Question

Express the following as a product: $\cos 2x-\cos 4x.$

Answer 1

We will use some known trigonometric identities to express the given difference as a product. We know the trigonometric identity $\cos C-\cos D=2\sin \dfrac{C+D}{2}\sin \dfrac{D-C}{2}.$ We will use this identity to express the given expression as a product.

Complete step by step solution:
Let us consider the given trigonometric expression $\cos 2x-\cos 4x.$
We need to write this expression as a product. So, for that we may use some familiar trigonometric identities.
We know the trigonometric identity given by $\cos C-\cos D=2\sin \dfrac{C+D}{2}\sin \dfrac{D-C}{2}.$
Let us compare the left-hand side of the identity with the given expression.
We will get $C=2x$ and $D=4x.$
When we apply these values, we will get the right-hand side of the above identity as $2\sin \dfrac{2x+4x}{2}\sin \dfrac{4x-2x}{2}.$
We can write this as $2\sin \dfrac{6x}{2}\sin \dfrac{2x}{2}.$
Since $\dfrac{6}{2}$ is $3$ and $\dfrac{2}{2}$ is $1,$ we will get $2\sin 3x\sin x.$
So, we will get $\cos 2x-\cos 4x=2\sin 3x\sin x.$
Hence, we can write the given expression which is a difference $\cos 2x-\cos 4x$ as a product $2\sin 3x\sin x.$

Note: Let us derive the trigonometric identity we have used to solve the problem, $\cos C-\cos D=2\sin \dfrac{C+D}{2}\sin \dfrac{D-C}{2}.$ Let us consider $C=\dfrac{A+B}{2}$ and $D=\dfrac{A-B}{2}.$ Now, we will use the addition and subtraction rules for Cosine function. We will get $\cos C-\cos D=\cos \dfrac{A+B}{2}-\cos \dfrac{A-B}{2}.$ We know that $\cos \left( x+y \right)=\cos x\cos y-\sin x\sin y.$ Similarly, we have $\cos \left( x-y \right)=\cos x\cos y+\sin x\sin y.$ Here we will take $x=\dfrac{A}{2}$ and $y=\dfrac{B}{2}.$ From this, we will get $\cos C-\cos D=\cos \dfrac{A}{2}\cos \dfrac{B}{2}-\sin \dfrac{A}{2}\sin \dfrac{B}{2}-\cos \dfrac{A}{2}\cos \dfrac{B}{2}-\sin \dfrac{A}{2}\sin \dfrac{B}{2}.$ We will cancel the similar terms with opposite signs. Then, we will get $\cos C-\cos D=-\sin \dfrac{A}{2}\sin \dfrac{B}{2}-\sin \dfrac{A}{2}\sin \dfrac{B}{2}.$ When we add this, we will get $\cos C-\cos D=-2\sin \dfrac{A}{2}\sin \dfrac{B}{2}.$ And we know that $\dfrac{C+D}{2}=\dfrac{1}{2}\left( \dfrac{A+B}{2}+\dfrac{A-B}{2} \right)=\dfrac{A}{2}.$ Similarly, we know that $\dfrac{D-C}{2}=\dfrac{1}{2}\left( \dfrac{A-B}{2}-\dfrac{A+B}{2} \right)=-\dfrac{B}{2}.$ Now, we can write the identity as $\cos C-\cos D=2\sin \dfrac{A}{2}\sin \left( -\dfrac{B}{2} \right).$ Therefore, we will get an important trigonometric identity as $\cos C-\cos D=2\sin \dfrac{C+D}{2}\sin \dfrac{D-C}{2}.$ In this way, we can derive all the trigonometric identities using the known basic identities