Question

arcsin(-1/root2)

Answer 1

-\pi/4

Answer 2

To find the value of $\arcsin(-1/\sqrt{2})$, we need to find an angle $\theta$ such that $\sin(\theta) = -1/\sqrt{2}$ and $\theta$ lies in the principal value branch of $\arcsin(x)$, which is $[-\pi/2, \pi/2]$.

1. Recall known values: We know that $\sin(\pi/4) = 1/\sqrt{2}$.
2. Consider the negative sign: Since we are looking for $\sin(\theta) = -1/\sqrt{2}$, and the principal value branch for $\arcsin$ is $[-\pi/2, \pi/2]$, the angle must be in the fourth quadrant (where sine is negative).
3. Apply the property of odd function: For the sine function, $\sin(-\theta) = -\sin(\theta)$. Therefore, $\sin(-\pi/4) = -\sin(\pi/4) = -1/\sqrt{2}$.
4. Check the range: The angle $-\pi/4$ lies within the principal value branch $[-\pi/2, \pi/2]$.

Thus, $\arcsin(-1/\sqrt{2}) = -\pi/4$.

The function $\arcsin(x)$ gives the principal value of the angle whose sine is $x$. The range of $\arcsin(x)$ is $[-\pi/2, \pi/2]$. We need to find an angle $\theta$ in this range such that $\sin(\theta) = -1/\sqrt{2}$. We know that $\sin(\pi/4) = 1/\sqrt{2}$. Since $\sin(-\theta) = -\sin(\theta)$, we have $\sin(-\pi/4) = -\sin(\pi/4) = -1/\sqrt{2}$. The angle $-\pi/4$ lies in the range $[-\pi/2, \pi/2]$.