solve-in-R-arcsin-x-arcsin-1-x-2-pi-2-

Question Number 168084 by mathocean1 last updated on 02/Apr/22

s o l v e i n R

a r c s i n (x) + a r c s i n (\sqrt{1 - x^{2}}) = \frac{π}{2}

Commented by MJS_new last updated on 02/Apr/22

x ⩾ 0

Commented by mathocean1 last updated on 03/Apr/22

b u t f o r x = 2 t h e e q u a t i o n {d o e s n}^{'} t

h a v e s e n s .

Commented by MJS_new last updated on 03/Apr/22

\arcsin a + \arcsin b =

= \arcsin (a \sqrt{1 - b^{2}} + b \sqrt{1 - a^{2}})

with a = x \land b = \sqrt{1 + x^{2}} we get

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

= \arcsin (x ∣ x ∣ - x^{2} + 1)

with x < 0 we get \arcsin (1 - 2 x^{2})

with x ⩾ 0 we get \arcsin 1 = \frac{π}{2}

{doesn}^{'} t matter if \arcsin x and \arcsin \sqrt{1 - x^{2}}

are not real, the imaginary part vanishes

\arcsin 2 \approx \frac{π}{2} - 1 . 31696 i

\arcsin \sqrt{1 - 2^{2}} \approx 1 . 31696 i

Answered by Tyller last updated on 04/Apr/22

s e n (a r c s e n x + a r c s e n (\sqrt{1 - x^{2}}) = 1

x \times \sqrt{1 - 1 + x^{2}} + \sqrt{1 - x^{2}} \times \sqrt{1 - x^{2}} = 1

x ∣ x ∣ + ∣ 1 - x^{2} ∣= 1

p a r a x ⩾ 0 \cap x \in [- 1, 1] \Rightarrow x \in [0, 1]

x \times x + 1 - x^{2} = 1 \Leftrightarrow x \in [0, 1]

p a r a x < 0 e (1 - x^{2}) ⩾ 0 \Rightarrow x \in [- 1, 0)

- x^{2} + 1 - x^{2} = 1 \Rightarrow x = 0 ⇈

p / x ⩾ 0 e (1 - x^{2}) < 0 \Rightarrow x \in (- \propto . - 1) u (1 . + \propto)

x^{2} + x^{2} - 1 = 1 \Rightarrow x = \pm 1 \notin x

s g = [0.1]

Answered by Mathspace last updated on 04/Apr/22

l e t φ (x) = a r c s i n x + a r c s i n \sqrt{1 - x^{2}}

φ^{^{'}} (x) = \frac{1}{\sqrt{1 - x^{2}}} - \frac{x}{\sqrt{1 - x^{2}}} \times \frac{1}{\sqrt{1 - (1 - x^{2})}}

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

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

s o i f x \in] 0, 1 [w e g e t φ^{^{'}} (x) = 0 \Rightarrow

Answered by Mathspace last updated on 04/Apr/22

\Rightarrow φ (x) = C

x = 0 \Rightarrow o + a r c s i n (1) = C \Rightarrow

C = \frac{π}{2} \Rightarrow s e t o f s o l u t i o n i s [0, 1]

Commented by MJS_new last updated on 04/Apr/22

nice . but φ^{'} (x) is zero for x \in (0; + \infty)

\Rightarrow φ (x) = C \in R for x ⩾ 0 .

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