1. ## could you please solve these problem

Write cos(for the power of 4 ) (x) as a function in terms of cos 2x and cos 4x

2. ## Re: could you please solve these problem

let's see...

$\cos(2x)=\cos^2(x) - \sin^2(x) =\cos^2(x) - (1-\cos^2(x))= 2\cos^2(x)-1$

so

$\cos^2(x) = \dfrac{1+\cos(2x)}{2}$

Spoiler:
\begin{align*} &\cos^4(x) = \\ \\ &(\cos^2(x))^2 = \\ \\ &\dfrac{(1+\cos(2x))^2}{4} = \\ \\ &\dfrac 1 4\left(\cos^2(2x)+2\cos(2x)+1\right) \end{align*}

From above letting $x \to 2x$

$\cos^2(2x) = \dfrac{1+\cos(4x)}{2}$

So combining these we get

\begin{align*} &\cos^4(x) = \\ \\ &\dfrac 1 4\left(\cos^2(2x)+2\cos(2x)+1\right) = \\ \\ &\dfrac 1 4\left(\dfrac{1+\cos(4x)}{2}+2\cos(2x) + 1\right) \end{align*}

and you can complete the remaining algebra