If $x$ takes a negative value, prove that $\sin^{-1}x=-\cos^{-1}\sqrt{1-x^2}$

Question

Let $\sin^{-1}x=y$

$$\sin y= x$$

Therefore $$\cos y =\sqrt {1-x^2}$$

$$y=\cos^{-1}\sqrt{1-x^2}$$ Now $x$ is negative, so $y$ will also be negative if we work in $[-\frac{\pi}{2},\frac{\pi}{2}]$

But then $\cos y$ will always be positive. Even if it isn’t, I don’t know why we need to explicitly use a $-$ before the answer. Please explain why we need to add the minus sign

Answer 1

If $x$ is negative, it will range from $-1$ to $0$, i.e. $\sin^{-1}x \in \left[-\frac{\pi}{2}, 0\right)$. Now, $$0\le \sqrt{1-x^2} \lt 1 \\ \implies 0\lt \cos^{-1} \sqrt{1-x^2} \le \frac{\pi}{2} $$ We clearly need to multiply this by $-1$ to ensure that it lies in the range of $\sin^{-1} x$.

Answer 2

Note that, for $t\in[0,\frac\pi2]$, we have $\sin^{-1}t=\cos^{-1}\sqrt{1-t^2}$. Then, for negative $x=-t$, we get $\sin^{-1}(-x)=\cos^{-1}\sqrt{1-(-x)^2}$, which is

$$\sin^{-1}x=-\cos^{-1}\sqrt{1-x^2}$$