Question

Решите пожалуйста 16.3 и 16.4

Answer 1

$\boxed {\ y=arcsinx\ \ ,\ \ x\in [-1\, ;\, 1\ ]\ \ ,\ \ y\in [\, -\dfrac{\pi}{2}\, ;\, \dfrac{\pi}{2}\ ]\ }\\\\\\16.4.\ \ 1)\ y=arcsinx\ \ ,\ \ x\in D(y)=[-1\, ;\, 1\ ]\\\\\\2)\ y=arcsin0,5x\ \ ,\ \ -1\leq 0,5x\leq 1\ \ ,\ \ -2\leq x\leq 2\ \ ,\ \ x\in D(y)=[-2\, ;\, 2\ ]\\\\\\3)\ y=arcsin(3-2x)\ \ ,\ \ -1\leq 3-2x\leq 1\ \ ,\ \ -4\leq -2x\leq -2\ \ ,\\\\2\leq 2x\leq 4\ \ ,\ \ 1\leq x\leq 2\ \ ,\ \ x\in D(y)=[\, 1\, ;\, 2\ ]$

$4)\ y=arcsin(3x+1)\ \ ,\ \ -1\leq 3x+1\leq 1\ \ ,\ \ -2\leq 3x\leq 0\ \ ,\\\\-\dfrac{2}{3} \leq x\leq 0\ \ ,\ \ x\in D(y)=[\, -\dfrac{2}{3}\, ;\, 0\ ]\\\\\\5)\ y=arcsin(x^2-1)\ \ ,\ \ -1\leq x^2-1\leq 1\ \ ,\ \ 0\leq x^2\leq 2\ \ ,\\\\-\sqrt2\leq x\leq \sqrt2\ \ ,\ \ x\in D(y)=[\, -\sqrt2\, ;\, \sqrt2\ ]\\\\\\6)\ y=arcsin(x^2-3)\ \ ,\ \ -1\leq x^2-3\leq 1\ \ ,\ \ -2\leq x^2\leq 4\ \ ,\\\\0\leq x^2\leq 4\ \ ,\ \ -2\leq x\leq 2\ \ ,\ \ x\in D(y)=[\, -2\, ;\, 2\ ]$

$16.3.\ \ arcsin(sin\dfrac{\pi }{8})=\dfrac{\pi}{8}\ \ ,\ \ tak\ kak\ \ -1\leq sin\dfrac{\pi}{8}\leq 1\\\\arcsin(sin\dfrac{\pi}{12})=\dfrac{\pi }{12}\ \ ,\ \ tak\ kak\ \ -1\leq sin\dfrac{\pi}{12}\leq 1\\\\arcsin(cos\pi )=arcsin (-1)=-arcsin1=-\dfrac{\pi}{2}\\\\arcsin(cos\dfrac{\pi}{4})=arcsin\dfrac{\sqrt2}{2}=\dfrac{\pi}{4}\\\\arcsin(cos(-\dfrac{\pi}{6}))=arcsin(cos\dfrac{\pi}{6})=arcsin\dfrac{\sqrt3}{2}=\dfrac{\pi}{3}\\\\arcsin(cos\dfrac{2\pi}{3})=arcsin(-\dfrac{1}{2})=-arcsin\dfrac{1}{2}=-\dfrac{\pi}{6}$