$\sin^{-1}(\cos\frac{3\pi}{5})$ equals

The expression is $\sin^{-1}(\cos\frac{3\pi}{5})$. To simplify, convert cosine into sine.

Using the identity $\cos\theta = \sin\left(\frac{\pi}{2} - \theta\right)$ with $\theta = \frac{3\pi}{5}$:

$\cos\frac{3\pi}{5} = \sin\left(\frac{\pi}{2} - \frac{3\pi}{5}\right)$

To subtract the fractions, find a common denominator:

$\frac{\pi}{2} - \frac{3\pi}{5}$

$= \frac{5\pi}{10} - \frac{6\pi}{10}$

$= \frac{-\pi}{10}$

Therefore:

$\cos\frac{3\pi}{5} = \sin\left(\frac{-\pi}{10}\right)$

Substituting back into the original expression:

$\sin^{-1}(\cos\frac{3\pi}{5})$

$= \sin^{-1}\left(\sin\frac{-\pi}{10}\right)$

For $\sin^{-1}(\sin x) = x$, the value $x$ must be in the range $\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]$.

Since $-\frac{\pi}{2} < \frac{-\pi}{10} < \frac{\pi}{2}$, the value is in the valid range.

Therefore:

$\sin^{-1}\left(\sin\frac{-\pi}{10}\right) = \frac{-\pi}{10}$

The answer is $\frac{-\pi}{10}$.