If x ∈((3 π/2), 2 π), then the value of the expression sin -1[ cos cos -1( cos x) sin -1( sin x)], is

Question

If x ∈((3 π/2), 2 π), then the value of the expression sin -1[ cos cos -1( cos x) sin -1( sin x)], is (A) -(π/2) (B) (π/2) (C) 0 (D) π

Tardigrade Admin · Accepted Answer

x ∈((3 π/2), 2 π) Then, cos -1( cos x)=2 π-x and sin -1( sin x)=x-2 π ∴ cos -1( cos x)+ sin -1( sin x)=0 Therefore, sin -1[ cos cos -1( cos x)+ sin -1( sin x) ] = sin -1 cos (0) = sin -1(1)=(π/2)

Q. If $x \in (\frac{3 π}{2}, 2 π)$ , then the value of the expression $sin^{- 1} [cos {cos^{- 1} (cos x)} sin^{- 1} (sin x)]$ , is

$- \frac{π}{2}$

$\frac{π}{2}$

$0$

$π$

$x \in (\frac{3 π}{2}, 2 π)$
Then, $cos^{- 1} (cos x) = 2 π - x$
and $sin^{- 1} (sin x) = x - 2 π$
$∴ cos^{- 1} (cos x) + sin^{- 1} (sin x) = 0$
Therefore, $sin^{- 1} [cos {cos^{- 1} (cos x) + sin^{- 1} (sin x)}]$
$= sin^{- 1} {cos (0)} = sin^{- 1} (1) = \frac{π}{2}$

Q. If x∈(23π​,2π), then the value of the expression sin−1[cos{cos−1(cosx)}sin−1(sinx)], is

−2π​

2π​

0

π

x∈(23π​,2π) Then, cos−1(cosx)=2π−x and sin−1(sinx)=x−2π ∴cos−1(cosx)+sin−1(sinx)=0 Therefore, sin−1[cos{cos−1(cosx)+sin−1(sinx)}] =sin−1{cos(0)}=sin−1(1)=2π​

Q. If $x \in (\frac{3 π}{2}, 2 π)$ , then the value of the expression $sin^{- 1} [cos {cos^{- 1} (cos x)} sin^{- 1} (sin x)]$ , is

$- \frac{π}{2}$

$\frac{π}{2}$

$0$

$π$

$x \in (\frac{3 π}{2}, 2 π)$
Then, $cos^{- 1} (cos x) = 2 π - x$
and $sin^{- 1} (sin x) = x - 2 π$
$∴ cos^{- 1} (cos x) + sin^{- 1} (sin x) = 0$
Therefore, $sin^{- 1} [cos {cos^{- 1} (cos x) + sin^{- 1} (sin x)}]$
$= sin^{- 1} {cos (0)} = sin^{- 1} (1) = \frac{π}{2}$