Show that $2\cos(mp-nq)$ is one of the values of $\left( \frac{x^m}{y^n}+\frac{y^n}{x^m} \right)$

Question

Q:If $2\cos p=x+\frac{1}{x}$ and $2\cos q=y+\frac{1}{y}$ then show that $2\cos(mp-nq)$ is one of the values of $\left( \frac{x^m}{y^n}+\frac{y^n}{x^m} \right)$
My Approach:$2\cos p=x+\frac{1}{x}\Rightarrow x^2-2\cos px+1=0$ solving this equation i get $$x=\cos p\pm i\sin p$$ and similarly,$$y=\cos q\pm i\sin q$$Because somehow i guess $$x^m=\cos mp\pm i\sin mp,y^n=\cos nq\pm i\sin nq$$ maybe needed.But now i get stuck. Any hints or solution will be appreciated.
Thanks in advance.

Answer 1

Hint:

Use Euler's formula to write

$$x=e^{\pm ip},y=e^{\pm iq}.$$

Then see what

$$\frac{x^m}{y^n}+\frac{y^n}{x^m}$$

simplifies to.

Answer 2

As we know form before$$x+{1\over x}\ge 2\quad,\quad x>0$$and $$x+{1\over x}\le -2\quad,\quad x<0$$therefore$$2\cos p\ge 2\text{ or }2\cos p\le -2$$which leads to$$x=\pm 1$$similarly$$y=\pm1$$case $1$: $x=y=1$
$$2\cos(mp-nq)=2\cos(m2k_1\pi-n2k_2\pi)=2$$which is valid.
case $2$: $x=y=-1$ $$2\cos(mp-nq)=2\cos(m(2k_1\pi+\pi)-n(2k_2\pi+\pi))=2\cos((m-n)\pi)=2\cdot(-1)^{m-n}=(-1)^{m-n}+(-1)^{n-m}={x^m\over y^n}+{y^n\over x^m}$$
the 2 other cases hold either then the statement is true.