Expand cos7θ in a series of cosine of multiple of θ

Let $e^{i\theta}=x$

$\therefore e^{i\theta}=\dfrac{1}{x}$

$\therefore \cos\theta=\dfrac{e^{i\theta}+e^{-i\theta}}{2}=\dfrac{1}{2}\left ( x+\dfrac{1}{x} \right )\rightarrow (1)$

Now, $\cos^{7}\theta=\dfrac{1}{2^{7}}\left ( x+\dfrac{1}{x} \right )^{7}$ (From 1)

$=\dfrac{1}{128}\left ( 1x^{7}+7x^{6}\cdot\dfrac{1}{x}+21x^{5}\cdot\dfrac{1}{x^{2}}+35x^{4}\cdot \dfrac{1}{x^{3}}+35x^{3}\cdot\dfrac{1}{x^{4}}+21x^{2}\cdot\dfrac{1}{x^{5}}+7x\cdot\dfrac{1}{x^{6}}+\dfrac{1}{x^{7}}\right )$

$=\dfrac{1}{128}\left ( x^{7}+7x^{5}+21x^{3}+35x +\dfrac{35}{x}+\dfrac{21}{x^{3}}+\dfrac{7}{x^{5}}+\dfrac{1}{x^{7}}\right )$

$=\dfrac{1}{128}\left [ \left ( x^{7}+\dfrac{1}{x^{7}} \right ) +7\left ( x^{5}+\dfrac{1}{x^{5}} \right )+21\left ( x^{3}+\dfrac{1}{x^{3}} \right )+35\left ( x+\dfrac{1}{x} \right )\right ]$ …