when to use restrictions (domain and range) on trig functions I'm doing math practice problems from "Precalculus for Dummies 1,000 Practice Problems Book" and I'm confused about when to apply restrictions to trig function questions. This book has all the solutions step by step in the back so I know how the problem is solved. What confuses me is why the restrictions are used in some problems and not others. 
The restrictions that I'm talking about are: 
The restrictions for the regular trig functions should be the the same but the range and domain's values are switched if I'm not mistaken.
These were the main problems that made me confused:


*

*Find an exact value of $y$, $y=\arcsin\left(\frac{\sqrt{3}}{2}\right)$


*

*if I solve for $y$ it equals $2\pi/3$ and $4\pi/3$, but since $\cos(x)$ has the restrictions where domain is $[-1,1]$ and the range is $[0,\pi]$ the final answer is just $2\pi/3$


*Find an exact value of $y$, $y= \cos(\arctan(-1))$


*

*in this case, while solving for y, I get to a step that looks like this: 
$y=\cos(7\pi/4)$. The answers/solutions in the back then show $y= \sqrt{2}/2$ is the answer. 

*However, I thought $\cos(x)$'s domain should have been restricted to $[-1,1]$. Why is the $7\pi/4$ allowed?

*At this point I was thinking, then do the restrictions only apply to inverse trig functions? But that didn't make sense either because I should be able to rewrite $y=\cos(7\pi/4)$ as $\arccos(y)=7\pi/4$; the "$7\pi/4$" part is not part of $\arccos$'s range restriction, $[0,\pi]$


*Find all solutions of the equation in the interval $[0^\circ,360^\circ)$. $2cos^2x-2=3\cos x$ 


*

*I get to a step where $\cos x= -1/2$ and $2$. I solve the $\cos x= -1/2$ part which equals $120^\circ$ and $240^\circ$; however, the book says I can't solve for $\cos x = 2$ because "There are no solutions for the second factor, $\cos x = 2$, because $\cos(x)$ is $[-1,1]$"



This confuses me because question 2 allows for $\cos x$'s domain to be outside of the restriction while question 3 doesn't allow for $\cos x$'s range to be outside of the restriction.
 A: The reason for domain restrictions is mainly because we want the "trig functions" to truly be functions in the strict mathematical sense.
That means for every element in the domain the function must produce exactly one function value.
Now one thing about functions is they don't always work equally well in both directions.
The definition of a function says you can get from any point in the domain to a unique point in the range; it says nothing about going from the range to the domain.
As a real-life analogy, there are machines that can turn standing trees into wood chips, but not (yet) any machine that can turn wood chips into a standing tree.
When a function $f$ has a genuine inverse function $f^{-1},$ the inverse really does reverse the effect of the original function:
$f^{-1}(f(x)) = x$ for any $x$ in the domain of $f.$
This works only for a function that is one-to-one, that is, when each value in the range comes from just one unique value in the domain.
As soon as you find even two values in the domain of a function that give the same output value when you use them as input to your function, you know there will not be a genuine inverse function.
The best we can do in cases like that is choose part of the domain of the original function, throw the rest away, and construct a limited "inverse" function that maps the range back to that part of the original function's domain.
So it is reasonable to expect that the domain of the inverse function will be the range of the original function (it certainly cannot be more than that), but it is not reasonable to expect in general that the range of the inverse function will be the domain of the original function. That only works when the original function is one-to-one.
The sine and cosine functions are not one-to-one, so it will not be the case that the range of $\arcsin(y)$ will be the domain of $\sin(x)$
or that you will have $\arccos(\cos(x)) = x$ for every number $x.$
The solution is not to redefine the original function so that it is invertible,
such as by making $\cos(x)$ undefined except when $0 \leq x \leq \pi$
(the range of $\arccos$).
Functions like the sine and cosine are far too useful when we allow them to be many-to-one (multiple input values $x$ that can produce the same output result $f(x)$);
we'd be giving up too much to make them truly invertible.
So we put up with an asymmetric situation: $\cos(x)$ is defined for all $x,$
but $\arccos(y)$ can only "get back" to a small subset of the values for which
$\cos(x)$ is defined, and we choose that subset to be $[0,\pi].$

A: 
Find an exact value of $y$, $y = \arcsin\left(\frac{\sqrt{3}}{2}\right)$.

As you can see in the table, $\arcsin x$ has domain $[-1, 1]$ and range $\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]$.
Therefore, $y = \arcsin\left(\frac{\sqrt{3}}{2}\right)$ is the unique value of $y$ in the interval $\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]$ such that $\sin y = \frac{\sqrt{3}}{2}$.  That value is $y = \pi/3$.  Hence, 
$$y = \arcsin\left(\frac{\sqrt{3}}{2}\right) = \frac{\pi}{3}$$
It is true that $\sin\left(\frac{2\pi}{3}\right) = \frac{\sqrt{3}}{2}$.  However, $2\pi/3$ is not a valid solution since $\frac{2\pi}{3} \notin \left[-\frac{\pi}{2}, \frac{\pi}{2}\right]$.
Observe that $\sin\left(\frac{4\pi}{3}\right) = -\frac{\sqrt{3}}{2}$ since $4\pi/3$ is a third-quadrant angle.  Thus, $4\pi/3$ is not a valid solution both because $\sin\left(\frac{4\pi}{3}\right) \neq \frac{\sqrt{3}}{2}$ and because $\frac{4\pi}{3} \notin \left[-\frac{\pi}{2}, \frac{\pi}{2}\right]$.
It is not clear to me why you referred to the cosine function in this problem unless you meant to solve for $y = \arccos\left(\frac{\sqrt{3}}{2}\right)$.  As you can see in the table, $\arccos x$ had domain $[-1, 1]$ and range $[0, \pi]$.  Thus, $y = \arccos x$ is the unique value of $y$ in the interval $[0, \pi]$ such that $\cos y = \frac{\sqrt{3}}{2}$, which is $y = \pi/6$.  Hence, 
$$y = \arccos\left(\frac{\sqrt{3}}{2}\right) = \frac{\pi}{6}$$

Find an exact value of $y = \cos[\arctan(-1)]$.

As you can see in the table, $\arctan x$ has domain $(-\infty, \infty)$ and  range $\left(-\frac{\pi}{2}, \frac{\pi}{2}\right)$.  Thus, $\arctan(-1)$ is the unique value of $u$ in the interval $\left(-\frac{\pi}{2}, \frac{\pi}{2}\right)$ such that $\tan u = -1$, which is $-\pi/4$.  Hence, 
$$\cos[\arctan(-1)] = \cos\left(-\frac{\pi}{4}\right) = \cos\left(\frac{\pi}{4}\right) = \frac{\sqrt{2}}{2}$$
where we have used the fact that $\cos(-x) = \cos x$. 
The domain of the cosine function is $(-\infty, \infty)$.  Its range is $[-1, 1]$.  Therefore, there are no restrictions on the domain of the cosine function.  
Since $f(x) = \cos x$ is periodic, to define an inverse function, we must first restrict its domain so that there is a unique value of $x$ for each value of $y = \cos x$. By convention, the arccosine function is the inverse of the restricted cosine function 
$$g(x) = \cos x, 0 \leq x \leq \pi$$
which has domain $[0, \pi]$ and range $[-1, 1]$.  Thus, as stated above, the arccosine function has domain $[-1, 1]$ and range $[0, \pi]$.   
Be careful not to confuse the domain and range of $f(x) = \cos x$ with the domain and range of $g^{-1}(x) = \arccos x$.

Find all solutions of the equation $2\cos^2x - 2 = 3\cos x$ in the interval $[0^\circ, 360^\circ)$.  

As stated above, the function $f(x) = \cos x$ has domain $(-\infty, \infty)$ and range $[-1, 1]$.  The equation $\cos x = 2$ has no solution since there is no $y$ in the range $[-1, 1]$ such that $y = \cos x = 2$. 
