Problem in the derivation of the relation between $\sin^{-1}(x)$ & $\cos^{-1}(x)$ My book's derivation:
Let, $\sin^{-1}x=\theta\implies \sin\theta=x$
Now, $\cos\theta=\sqrt{1-\sin^2\theta}=\sqrt{1-x^2}\implies\theta=\cos^{-1}\sqrt{1-x^2}$
So, $$\fbox{$\theta=\sin^{-1}x=\cos^{-1}\sqrt{1-x^2}$}$$
My problem:
My problem with this derivation is that $\sqrt{1-x^2}$ is always positive. So, the principal value of $\theta$ will always remain in the 1st quadrant. So, for negative values of $\cos\theta$, we will not be able to determine $\theta$ accurately. So, how will I find the correct value of $\theta$ when $\cos\theta$ is negative? So, how will I find the correct value of $\theta$ when $\cos\theta$ is negative?
Example of my problem:
Let $\theta_1=\cos^{-1}(\frac{-4}{5})=\sin^{-1}(\frac{3}{5})$ is in the 2nd quadrant, so $\theta_1=143.1301^{\circ}$, but according to the derived formula $\theta_2=\cos^{-1}(\sqrt{1-(\frac{3}{5})^2})=36.869^{\circ}$, so we can see that $\theta_1\neq \theta_2$ when they should've been equal. This formula can't differentiate between when $\cos^{-1}(x)$ is in the 1st quadrant or when it is in the second quadrant. How do I find the value of $\theta$ correctly when $\cos\theta$ is negative or is the formula so that I just can't?

In fact, to work with the inverse of the cosine function, we must devise a local concept that guarantees that such a function is bijective. Usually, for classic reasons, we define
$$\cos: [0,\pi] \longrightarrow [-1,1]$$
so that it makes sense to define $\cos^{-1}:[-1,1] \longrightarrow [0,\pi]$. That's why the principal value of the $\theta$ angle you mentioned is always in the first quadrant.
