Question

What is the formula for \[\cos 3x\]?

Answer 1

Hint: These types of problems are pretty straight forward and are very simple to solve. This particular question is a great example of trigonometric functions and equations. For solving this given problem, we need to have a knowledge of some of the basic formulae of trigonometry along with the values of the angles of different trigonometric functions. We must remember the formula\[\cos \left( A+B \right)=\cos A\cos B-\sin A\sin B\] . From this given formula we first need to find the value of \[\cos 2x\] and then from this, we can very easily find out our required value of \[\cos 3x\] .

Complete step by step solution:
Now, we start off with the solution to the given problem by saying that, we find the value of \[\cos 2x\] using the formula of \[\cos \left( A+B \right)=\cos A\cos B-\sin A\sin B\] , where \[A=B=x\] . Now, evaluating this we get,
\[\begin{align}
  & \Rightarrow \cos 2x=\cos \left( x+x \right)=\cos x\cos x-\sin x\sin x \\
 & \Rightarrow \cos 2x=\cos \left( x+x \right)={{\cos }^{2}}x-{{\sin }^{2}}x \\
\end{align}\]
Now we find the value of \[\sin 2x\] , using the formula \[\sin \left( A+B \right)=\sin A\cos B+\cos A\sin B\] by putting \[A=B=x\] . Now, evaluating this we get,
\[\begin{align}
  & \Rightarrow \sin 2x=\sin \left( x+x \right)=\sin x\cos x+\cos x\sin x \\
 & \Rightarrow \sin 2x=\sin \left( x+x \right)=2\sin x\cos x \\
\end{align}\]
Now we can write \[\cos 3x\] as \[\cos \left( 2x+x \right)\] . Now applying the above formula we get,
\[\cos 3x=\cos \left( 2x+x \right)=\cos 2x\cos x-\sin 2x\sin x\]
Now for this, using the value of \[\cos 2x\] and \[\sin 2x\] from the above we get,
\[\begin{align}
  & \Rightarrow \cos 3x=\cos \left( 2x+x \right)=\left( {{\cos }^{2}}x-{{\sin }^{2}}x \right)\cos x-\left( 2\sin x\cos x \right)\sin x \\
 & \Rightarrow \cos 3x={{\cos }^{3}}x-{{\sin }^{2}}x\cos x-2{{\sin }^{2}}x\cos x \\
 & \Rightarrow \cos 3x={{\cos }^{3}}x-3{{\sin }^{2}}x\cos x \\
\end{align}\]
Now in the above formed equation, we write \[{{\sin }^{2}}x=1-{{\cos }^{2}}x\] as we don’t want to keep any ‘sin’ terms in the formulae of cosine. We then evaluate it as,
\[\begin{align}
  & \Rightarrow \cos 3x={{\cos }^{3}}x-3\left( 1-{{\cos }^{2}}x \right)\cos x \\
 & \Rightarrow \cos 3x={{\cos }^{3}}x-3\cos x+3{{\cos }^{3}}x \\
 & \Rightarrow \cos 3x=4{{\cos }^{3}}x-3\cos x \\
\end{align}\]

Note: For such problems, we need to keep in mind the fundamental formula of trigonometry which are very useful while solving problems like these. We also need to be very careful while substituting the values of the functions that come in our way of solving. After putting the values correctly we need to add or subtract the terms of the equation accordingly.