## Trigonometry (11th Edition) Clone

$$y=\pi$$
$$y=\cos^{-1}(-1)$$ First, we see that the domain of inverse cosine function is $[-1,1]$. $-1$ lies in this range, so $\cos^{-1}(-1)$ does exist. Also, it should be noted that the range of inverse cosine function is $[0,\pi]$. In other words, $y\in[0,\pi]$. We can rewrite $y=\cos^{-1}(-1)$ into $\cos y=-1$ In the range $[0,\pi]$, we find that only $\cos\pi=-1$ That means, the exact value of $y$ here is $$y=\pi$$