prove-that-1-2-tan-1-x-cos-1-1-1-x-2-2-1-x-2-using-substitution-x-cos-2-

Question Number 74526 by Kunal12588 last updated on 25/Nov/19

p r o v e t h a t

\frac{1}{2} {t a n}^{- 1} x = {c o s}^{- 1} (\sqrt{\frac{1 + \sqrt{1 + x^{2}}}{2 \sqrt{1 + x^{2}}}})

u s i n g s u b s t i t u t i o n x = c o s 2 θ

Answered by mind is power last updated on 25/Nov/19

m i s s c l i c k

i t h i n k i s x = t g (2 θ)

t r u e f o r x ⩾ 0

x = - 1 {t a n}^{-} (- 1) = - \frac{π}{4}

w e g e t - \frac{π}{8}

{c o s}^{-} (x) \in [0, π]

s i n c e t \to t g (2 θ) i s b j e c t i o n

I =] 0, \frac{π}{4} [\to R^{+}

\Rightarrow \forall x \in R \exists! θ \in I ∣ x = t g (2 θ)

\tan^{- 1} (t g (2 θ)) = 2 θ s i n c e 2 θ \in] 0, \frac{π}{2} [

l H s = θ

1 + x^{2} = \frac{1}{{c o s}^{2} (2 θ)} \Rightarrow \sqrt{1 + x^{2}} = \frac{1}{c o s (2 θ)}, s i n c e c o s (2 θ) ⩾ 0 o v e r I

\sqrt{\frac{1 + \sqrt{1 + x^{2}}}{2 \sqrt{1 + x^{2}}}} = \sqrt{\frac{1 + \frac{1}{c o s (2 θ)}}{\frac{2}{c o s (2 θ)}}} = \sqrt{\frac{1 + c o s (2 θ)}{2}}

1 + c o s (2 θ) = 2 {c o s}^{2} (θ) \Rightarrow \sqrt{\frac{2 {c o s}^{2} (θ)}{2}} =∣ c o s (θ) ∣= c o s (θ)

{c o s}^{-} (c o s (θ)) = θ c a u s e θ \in] 0, \frac{π}{4} [

s o θ = θ t h i s \Rightarrow \forall x ⩾ 0 \frac{{t a n}^{-} (x)}{2} = \cos^{- 1} (\frac{\sqrt{1 + \sqrt{1 + x^{2}}}}{2 \sqrt{1 + x^{2}}})