Question and Answers Forum

All Questions      Topic List

None Questions

Previous in All Question      Next in All Question      

Previous in None      Next in None      

Question Number 128464 by n0y0n last updated on 07/Jan/21

sin^(−1) (sin92°)= ?

$$\mathrm{sin}^{−\mathrm{1}} \left(\mathrm{sin92}°\right)=\:? \\ $$

Commented by mr W last updated on 07/Jan/21

88°

$$\mathrm{88}° \\ $$

Answered by Olaf last updated on 08/Jan/21

  92° = (π/(180))×92 rad = ((23π)/(45)) rad  ∀x∈R, arcsin(sinx) = (−1)^(⌊(x/π)+(1/2)⌋) (x−⌊(x/π)+(1/2)⌋π)  (see another question above)  ⌊(((23π)/(45))/π)+(1/2)⌋ ≈ ⌊((91)/(90))⌋ = 1  ⇒ arcsin(sin92°) = (−1)^1 (((23π)/(45))−π) = ((22π)/(45))

$$ \\ $$$$\mathrm{92}°\:=\:\frac{\pi}{\mathrm{180}}×\mathrm{92}\:\mathrm{rad}\:=\:\frac{\mathrm{23}\pi}{\mathrm{45}}\:\mathrm{rad} \\ $$$$\forall{x}\in\mathbb{R},\:\mathrm{arcsin}\left(\mathrm{sin}{x}\right)\:=\:\left(−\mathrm{1}\right)^{\lfloor\frac{{x}}{\pi}+\frac{\mathrm{1}}{\mathrm{2}}\rfloor} \left({x}−\lfloor\frac{{x}}{\pi}+\frac{\mathrm{1}}{\mathrm{2}}\rfloor\pi\right) \\ $$$$\left(\mathrm{see}\:\mathrm{another}\:\mathrm{question}\:\mathrm{above}\right) \\ $$$$\lfloor\frac{\frac{\mathrm{23}\pi}{\mathrm{45}}}{\pi}+\frac{\mathrm{1}}{\mathrm{2}}\rfloor\:\approx\:\lfloor\frac{\mathrm{91}}{\mathrm{90}}\rfloor\:=\:\mathrm{1} \\ $$$$\Rightarrow\:\mathrm{arcsin}\left(\mathrm{sin92}°\right)\:=\:\left(−\mathrm{1}\right)^{\mathrm{1}} \left(\frac{\mathrm{23}\pi}{\mathrm{45}}−\pi\right)\:=\:\frac{\mathrm{22}\pi}{\mathrm{45}} \\ $$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com