Finding domain, codomain and range of $h(x)= \cos\big(\pi\frac{7x}{6}\big)$ I mainly need help with deciding the range of this function, I don't really understand it when I read my book. The only thing I know which I think I should use is that the value of a cos function is between -1 and 1.
$f: \mathbb{Q} \to \mathbb{R}$,  $ f(x)$= cos(π$x$)
$g: \mathbb{Z} \to \mathbb{Q}$,  $ g(x)= \frac{7x}{6}$ 
$h(x) = f(g(x)) $
$h(x) $ = cos$( π \frac{7x}{6} )   $
$h: \mathbb{Z} \to \mathbb{R}$
The domain for $h$ is $\mathbb{Z}$, and the codomain for $h$ is $\mathbb{R}$, is this correct$?$
But how do I find the range for $h?$
And also, is $h$ a surjective or injective function$?$
Many thanks in advance!
 A: Domain and codomain look okay.
Observe that $\cos{\pi x}$ ranges in values between $-1$ and $1$, and changing the inside to $\frac{ 7\pi x}{6}$ will only affect the period of the function. It will have no affect on ranging between $-1$ and $1$.
However, don't forget about your domain being $\mathbb{Z}$. This will mean that the inside of the cosine for $h(x)$ will always be a multiple of $\frac{7\pi }{6}$, which will limit some of the values that cosine can normally achieve.
For example, $h(12)=1$ and $h(7) = \frac{\sqrt{3}}{2}$. However, there is no $x$ value in the domain that could give us a solution to, say, $h(x) = 0.3$ or $h(x) = \frac{\sqrt{2}}{2}$.
In fact, the only possible values for $h(x)$ are $0, \pm \frac{1}{2}, \pm \frac{\sqrt{3}}{2}, \pm 1$. This has to do with the $6$ in the denominator as well as the fact that  $(7n) \div 6$ for any integer $n$ can have all of the possible remainders between $0$ and $5$. This indicates that the range is actually all the possible values for $\cos(\pi n/6)$ for any integer $n$, which happen to be the same list of values from earlier. (Anyone that thinks they can word this better please let me know below, as I'm not sure if this explanation is valid)  Therefore the set including these values would be your range.
Range = $\{0, \pm \frac{1}{2}, \pm \frac{\sqrt{3}}{2}, \pm 1\}$
