Inverse of cosine

Question

Always we talk about inverse of cosine in $[0 , \pi]$ interval but the cosine function also is one-to-one in $[\pi , 2\pi] , [2\pi , 3\pi] , \dots$ . If we consider $[\pi , 2\pi]$ interval the graph of $\arccos$ will be different from $[0 , \pi]$. Is this true ? And if it is true many formulas will change !

Answer 1

Let us use notation $$\cos_n\colon [n\pi,(n+1)\pi]\to[-1,1]$$ and observe the appropriate inverses $$\cos_n^{-1}\colon[-1,1]\to[n\pi,(n+1)\pi].$$ Define $\arccos = \cos_0^{-1}$ as usual. We want to inspect how $\cos_n^{-1}$ and $\arccos$ are related.

First of all, draw the graph of $\cos\colon\Bbb R\to [-1,1]$ and reflect it along the line $y = x$. This gives you graphs of all $\cos_n^{-1}$ in one plot. It is easy to observe that

$$\cos_n^{-1}(x)-n\pi - \frac\pi 2 = \begin{cases}\arccos x - \frac\pi 2, \ \text{$n$ even}\\ -(\arccos x - \frac\pi 2),\ \text{$n$ odd}\end{cases}$$

or combining it into single formula: $$\cos_n^{-1}x = (-1)^n(\arccos x - \frac\pi 2) + n\pi + \frac\pi 2$$

Thus, there is no essential difference between $\cos_n^{-1}$ for different $n$ (the graphs are just translated and/or reflected), all the analytic properties can be obtained from $\arccos$ and usually, you don't even want to be bothered with the above formula if you are, for example, solving some trigonometric equation. You just use periodicity to obtain all the solutions.

Answer 2

The graph will be different, no doubt. You have to realise that the inverse trigonometric functions are multivalued. So we have to specify a domain, so that they are one-to-one in that domain. Accordingly, if we change the domain, we will get a different graph. But the characteristics of the graph will be the same; it will only be laterally shifted along the $\theta$ axis.

$$\frac12=\sin 30^\circ=\sin 150^\circ= \ldots $$

So, $$\arcsin \frac12=n\pi + (-1)^n\cdot \frac{\pi}{6}=30^\circ=150^\circ= \ldots $$

Hence, you see what actually happens.