Chapter C, Problem 26E

To determine

To prove:

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

Explanation of Solution

Given information:

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

Formula Used:

Subtraction formula:

  $\cos \left(x-y\right)=\cos x\cdot \cos y+\sin x\sin y$

The double angle formulas:

  $\begin{array}{l}\cos 2x=2{\cos }^{2}x-1\\ \sin 2x=2\sin x\cdot \cos x\end{array}$

Proof:

The equation is given as:

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

The left hand side of the eq.:

  $\cos 3\theta =\cos \left(4\theta -\theta \right)$

  $\cos 3\theta =\cos 4\theta \cos \theta +\sin 4\theta \sin \theta$

  $\cos 3\theta =\left(2{\cos }^{2}2\theta -1\right)\cos \theta +\left(2\sin 2\theta \cos 2\theta \right)\sin \theta \left[\begin{array}{l}\because 2{\cos }^2\theta -1=\cos 2\theta \\ \sin 2\theta =2\sin \theta \cdot \cos \theta \end{array}\right]$

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

  $\cos 3\theta =\left(2\left(4{\cos }^4\theta -4{\cos }^2\theta +1\right)-1\right)\cos \theta +\left(2\sin 2\theta \cos 2\theta \right)\sin \theta \left[\because {\left(a-b\right)}^2=a^2-2ab+b^2\right]$

  $\cos 3\theta =\left(8{\cos }^4\theta -8{\cos }^2\theta +2-1\right)\cos \theta +\left(2\left(2\sin \theta \cdot \cos \theta \right)\left(2{\cos }^2\theta -1\right)\right)\sin \theta$

  $\cos 3\theta =\left(8{\cos }^5\theta -8{\cos }^3\theta +\cos \theta \right)+\left(4{\sin }^2\theta \cos \theta \right)\left(2{\cos }^2\theta -1\right)$

  $\cos 3\theta =\left(8{\cos }^5\theta -8{\cos }^3\theta +\cos \theta \right)+\left(4\left(1-{\cos }^2\theta \right)\cos \theta \right)\left(2{\cos }^2\theta -1\right)\left[\because {\sin }^2\theta +{\cos }^2\theta =1\right]$

  $\cos 3\theta =\left(8{\cos }^5\theta -8{\cos }^3\theta +\cos \theta \right)+\left(4\cos \theta \right)\left(1-{\cos }^2\theta \right)\left(2{\cos }^2\theta -1\right)$

  $\cos 3\theta =\left(8{\cos }^5\theta -8{\cos }^3\theta +\cos \theta \right)+\left(4\cos \theta \right)\left(2{\cos }^2\theta -1-2{\cos }^4\theta +{\cos }^2\theta \right)$

  $\cos 3\theta =\left(8{\cos }^5\theta -8{\cos }^3\theta +\cos \theta \right)+\left(4\cos \theta \right)\left(3{\cos }^2\theta -1-2{\cos }^4\theta \right)$

  $\cos 3\theta =8{\cos }^5\theta -8{\cos }^3\theta +\cos \theta +12{\cos }^3\theta -4\cos \theta -8{\cos }^5\theta$

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

Since, left hand side $\left(\cos 3\theta \right)$ equals to right hand side $\left(4{\cos }^3\theta -3\cos \theta \right)$ for all values $\theta$ . So, $\cos 3\theta =4{\cos }^3\theta -3\cos \theta$ is an identity.

Hence, proved.