find-a-simple-expression-for-y-sin-1-sin-x-x-R-and-x-in-rad-

Question Number 128546 by mr W last updated on 08/Jan/21

$${find}\:{a}\:{simple}\:{expression}\:{for} \\ $$$${y}=\mathrm{sin}^{−\mathrm{1}} \left(\mathrm{sin}\:{x}\right) \\ $$$${x}\in\mathbb{R}\:{and}\:{x}\:{in}\:{rad}. \\ $$

Commented by liberty last updated on 08/Jan/21

sin x = sin y ; y = x+2kπ where −(π/2)<y<(π/2)

$$\mathrm{sin}\:\mathrm{x}\:=\:\mathrm{sin}\:\mathrm{y}\:;\:\mathrm{y}\:=\:\mathrm{x}+\mathrm{2k}\pi\: \\ $$$$\mathrm{where}\:−\frac{\pi}{\mathrm{2}}<\mathrm{y}<\frac{\pi}{\mathrm{2}} \\ $$$$ \\ $$

Commented by mr W last updated on 08/Jan/21

example: x=5 how do you get y=f(5) with your formula?

$${example}: \\ $$$${x}=\mathrm{5} \\ $$$${how}\:{do}\:{you}\:{get}\:{y}={f}\left(\mathrm{5}\right)\:{with}\:{your} \\ $$$${formula}? \\ $$

Commented by john_santu last updated on 08/Jan/21

$$\mathrm{do}\:\mathrm{you}\:\mathrm{meant}\:\mathrm{5}\:\mathrm{in}\:\mathrm{degress}?\:\mathrm{or}\:\mathrm{radiant}−? \\ $$

Commented by Olaf last updated on 08/Jan/21

f(5) = (−1)^(⌊(5/π)+(1/2)⌋) (5−⌊(5/π)+(1/2)⌋π) ⌊(5/π)+(1/2)⌋ ≈ ⌊2,091⌋ = 2 ⇒ f(5) = (−1)^2 (5−2π) = 5−2π

$${f}\left(\mathrm{5}\right)\:=\:\left(−\mathrm{1}\right)^{\lfloor\frac{\mathrm{5}}{\pi}+\frac{\mathrm{1}}{\mathrm{2}}\rfloor} \left(\mathrm{5}−\lfloor\frac{\mathrm{5}}{\pi}+\frac{\mathrm{1}}{\mathrm{2}}\rfloor\pi\right) \\ $$$$\lfloor\frac{\mathrm{5}}{\pi}+\frac{\mathrm{1}}{\mathrm{2}}\rfloor\:\approx\:\lfloor\mathrm{2},\mathrm{091}\rfloor\:=\:\mathrm{2} \\ $$$$\Rightarrow\:{f}\left(\mathrm{5}\right)\:=\:\left(−\mathrm{1}\right)^{\mathrm{2}} \left(\mathrm{5}−\mathrm{2}\pi\right)\:=\:\mathrm{5}−\mathrm{2}\pi \\ $$

Answered by Olaf last updated on 08/Jan/21

∀x∈R, ∃k∈Z \ x−kπ ∈[−(π/2),+(π/2)[ ⇔ kπ ≤ x+(π/2) < (k+1)π ⇔ k ≤ (x/π)+(1/2) < k+1 k = ⌊(x/π)+(1/2)⌋ (integer part) and ∀θ∈R, ∀q∈Z, sin(θ+qπ) = (−1)^q sinθ ⇒ sinx = sin(x−kπ+kπ) = (−1)^k sin(x−kπ) ⇒ arcsin(sinx) = arcsin((−1)^k sin(x−kπ)) = (−1)^k arcsin(sin(x−kπ)) and x−kπ ∈[−(π/2),+(π/2)[ ⇒ arcsin(sinx) =(−1)^k (x−kπ) Formula for arcsin(sinx) : arcsin(sinx) = (−1)^(⌊(x/π)+(1/2)⌋) (x−⌊(x/π)+(1/2)⌋π)

$$\forall{x}\in\mathbb{R},\:\exists{k}\in\mathbb{Z}\:\backslash\:{x}−{k}\pi\:\in\left[−\frac{\pi}{\mathrm{2}},+\frac{\pi}{\mathrm{2}}\left[\right.\right. \\ $$$$\Leftrightarrow\:{k}\pi\:\leqslant\:{x}+\frac{\pi}{\mathrm{2}}\:<\:\left({k}+\mathrm{1}\right)\pi \\ $$$$\Leftrightarrow\:{k}\:\leqslant\:\frac{{x}}{\pi}+\frac{\mathrm{1}}{\mathrm{2}}\:<\:{k}+\mathrm{1} \\ $$$${k}\:=\:\lfloor\frac{{x}}{\pi}+\frac{\mathrm{1}}{\mathrm{2}}\rfloor\:\left(\mathrm{integer}\:\mathrm{part}\right) \\ $$$$\mathrm{and}\:\forall\theta\in\mathbb{R},\:\forall{q}\in\mathbb{Z},\:\mathrm{sin}\left(\theta+{q}\pi\right)\:=\:\left(−\mathrm{1}\right)^{{q}} \mathrm{sin}\theta \\ $$$$\Rightarrow\:\mathrm{sin}{x}\:=\:\mathrm{sin}\left({x}−{k}\pi+{k}\pi\right)\:=\:\left(−\mathrm{1}\right)^{{k}} \mathrm{sin}\left({x}−{k}\pi\right) \\ $$$$\Rightarrow\:\mathrm{arcsin}\left(\mathrm{sin}{x}\right)\:=\:\mathrm{arcsin}\left(\left(−\mathrm{1}\right)^{{k}} \mathrm{sin}\left({x}−{k}\pi\right)\right) \\ $$$$=\:\left(−\mathrm{1}\right)^{{k}} \mathrm{arcsin}\left(\mathrm{sin}\left({x}−{k}\pi\right)\right) \\ $$$$\mathrm{and}\:{x}−{k}\pi\:\in\left[−\frac{\pi}{\mathrm{2}},+\frac{\pi}{\mathrm{2}}\left[\right.\right. \\ $$$$\Rightarrow\:\mathrm{arcsin}\left(\mathrm{sin}{x}\right)\:=\left(−\mathrm{1}\right)^{{k}} \left({x}−{k}\pi\right) \\ $$$$ \\ $$$$\boldsymbol{\mathrm{Formula}}\:\boldsymbol{\mathrm{for}}\:\boldsymbol{\mathrm{arcsin}}\left(\boldsymbol{\mathrm{sin}{x}}\right)\:: \\ $$$$\mathrm{arcsin}\left(\mathrm{sin}{x}\right)\:=\:\left(−\mathrm{1}\right)^{\lfloor\frac{{x}}{\pi}+\frac{\mathrm{1}}{\mathrm{2}}\rfloor} \left({x}−\lfloor\frac{{x}}{\pi}+\frac{\mathrm{1}}{\mathrm{2}}\rfloor\pi\right) \\ $$

Commented by mr W last updated on 08/Jan/21

$${thanks}\:{alot}! \\ $$

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