if f(x)=sin^(−1) (cos[x]) find Df and Rf the function notice/ [...] is floor


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Question Number 87625 by M±th+et£s last updated on 05/Apr/20

$i f f (x) = {s i n}^{- 1} (c o s [x])$ $f i n d D f a n d R f t h e f u n c t i o n$ $n o t i c e / [. . .] i s f l o o r$
	Answered by mind is power last updated on 05/Apr/20

	${s i n}^{-} i s d e f i n d o v e r / [- 1, 1]$ $v o s [x] \in [- 1, 1] f i s d e f i n d o v e r R$ $c o s (n) i s d e n s e i n [- 1, 1]$ $\Rightarrow R f =] - \frac{π}{2}; \frac{π}{2} [$
	Commented by M±th+et£s last updated on 05/Apr/20

	$t h a n k y o u s i r$
	Commented by mind is power last updated on 05/Apr/20

	$c a n y o u s i r p l e a s r r e p o s t ? \int \frac{d x}{{((x + 1) . . . . (x + n))}^{2}}$
	Answered by mr W last updated on 06/Apr/20

	$i^{'} l l t r y t o a n s w e r t h e s e c o n d p a r t o f t h e$ $q u e s t i o n : r a n g e o f t h e f u n c t i o n = ?$ $\sin^{- 1} (\cos [x]) = y$ $a c c . t o d e f i n i t i o n : - \frac{π}{2} ⩽ y ⩽ \frac{π}{2}$ $\Rightarrow \cos [x] = \sin y = \cos (\frac{π}{2} - y)$ $\Rightarrow [x] = 2 n π \pm (\frac{π}{2} - y) = m$ $\Rightarrow y = \frac{π}{2} \pm (m - 2 n π)$ $\Rightarrow - \frac{π}{2} ⩽ \frac{π}{2} \pm (m - 2 n π) ⩽ \frac{π}{2}$ $\Rightarrow - π ⩽ \pm (m - 2 n π) ⩽ 0$ $\Rightarrow (2 n - 1) π ⩽ m ⩽ 2 n π$ $\Rightarrow ⌈ (2 n - 1) π ⌉ ⩽ m ⩽ ⌊ 2 n π ⌋$ $\Rightarrow 0 ⩽ m - 2 n π ⩽ π$ $\Rightarrow 2 n π ⩽ m ⩽ (2 n + 1) π$ $\Rightarrow ⌈ 2 n π ⌉ ⩽ m ⩽ ⌊ (2 n + 1) π ⌋$ $\Rightarrow y = \frac{π}{2} + m - 2 n π w i t h ⌈ (2 n - 1) π ⌉ ⩽ m ⩽ ⌊ 2 n π ⌋$ $\Rightarrow y = \frac{π}{2} - m + 2 n π w i t h ⌈ 2 n π ⌉ ⩽ m ⩽ ⌊ (2 n + 1) π ⌋$
	Commented by M±th+et£s last updated on 06/Apr/20

	$g o d b l e s s y o u s i r$
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