Question

How do you rewrite $\cos 3\theta$ in terms of only $\cos \theta$?

Answer 1

Rewrite $\mathbf{cos}\mathbf{3}\mathbf{\theta }$ in terms of $\mathbf{cos\theta }$:

$\cos 3\theta$ can also be written as,

$\cos 3\theta =\cos \left(2\theta +\theta \right)$

$=\cos 2\theta \cos \theta -\sin 2\theta \sin \theta$ $\left[\because \cos \left(a+b\right)=\mathrm{cosa}\mathrm{cosb}-\mathrm{sina}\mathrm{sinb}\right]$

$=\left(2{\cos }^2\theta -1\right)\cos \theta -\left(2\cos \theta \sin \theta \right)\sin \theta$ $[\because \cos 2a=2{\cos }^{2}a-1;\sin 2a=2\mathrm{cosa}\mathrm{sina}]$

$=2{\cos }^3\theta -\cos \theta -2{\sin }^2\theta \cos \theta$

$=2{\cos }^3\theta -\cos \theta -2\left(1-{\cos }^2\theta \right)\cos \theta$ $\left[\because {\sin }^{2}a+{\cos }^{2}a=1\right]$

$=2{\cos }^3\theta -\cos \theta -2\cos \theta +2{\cos }^3\theta$

$=4{\cos }^3\theta -3\cos \theta$

Hence, $\mathbf{cos}\mathbf{}\mathbf{3}\mathbf{\theta }$ can be written in terms of $\mathbf{cos}\mathbf{}\mathbf{\theta }$ as $\mathbf{4}{\mathbf{cos}}^{\mathbf{3}}\mathbf{\theta }\mathbf{-}\mathbf{3}\mathbf{cos}\mathbf{}\mathbf{\theta }$.