Question

Find an expression for $\cos 3x$ in terms of $\cos x$.

Answer 1

Solve for the value of $\cos 3x$ in terms of $\cos x$.

Expand $3x$as $2x+x$ in the expression.

$\Rightarrow \cos \left(3x\right)=\cos \left(2x+x\right)$

Use trignometric identity $\mathit{c}\mathit{o}\mathit{s}\left(a+b\right)\mathbf{=}\mathit{c}\mathit{o}\mathit{s}\mathit{a}\mathbf{.}\mathit{c}\mathit{o}\mathit{s}\mathit{b}\mathbf{-}\mathit{s}\mathit{i}\mathit{n}\mathit{a}\mathbf{.}\mathit{s}\mathit{i}\mathit{n}\mathit{b}$

Applying the identity in the above expression.

$\begin{array}{rcl}& \Rightarrow & \cos 3x=\cos 2x.\cos x-\sin 2x.\sin x\\ & =& \left(-1+2{\cos }^{2}x\right)\cos x-2\sin x.\cos x.\sin x\left[\because 1+cos2x=2co{s}^{2}x,\sin 2x=2\sin x\cos x\right]\\ & =& 2{\cos }^{3}x-2{\sin }^{2}x\cos x-\cos x\\ & =& 2{\cos }^{3}x-2\left(1-{\cos }^{2}x\right)\cos x-\cos x\\ & =& 4{\cos }^{3}x-3\cos x\end{array}$

Hence, $\mathit{c}\mathit{o}\mathit{s}\mathbf{3}\mathit{x}\mathbf{=}\mathbf{4}\mathit{c}\mathit{o}{\mathit{s}}^{\mathbf{3}}\mathit{x}\mathbf{-}\mathbf{3}\mathit{c}\mathit{o}\mathit{s}\mathit{x}$.