Express \(cos^4 \theta\) in terms of \(cos\,n\theta\)

\begin{align} & We\ know\ that\ for\ a\ complex\ number\ z \\ 2cos\theta & = \left(z+\frac{1}{z}\right) \quad \quad \enclose{circle}{\color{black}{1}}\\ \therefore\ (2cos\theta)^4 & = \left(z+\frac{1}{z}\right)^4\quad \quad \enclose{circle}{\color{black}{2}} \\ Using\ the\ binomial\ & expansion\ of\ the\ RHS\ of\ equation\ \enclose{circle}{\color{black}{2}}\\ \left(z+\frac{1}{z} \right)^4 & = z^4 + 4z^3 \left(\frac{1}{z}\right) + 6z^2 \left(\frac{1}{z}\right)^2 + 4z\left(\frac{1}{z}\right)^3 + \left(\frac{1}{z}\right)^4 \\ & =z^4 + 4z^2 + 6z^0 + 4\left(\frac{1}{z^2}\right) + \left(\frac{1}{z^4}\right)\\ & = \left(z^4+ \frac{1}{z^4}\right) + 4\left(z^2+ \frac{1}{z^2}\right) + 6 \\ \\ & = 2cos4\theta + 8cos2\theta + 6 \quad \quad (by\ deMoivre's\ Theorem) \\ \\ Therefore,\ from\ equation\ \enclose{circle}{\color{black}{2}} \\ \\ 2^4cos^4\theta & = 2cos4\theta + 8cos2\theta + 6 \\ cos^4 \theta & = \frac{1}{16}\left(2cos4\theta + 8cos2\theta + 6 \right) \\ & = \frac{1}{8}\left(cos4\theta + 4cos2\theta + 3\right) \end{align}