Vtyjkyuk

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:

1. 
   

   Find an exact value of $y$, $y=arcsinleft(fracsqrt32right)$

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


2. 
   

   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:
     $y=cos(7pi/4)$. The answers/solutions in the back then show $y= sqrt2/2$ is the answer.

   
   

   • However, I thought $cos(x)$'s domain should have been restricted to $[-1,1]$. Why is the $7pi/4$ 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 $y=cos(7pi/4)$ as $arccos(y)=7pi/4$; the "$7pi/4$" part is not part of $arccos$'s range restriction, $[0,pi]$

   
   


3. 
   

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

   
   
   
   • 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.

Find an exact value of $y$, $y = arcsinleft(fracsqrt32right)$.

As you can see in the table, $arcsin x$ has domain $[-1, 1]$ and range $left[-fracpi2, fracpi2right]$.
Therefore, $y = arcsinleft(fracsqrt32right)$ is the unique value of $y$ in the interval $left[-fracpi2, fracpi2right]$ such that $sin y = fracsqrt32$. That value is $y = pi/3$. Hence,
$$y = arcsinleft(fracsqrt32right) = fracpi3$$

It is true that $sinleft(frac2pi3right) = fracsqrt32$. However, $2pi/3$ is not a valid solution since $frac2pi3 notin left[-fracpi2, fracpi2right]$.

Observe that $sinleft(frac4pi3right) = -fracsqrt32$ since $4pi/3$ is a third-quadrant angle. Thus, $4pi/3$ is not a valid solution both because $sinleft(frac4pi3right) neq fracsqrt32$ and because $frac4pi3 notin left[-fracpi2, fracpi2right]$.

It is not clear to me why you referred to the cosine function in this problem unless you meant to solve for $y = arccosleft(fracsqrt32right)$. 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 $cos y = fracsqrt32$, which is $y = pi/6$. Hence,
$$y = arccosleft(fracsqrt32right) = fracpi6$$

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

As you can see in the table, $arctan x$ has domain $(-infty, infty)$ and range $left(-fracpi2, fracpi2right)$. Thus, $arctan(-1)$ is the unique value of $u$ in the interval $left(-fracpi2, fracpi2right)$ such that $tan u = -1$, which is $-pi/4$. Hence,
$$cos[arctan(-1)] = cosleft(-fracpi4right) = cosleft(fracpi4right) = fracsqrt22$$
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 $2cos^2x - 2 = 3cos x$ 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 $y = cos x = 2$.