For the principal value branch, the value of $\sin\left(\frac{\pi}{2} - \sin^{-1}\left(-\frac{\sqrt{3}}{2}\right)\right)$ is

The value of $\sin\left(\frac{\pi}{2} - \sin^{-1}\left(-\frac{\sqrt{3}}{2}\right)\right)$ needs to be found.

---

$\sin^{-1}(x)$ gives the angle whose sine is $x$, where the angle is between $-\frac{\pi}{2}$ and $\frac{\pi}{2}$ (principal value branch).

Since $\sin\left(\frac{\pi}{3}\right) = \frac{\sqrt{3}}{2}$ and sine is negative in the fourth quadrant:

$\sin^{-1}\left(-\frac{\sqrt{3}}{2}\right) = -\frac{\pi}{3}$

---

Using the complementary angle formula $\sin\left(\frac{\pi}{2} - \theta\right) = \cos(\theta)$:

$\sin\left(\frac{\pi}{2} - \sin^{-1}\left(-\frac{\sqrt{3}}{2}\right)\right) = \cos\left(\sin^{-1}\left(-\frac{\sqrt{3}}{2}\right)\right)$

This asks for the cosine of the angle whose sine is $-\frac{\sqrt{3}}{2}$.

---

Let $\theta = \sin^{-1}\left(-\frac{\sqrt{3}}{2}\right)$, so $\sin(\theta) = -\frac{\sqrt{3}}{2}$.

Using $\sin^2(\theta) + \cos^2(\theta) = 1$:

$\cos^2(\theta) = 1 - \sin^2(\theta)$

$\cos^2(\theta) = 1 - \left(-\frac{\sqrt{3}}{2}\right)^2$

$\cos^2(\theta) = 1 - \frac{3}{4}$

$\cos^2(\theta) = \frac{1}{4}$

$\cos(\theta) = \pm\frac{1}{2}$

---

Since $\theta$ is in the range $\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]$ (principal value branch), cosine is always non-negative in this range.

Therefore $\cos(\theta) = \frac{1}{2}$

The value of $\sin\left(\frac{\pi}{2} - \sin^{-1}\left(-\frac{\sqrt{3}}{2}\right)\right) = \frac{1}{2}$