Question

What is the formula of $\cos \left(4x\right)$?

Answer 1

Find the formula for $\cos \left(4x\right)$.

$$\begin{gathered} \begin{array}{rcl}& & \cos 4x=\cos \left(2x+2x\right)\\ & & \Rightarrow \cos \left(4x\right)=\cos 2x.\cos 2x-\sin 2x.\sin 2x\mathbf{[}\mathbf{\because }\mathbf{c}\mathbf{o}\mathbf{s}\mathbf{}\mathbf{(}\mathbf{a}\mathbf{+}\mathbf{b}\mathbf{)}\mathbf{=}\mathbf{c}\mathbf{o}\mathbf{s}\mathbf{}\mathbf{a}\mathbf{.}\mathbf{c}\mathbf{o}\mathbf{s}\mathbf{}\mathbf{b}\mathbf{-}\mathbf{s}\mathbf{i}\mathbf{n}\mathbf{}\mathbf{a}\mathbf{.}\mathbf{s}\mathbf{i}\mathbf{n}\mathbf{}\mathbf{b}\mathbf{]}\\ & & \Rightarrow \cos \left(4x\right)=\left(2{\cos }^{2}x-1\right).\left(2{\cos }^{2}x-1\right)-(2\sin x.\cos x)(2\sin x.\cos x)\mathbf{[}\mathbf{\because }\mathbf{s}\mathbf{i}\mathbf{n}\mathbf{}\mathbf{2}\mathbf{x}\mathbf{=}\mathbf{2}\mathbf{s}\mathbf{i}\mathbf{n}\mathbf{}\mathbf{x}\mathbf{.}\mathbf{c}\mathbf{o}\mathbf{s}\mathbf{}\mathbf{x}\mathbf{]} \\ \Rightarrow \cos \left(4x\right)={\left(2{\cos }^{2}x-1\right)}^2-{(2\sin x.\cos x)}^2\left[\mathbf{\because }\mathit{c}\mathit{o}\mathit{s}\mathbf{}\mathbf{2}\mathit{x}\mathbf{=}\mathbf{2}\mathbf{}\mathit{c}\mathit{o}{\mathit{s}}^{\mathbf{2}}\mathbf{}\mathit{x}\mathbf{-}\mathbf{1}\right] \\ \Rightarrow \cos \left(4x\right)=\left(4{\cos }^{4}x-4{\cos }^{2}x+1\right)-(4{\sin }^{2}x.{\cos }^{2}x) \\ \Rightarrow \cos \left(4x\right)=\left(4{\cos }^{4}x-4{\cos }^{2}x+1\right)-\left(4\left(1-{\cos }^{2}x\right).{\cos }^{2}x\right)\end{array} \\ \begin{array}{rcl}& \Rightarrow & \cos \left(4x\right)=\left(8{\cos }^{4}x-8{\cos }^{2}x+1\right)\end{array} \end{gathered}$$

Hence, the formula is $\begin{array}{rcl}\cos \left(2x+2x\right)& =& 8{\cos }^4\left(x\right)+1-8{\cos }^2\left(x\right)\end{array}$.