Question

Rewrite 9cos 4x in terms of cos x.

Answer 1

$\bf \qquad \textit{Quad identities}\\\\ sin(4\theta )= \begin{cases} 8sin(\theta )cos^3(\theta )-4sin(\theta )cos(\theta )\\ 4sin(\theta )cos(\theta )-8sin^3(\theta )cos(\theta ) \end{cases} \\\\\\ cos(4\theta)=8cos^4(\theta )-8cos^2(\theta )+1\\\\ -------------------------------\\\\ 9cos(4x)\implies 9[8cos^4(x)-8cos^2(x)+1] \\\\\\ 72cos^4(x)-72cos^2(x)+9$


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as far as the previous one on the 2tan(3x)

$\bf tan(2\theta)=\cfrac{2tan(\theta)}{1-tan^2(\theta)}\qquad tan({{ \alpha}} + {{ \beta}}) = \cfrac{tan({{ \alpha}})+ tan({{ \beta}})}{1- tan({{ \alpha}})tan({{ \beta}})}\\\\ -------------------------------$

$\bf 2tan(3x)\implies 2tan(2x+x)\implies 2\left[ \cfrac{tan(2x)+tan(x)}{1-tan(2x)tan(x)}\right] \\\\\\ 2\left[ \cfrac{\frac{2tan(x)}{1-tan^2(x)}+tan(x)}{1-\frac{2tan(x)}{1-tan^2(x)}tan(x)}\right]\implies 2\left[ \cfrac{\frac{2tan(x)+tan(x)-tan^3(x)}{1-tan^2(x)}}{\frac{1-tan(x)-2tan^3(x)}{1-tan^2(x)}} \right]$

$\bf 2\left[ \cfrac{2tan(x)+tan(x)-tan^3(x)}{1-tan^2(x)}\cdot \cfrac{1-tan^2(x)}{1-tan(x)-2tan^3(x)} \right] \\\\\\ 2\left[ \cfrac{3tan(x)-tan^3(x)}{1-tan^2(x)-2tan^3(x)} \right]\implies \cfrac{6tan(x)-2tan^3(x)}{1-tan^2(x)-2tan^3(x)}$