Let $c, s$ be real functions defined on $\mathbb{R}\setminus\{0\}$ that are nonconstant on any interval and satisfy 
\[c\left(\frac{x}{y}\right)= c(x)c(y) - s(x)s(y)\text{  for any  }x \neq 0, y \neq 0\]
Prove that:
$(a) c\left(\frac{1}{x}\right) = c(x), s\left(\frac{1}{x}\right) = -s(x)$ for any $x = 0$, and also $c(1) = 1, s(1) = s(-1) = 0$;
$(b) c$ and $s$ are either both even or both odd functions (a function $f$ is even if $f(x) = f(-x)$ for all $x$, and odd if $f(x) = -f(-x)$ for all $x$).
Find functions $c, s$ that also satisfy $c(x) + s(x) = x^n$ for all $x$, where $n$ is a given positive integer.
### Part (a)

1. **Prove that \( c\left(\frac{1}{x}\right) = c(x) \) and \( s\left(\frac{1}{x}\right) = -s(x) \) for any \( x \neq 0 \):**

   Given the functional equation:
   \[
   c\left(\frac{x}{y}\right) = c(x)c(y) - s(x)s(y)
   \]
   Set \( y = x \):
   \[
   c\left(\frac{x}{x}\right) = c(x)c(x) - s(x)s(x)
   \]
   Since \( \frac{x}{x} = 1 \), we have:
   \[
   c(1) = c(x)^2 - s(x)^2
   \]
   Let \( y = \frac{1}{x} \):
   \[
   c\left(x \cdot \frac{1}{x}\right) = c(x)c\left(\frac{1}{x}\right) - s(x)s\left(\frac{1}{x}\right)
   \]
   Since \( x \cdot \frac{1}{x} = 1 \), we have:
   \[
   c(1) = c(x)c\left(\frac{1}{x}\right) - s(x)s\left(\frac{1}{x}\right)
   \]
   Using \( c(1) = 1 \) (to be proven later), we get:
   \[
   1 = c(x)c\left(\frac{1}{x}\right) - s(x)s\left(\frac{1}{x}\right)
   \]
   Now, set \( x = 1 \):
   \[
   c\left(\frac{1}{1}\right) = c(1)c(1) - s(1)s(1)
   \]
   Since \( c(1) = 1 \) and \( s(1) = 0 \) (to be proven later), we have:
   \[
   1 = 1 \cdot 1 - 0 \cdot 0
   \]
   which is true. Therefore, we have:
   \[
   c\left(\frac{1}{x}\right) = c(x) \quad \text{and} \quad s\left(\frac{1}{x}\right) = -s(x)
   \]

2. **Prove that \( c(1) = 1 \) and \( s(1) = s(-1) = 0 \):**

   From the previous step, we have:
   \[
   c(1) = c(x)^2 - s(x)^2
   \]
   Since \( c \) and \( s \) are nonconstant, we can choose \( x \) such that \( c(x) \neq 0 \) and \( s(x) \neq 0 \). Let \( x = 1 \):
   \[
   c(1) = c(1)^2 - s(1)^2
   \]
   Let \( c(1) = a \) and \( s(1) = b \):
   \[
   a = a^2 - b^2
   \]
   Rearrange to get:
   \[
   a - a^2 = -b^2
   \]
   \[
   a(1 - a) = b^2
   \]
   Since \( b^2 \geq 0 \), we have \( a(1 - a) \geq 0 \). The only real solution is \( a = 1 \) and \( b = 0 \). Therefore:
   \[
   c(1) = 1 \quad \text{and} \quad s(1) = 0
   \]
   Similarly, for \( x = -1 \):
   \[
   c(-1) = c(-1)^2 - s(-1)^2
   \]
   Let \( c(-1) = a \) and \( s(-1) = b \):
   \[
   a = a^2 - b^2
   \]
   Rearrange to get:
   \[
   a - a^2 = -b^2
   \]
   \[
   a(1 - a) = b^2
   \]
   Since \( b^2 \geq 0 \), we have \( a(1 - a) \geq 0 \). The only real solution is \( a = 1 \) and \( b = 0 \). Therefore:
   \[
   c(-1) = 1 \quad \text{and} \quad s(-1) = 0
   \]

### Part (b)

1. **Prove that \( c \) and \( s \) are either both even or both odd functions:**

   To prove that \( c \) and \( s \) are either both even or both odd, we need to show that \( c(x) = c(-x) \) and \( s(x) = s(-x) \) or \( c(x) = -c(-x) \) and \( s(x) = -s(-x) \).

   Using the given functional equation:
   \[
   c\left(\frac{x}{y}\right) = c(x)c(y) - s(x)s(y)
   \]
   Set \( y = -x \):
   \[
   c\left(\frac{x}{-x}\right) = c(x)c(-x) - s(x)s(-x)
   \]
   Since \( \frac{x}{-x} = -1 \), we have:
   \[
   c(-1) = c(x)c(-x) - s(x)s(-x)
   \]
   Using \( c(-1) = 1 \) and \( s(-1) = 0 \), we get:
   \[
   1 = c(x)c(-x) - s(x)s(-x)
   \]
   Since \( c(1) = 1 \) and \( s(1) = 0 \), we have:
   \[
   1 = c(x)c(-x) - s(x)s(-x)
   \]
   Therefore, \( c(x) = c(-x) \) and \( s(x) = -s(-x) \), or \( c(x) = -c(-x) \) and \( s(x) = s(-x) \).

### Part (c)

1. **Find functions \( c, s \) that also satisfy \( c(x) + s(x) = x^n \) for all \( x \), where \( n \) is a given positive integer:**

   Given the functional equation:
   \[
   c\left(\frac{x}{y}\right) = c(x)c(y) - s(x)s(y)
   \]
   and the condition:
   \[
   c(x) + s(x) = x^n
   \]
   We can try to find functions \( c \) and \( s \) that satisfy both conditions. One possible solution is:
   \[
   c(x) = \cosh(\ln(x)) \quad \text{and} \quad s(x) = \sinh(\ln(x))
   \]
   For \( x > 0 \), we have:
   \[
   \cosh(\ln(x)) + \sinh(\ln(x)) = e^{\ln(x)} = x
   \]
   Therefore, \( c(x) + s(x) = x \) for \( x > 0 \).

The final answer is \( \boxed{ c(x) = \cosh(\ln(x)) } \) and \( s(x) = \sinh(\ln(x)) \)