We are asked to find the value of \( \sin^{-1}\left(\sin \left(\frac{5\pi}{6}\right)\right) \). First, recall that the function \( \sin^{-1}(x) \) (inverse sine) is defined on the principal range \( \left[ -\frac{\pi}{2}, \frac{\pi}{2} \right] \). This means that for any angle \( \theta \), \( \sin^{-1}(\sin \theta) \) will return the value of \( \theta \) within this range.
Now, consider the angle \( \frac{5\pi}{6} \). This angle is greater than \( \frac{\pi}{2} \), so we need to adjust it to fall within the principal range.
Since \( \sin \left( \frac{5\pi}{6} \right) = \sin \left( \frac{\pi}{6} \right) \), we have:
\[ \sin^{-1} \left( \sin \left( \frac{5\pi}{6} \right) \right) = \sin^{-1} \left( \sin \left( \frac{\pi}{6} \right) \right) \] Therefore, the result is \( \frac{\pi}{6} \), as \( \frac{\pi}{6} \) is within the principal range of \( \sin^{-1} \).
Thus, the correct answer is option (E), \( \frac{\pi}{6} \).
Let \( f(x) = \frac{x^2 + 40}{7x} \), \( x \neq 0 \), \( x \in [4,5] \). The value of \( c \) in \( [4,5] \) at which \( f'(c) = -\frac{1}{7} \) is equal to:
The general solution of the differential equation \( \frac{dy}{dx} = xy - 2x - 2y + 4 \) is:
The minimum value of the function \( f(x) = x^4 - 4x - 5 \), where \( x \in \mathbb{R} \), is:
The critical points of the function \( f(x) = (x-3)^3(x+2)^2 \) are:
For the reaction:
\[ 2A + B \rightarrow 2C + D \]
The following kinetic data were obtained for three different experiments performed at the same temperature:
\[ \begin{array}{|c|c|c|c|} \hline \text{Experiment} & [A]_0 \, (\text{M}) & [B]_0 \, (\text{M}) & \text{Initial rate} \, (\text{M/s}) \\ \hline I & 0.10 & 0.10 & 0.10 \\ II & 0.20 & 0.10 & 0.40 \\ III & 0.20 & 0.20 & 0.40 \\ \hline \end{array} \]
The total order and order in [B] for the reaction are respectively: