Question

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

Answer 1

Hint: We first try to use the associative formula of $\cos \left( x+y \right)=\cos x\cos y-\sin x\sin y$ for $\cos 3\theta $. We convert all the ratios to cos form by using $\cos 2\theta =2{{\cos }^{2}}\theta -1$ and $\sin 2\theta =2\sin \theta \cos \theta $. Change of identity formula of ${{\sin }^{2}}\theta =1-{{\cos }^{2}}\theta $ gives the solution.

Complete step-by-step answer:
We know the multiple angle formula of $\cos 2\theta =2{{\cos }^{2}}\theta -1$ and $\sin 2\theta =2\sin \theta \cos \theta $.
We now break the given $\cos 3\theta $ as $\cos \left( 2\theta +\theta \right)$.
We now use the associative angle formula of $\cos \left( x+y \right)=\cos x\cos y-\sin x\sin y$.
Placing the values $x=2\theta ,y=\theta $ we get
$\begin{align}
  & \cos \left( 2\theta +\theta \right) \\
 & =\cos \left( 2\theta \right)\cos \theta -\sin \left( 2\theta \right)\sin \theta \\
\end{align}$
Now we replace the values and get
$\begin{align}
  & \cos \left( 2\theta +\theta \right) \\
 & =\left( 2{{\cos }^{2}}\theta -1 \right)\cos \theta -\left( 2\sin \theta \cos \theta \right)\sin \theta \\
 & =2{{\cos }^{3}}\theta -\cos \theta -2{{\sin }^{2}}\theta \cos \theta \\
\end{align}$
Now we try to convert the ratio sin into ratio of cos. We use ${{\sin }^{2}}\theta =1-{{\cos }^{2}}\theta $.
Therefore,
$\begin{align}
  & \cos \left( 2\theta +\theta \right) \\
 & =2{{\cos }^{3}}\theta -\cos \theta -2{{\sin }^{2}}\theta \cos \theta \\
 & =2{{\cos }^{3}}\theta -\cos \theta -2\left( 1-{{\cos }^{2}}\theta \right)\cos \theta \\
 & =2{{\cos }^{3}}\theta -\cos \theta -2\cos \theta +2{{\cos }^{3}}\theta \\
 & =4{{\cos }^{3}}\theta -3\cos \theta \\
\end{align}$
Therefore, expressing $\cos 3\theta $ in terms of only $\cos \theta $, we get $\cos 3\theta =4{{\cos }^{3}}\theta -3\cos \theta $.
Now we try to convert them into ratio sin.
$\begin{align}
  & \cos 3\theta \\
 & =\cos \theta \left( 4{{\cos }^{2}}\theta -3 \right) \\
 & =\cos \theta \left( 4-4{{\sin }^{2}}\theta -3 \right) \\
 & =\cos \theta \left( 1-4{{\sin }^{2}}\theta \right) \\
\end{align}$
So, the correct answer is “$\cos \theta \left( 1-4{{\sin }^{2}}\theta \right)$”.

Note: We cannot convert the whole expression into expression of $\sin \theta $ as that brings the root form of \[\cos \theta =\pm \sqrt{1-{{\sin }^{2}}\theta }\]. This form is not a proper expression to get the exact value we need to know about the position of the angle to find the sign.