A Negative Inside

A negative input moves the inner angle out of quadrant I, and the range decides where. θ = arcsin(−3/5) is between −90° and 0°, in quadrant IV, where the point is (4, −3) on a circle of radius 5. So $\tan\left(\arcsin\left(\frac{-3}{5}\right)\right) = \frac{-3}{4}$, but $\cos\left(\arcsin\left(\frac{-3}{5}\right)\right) = \frac{4}{5}$ is positive: cosine is positive in quadrant IV.

Arccos of a negative number is in quadrant II, between 90° and 180°, where x is negative and y positive. Work the sizes out on a triangle as before, then give each side the sign of its quadrant.