Prove the following expression
$\sin x+\sin 3x+\sin 5x+\sin 7x=4\cos x\cos 2x\sin 4x$

Hint: To solve this question, recall all the formulas that you have studied in trigonometry. There is a formula in trigonometry for addition of two $\sin $ functions with different arguments. The formula is $\sin A+\sin B=2\sin \left( \dfrac{A+B}{2} \right)\cos \left( \dfrac{A-B}{2} \right)$. Use this formula to solve this question.

Complete step-by-step answer:

Before proceeding with the question, one must know all the formulas that will be required to solve this question. In trigonometry, we have a formula that can be applied to the sum of two $\sin $ functions with different arguments. That formula is,
$\sin A+\sin B=2\sin \left( \dfrac{A+B}{2} \right)\cos \left( \dfrac{A-B}{2} \right)..............\left( 1 \right)$
Also, in trigonometry, there is a formula for $\cos $ function. That formula is,
$\cos \left( -A \right)=\cos \left( A \right)..............\left( 2 \right)$
In this question, we have to prove $\sin x+\sin 3x+\sin 5x+\sin 7x=4\cos x\cos 2x\sin 4x$.
Let us start from the left side of the expression. On the left side, we have,
\[\sin x+\sin 3x+\sin 5x+\sin 7x\]
We have to prove this equal to the right side of the expression i.e. $4\cos x\cos 2x\sin 4x$.
Let us club some terms on the left side with the use of brackets.
$\left( \sin x+\sin 3x \right)+\left( \sin 5x+\sin 7x \right)....................\left( 3 \right)$
Inside the individual bracket, we can see that the term is the sum of two $\sin $ functions with different arguments. Hence, we can apply formula $\left( 1 \right)$ in both the brackets.
Applying formula $\left( 1 \right)$ to the first bracket i.e. $\sin x+\sin 3x$, we get,
\[\begin{align}
  & \sin x+\sin 3x=2\sin \left( \dfrac{x+3x}{2} \right)\cos \left( \dfrac{x-3x}{2} \right) \\
 & \Rightarrow \sin x+\sin 3x=2\sin \left( \dfrac{4x}{2} \right)\cos \left( \dfrac{-2x}{2} \right) \\
 & \Rightarrow \sin x+\sin 3x=2\sin \left( 2x \right)\cos \left( -x \right) \\
\end{align}\]
From the formula $\left( 2 \right)$, we can say $\cos \left( -x \right)=\cos x$.
\[\Rightarrow \sin x+\sin 3x=2\sin 2x\cos x............\left( 4 \right)\]
Applying formula $\left( 1 \right)$ to the second bracket i.e. $\sin 5x+\sin 7x$, we get,
\[\begin{align}
  & \sin 5x+\sin 7x=2\sin \left( \dfrac{5x+7x}{2} \right)\cos \left( \dfrac{5x-7x}{2} \right) \\
 & \Rightarrow \sin 5x+\sin 7x=2\sin \left( \dfrac{12x}{2} \right)\cos \left( \dfrac{-2x}{2} \right) \\
 & \Rightarrow \sin 5+\sin 7x=2\sin \left( 6x \right)\cos \left( -x \right) \\
\end{align}\]
From the formula $\left( 2 \right)$, we can say $\cos \left( -x \right)=\cos x$.
\[\Rightarrow \sin 5x+\sin 7x=2\sin 6x\cos x............\left( 5 \right)\]
Substituting equation $\left( 4 \right)$ and equation $\left( 5 \right)$ in the expression $\left( 3 \right)$, we get,
 $\begin{align}
  & \left( \sin x+\sin 3x \right)+\left( \sin 5x+\sin 7x \right)=2\sin 2x\cos x+2\sin 6x\cos x \\
 & \Rightarrow \left( \sin x+\sin 3x \right)+\left( \sin 5x+\sin 7x \right)=2\cos x\left( \sin 2x+\sin 6x \right)..................\left( 6 \right) \\
\end{align}$
Applying formula $\left( 1 \right)$ to $\sin 2x+\sin 6x$ in the above expression, we get,
$\begin{align}
  & \sin 2x+\sin 6x=2\sin \left( \dfrac{2x+6x}{2} \right)\cos \left( \dfrac{2x-6x}{2} \right) \\
 & \Rightarrow \sin 2x+\sin 6x=2\sin \left( \dfrac{8x}{2} \right)\cos \left( \dfrac{-4x}{2} \right) \\
 & \Rightarrow \sin 2x+\sin 6x=2\sin \left( 4x \right)\cos \left( -2x \right) \\
\end{align}$
From the formula $\left( 2 \right)$, we can say $\cos \left( -2x \right)=\cos 2x$.
$\Rightarrow \sin 2x+\sin 6x=2\sin 4x\cos 2x$
Substituting $\sin 2x+\sin 6x=2\sin 4x\cos 2x$ in equation$\left( 6 \right)$, we get,
$\begin{align}
  & \left( \sin x+\sin 3x \right)+\left( \sin 5x+\sin 7x \right)=2\cos x\left( 2\sin 4x\cos 2x \right) \\
 & \Rightarrow \left( \sin x+\sin 3x \right)+\left( \sin 5x+\sin 7x \right)=4\cos x\cos 2x\sin 4x \\
\end{align}$
Hence, we have proved the left side of the expression in the question to it’s right side.

Note: One can also solve this question by simplifying the right side term using the formula $2\cos a\cos b=\cos \left( a+b \right)+\cos \left( a-b \right)$ and $2\sin a\cos b=\sin \left( a+b \right)+\sin \left( a-b \right)$. Simplifying the right side using these formulas, we will get the left side.