Question

How do you use the power reducing formulas to rewrite the expression ${\cos }^{4}x$ in terms of the first power of cosine?

Answer 1

Step-1: Rearrange the trigonometric identities:

We know the trigonometric identity $\cos 2x=2{\cos }^{2}x-1$.

Rearranging the terms, we have

$$\begin{gathered} \Rightarrow 2{\cos }^{2}x=\cos 2x+1 \\ \therefore {\cos }^{2}x=\frac{1}{2}\left(\cos 2x+1\right)\cdots \left(1\right) \end{gathered}$$

Substituting $2x$ for $x$ in equation $\left(1\right)$, we get

$$\begin{gathered} \Rightarrow {\cos }^{2}2x=\frac{1}{2}\left(\cos 2\left(2x\right)+1\right) \\ \therefore {\cos }^{2}2x=\frac{1}{2}\left(\cos 4x+1\right)\cdots \left(2\right) \end{gathered}$$

Now, ${\cos }^{4}x$ can be rewritten as,

$\Rightarrow {\cos }^{4}x={\left({\cos }^{2}x\right)}^2\dots \left(3\right)$

Step-2: Express ${\cos }^{4}x$ in terms of $\cos x$:

Substituting $\frac{1}{2}\left(\cos 2x+1\right)$ for ${\cos }^{2}x$ in equation $\left(3\right)$, we get

$$\begin{gathered} \Rightarrow {\cos }^{4}x={\left(\frac{1}{2}\left(\cos 2x+1\right)\right)}^2 \\ \Rightarrow {\cos }^{4}x=\frac{1}{4}{\left(\cos 2x+1\right)}^2 \\ \Rightarrow {\cos }^{4}x=\frac{1}{4}\left({\cos }^2\left(2x\right)+1^2+2\left(\cos \left(2x\right)\right)\left(1\right)\right)\left[\because {\left(a+b\right)}^2=a^2+b^2+2ab\right] \\ \therefore {\cos }^{4}x=\frac{1}{4}\left({\cos }^2\left(2x\right)+1+2\cos 2x\right)\dots \left(4\right) \end{gathered}$$

Substituting $\frac{1}{2}\left(\cos 4x+1\right)$ for ${\cos }^{2}2x$ in equation $\left(4\right)$, we get

$$\begin{gathered} \Rightarrow {\cos }^{4}x=\frac{1}{4}\left[\left(\frac{1}{2}\left(\cos 4x+1\right)\right)+1+2\cos 2x\right] \\ \Rightarrow {\cos }^{4}x=\frac{1}{4}\left[\frac{1}{2}\cos 4x+\frac{1}{2}+1+2\cos 2x\right]\left[\mathrm{Removing}\mathrm{brackets}\right] \\ \Rightarrow {\cos }^{4}x=\frac{1}{4}\left[\frac{1}{2}\left(\cos 4x+1+2+2\left(2\cos 2x\right)\right)\right]\left[\mathrm{Taking}\mathrm{LCM}\right] \\ \Rightarrow {\cos }^{4}x=\frac{1}{8}\left[\cos 4x+3+4\cos 2x\right]\left[\mathrm{Removing}\mathrm{brackets}\right] \\ \therefore {\cos }^{4}x=\frac{1}{8}\left[\cos 4x+4\cos 2x+3\right] \end{gathered}$$

Hence, ${\cos }^{4}x$ in terms of $\cos x$ is $\frac{1}{8}\left[\cos 4x+4\cos 2x+3\right]$.