Question

What is $\cos x$ times $\cos 2x$?

Answer 1

Hint: Here we will first find the value of the function $\cos 2x$ in terms of $\cos x$. For doing this we will use the trigonometric identity $\cos \left( a+b \right)=\cos a\cos b-\sin a\sin b$ where we will substitute a = b = x. Further we will use the identity ${{\cos }^{2}}x+si{{n}^{2}}x=1$ to replace the sine function. Once the value of $\cos 2x$is found we will take its product with $\cos x$ to get the answer.

Complete step by step answer:
Here we have been asked to simplify the expression given by the sentence $\cos x$ times $\cos 2x$. That means we need to take the product of the cosine functions $\cos x$ and $\cos 2x$. Let us assume its mathematical expression as E, so we have,
$\Rightarrow E=\cos x\times \cos 2x$
Now, first we will find the value of $\cos 2x$ in terms of $\cos x$ and then we will consider the required product. So we can write the argument of $\cos 2x$ as:
$\Rightarrow \cos 2x=\cos \left( x+x \right)$
Using the trigonometric identity $\cos \left( a+b \right)=\cos a\cos b-\sin a\sin b$ and substituting a = b = x we get,
$\begin{align}
  & \Rightarrow \cos 2x=\cos x\times \cos x-\sin x\times \sin x \\
 & \Rightarrow \cos 2x={{\cos }^{2}}x-{{\sin }^{2}}x \\
\end{align}$
Further simplifying the above relation by using the formula ${{\cos }^{2}}x+si{{n}^{2}}x=1$ to replace the sine function we get,
\[\begin{align}
  & \Rightarrow \cos 2x={{\cos }^{2}}x-\left( 1-{{\cos }^{2}}x \right) \\
 & \Rightarrow \cos 2x={{\cos }^{2}}x-1+{{\cos }^{2}}x \\
 & \Rightarrow \cos 2x=2{{\cos }^{2}}x-1 \\
\end{align}\]
Substituting the above value in expression E we get,
\[\begin{align}
  & \Rightarrow E=\cos x\times \left( 2{{\cos }^{2}}x-1 \right) \\
 & \therefore E=2{{\cos }^{3}}x-\cos x \\
\end{align}\]
Hence the above relation is our answer.

Note: You can also get the answer in a different format. What you can do is write the product \[\cos x\times \cos 2x\] as $\dfrac{1}{2}\left( 2\cos x\times \cos 2x \right)$ and then use the use the identity $2\cos a\cos b=\cos \left( a+b \right)+\cos \left( a-b \right)$ by considering a = 2x and b = x. In this case you will get the answer $\dfrac{1}{2}\left( \cos 3x+\cos x \right)$ and it will also be the correct answer only the form will be different. You must remember all the trigonometric identities as they are used in other chapters and subjects also.