution

Answer & Explanation

John Koga

Beginner2021-12-31Added 33 answers

Setting

x = \cos 2 θ

and remembering that

\cos 2 θ = 2 \cos 2 θ - 1

leads to:

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

= \cos^{- 1} \sqrt{\frac{1 + 2 \cos^{2} θ - 1}{2}}

= θ

And so one can use:

\frac{d}{d x} \cos^{- 1} \sqrt{\frac{1 + x}{2}} = \frac{d θ}{d x} = \frac{1}{\frac{d x}{d θ}}

Thomas Nickerson

Beginner2022-01-01Added 32 answers

Let

x = \cos (2 θ)

then

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

and

\frac{d x}{d θ} = - 2 \sin (2 θ)

\frac{d}{d x} (\cos^{- 1} \sqrt{\frac{1 + x}{2}}) = - \frac{12}{\frac{1}{\sin (2 θ)}} = - \frac{12}{\frac{1}{\sqrt{1 - x^{2}}}}

nick1337

Expert2022-01-08Added 777 answers

I think the natural way should be:
Let $\cos^{- 1} \sqrt{\frac{1 + x}{2}} = y$
$\Rightarrow 0 \leq y \leq \frac{π}{2}$ for real $1 \geq x \geq - 1$
$\Rightarrow x = \dots = \cos 2 y$
and consequently $2 y = \cos^{- 1} x a s 0 \leq \cos^{- 1} x \leq π$

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