Question

How do you prove $\cos 3\theta =4{{\cos }^{3}}\theta -3\cos \theta $ ? \[\]

Answer 1

Hint: We begin from left hand side of given identity by using compound angle formula of cosine $\cos \left( A+B \right)=\cos A\cos B-\sin A\sin B$ for $A=2\theta ,B=\theta $. We then use double formula for cosine $\cos 2A={{\cos }^{2}}A-{{\sin }^{2}}A$ and cosine $\sin 2A=2\sin A\cos A$. We convert any $\sin \theta $ if exists using the Pythagorean trigonometric identity ${{\sin }^{2}}\theta =1-{{\cos }^{2}}\theta $ to convert it into cosine. \[\]

Complete step by step answer:
We know that the cosine trigonometric ratio in a right angled triangle is the ratio of length of the side adjacent to the angle to the length of the hypotenuse. \[\]
We are asked prove the following identity
\[\cos 3\theta =4{{\cos }^{3}}\theta -3\cos \theta \]
We begin left hand side of the above identity and have;
\[\begin{align}
  & \cos \left( 3\theta \right) \\
 & \Rightarrow \cos \left( 2\theta +\theta \right) \\
\end{align}\]
We use cosine compound angle or sum of angles formula $\cos \left( A+B \right)=\cos A\cos B-\sin A\sin B$ for $A=2\theta ,B=\theta $ in the above step to have;
\[\Rightarrow \cos 2\theta \cos \theta -\sin 2\theta \sin \theta \]
We use cosine double angle formula $\cos 2A={{\cos }^{2}}A-{{\sin }^{2}}A$ and sine double angle formula $\sin 2A=2\sin A\cos A$ for $A=\theta $ in the above step to get;
\[\begin{align}
  & \Rightarrow \left( {{\cos }^{2}}\theta -{{\sin }^{2}}\theta \right)\cos \theta -2\sin \theta \cos \theta \sin \theta \\
 & \Rightarrow {{\cos }^{3}}\theta -{{\sin }^{2}}\theta \cos \theta -2{{\sin }^{2}}\theta \cos \theta \\
\end{align}\]
We take $\cos \theta $ common in the above step to collect the ${{\sin }^{2}}\theta $ to have;
\[\begin{align}
  & \Rightarrow \cos \theta \left( {{\cos }^{2}}\theta -{{\sin }^{2}}\theta -2{{\sin }^{2}}\theta \right) \\
 & \Rightarrow \cos \theta \left( {{\cos }^{2}}\theta -3{{\sin }^{2}}\theta \right) \\
\end{align}\]
We use the Pythagorean trigonometric identity ${{\sin }^{2}}\theta +{{\cos }^{2}}\theta =1\Rightarrow {{\sin }^{2}}\theta =1-{{\cos }^{2}}\theta $ in the above step to have;
\[\begin{align}
  & \Rightarrow \cos \theta \left( {{\cos }^{2}}\theta -3\left( 1-{{\cos }^{2}}\theta \right) \right) \\
 & \Rightarrow \cos \theta \left( {{\cos }^{2}}\theta -3+3{{\cos }^{2}}\theta \right) \\
 & \Rightarrow \cos \theta \left( 4{{\cos }^{2}}\theta -3 \right) \\
 & \Rightarrow 4{{\cos }^{3}}\theta -3\cos \theta \\
\end{align}\]
We see that the above obtained expression is at the right hand side of the given identity. Hence the identity is proved. \[\]

Note: We note that our primary goal in this proof was to convert all the sines and cosines. The other double angle formulas of cosines are $\cos 2\theta =1-2{{\sin }^{2}}\theta =2{{\cos }^{2}}\theta -1$. We cannot use $\cos 2\theta =1-2{{\sin }^{2}}\theta $ since we require all angles in sines. We can alternatively use $\cos 2\theta =2{{\cos }^{2}}\theta -1$ ,simplify and take $\cos \theta $ common from the second two terms to proceed towards proof. We must be careful about the confusion of cosine sum of angles formula from cosine difference of angle formula which is given by $\cos \left( A-B \right)=\cos A\cos B+\sin A\sin B$.