How do you solve the problem \[sin5x - sinx\] using a sum to product formula?

Hint: We will first describe the concept and formula of compound angles for the trigonometric ratio of sin. We use the formula $\sin a - \sin b = 2\cos \left( {\dfrac{{a + b}}{2}} \right)\sin \left( {\dfrac{{a - b}}{2}} \right)$ and formula ${\cos ^2}\theta = 1 - {\sin ^2}\theta $ to minimize and simplify our equation.

Complete answer:
We have given the equation \[sin5x - sinx\] to simplify using sum to product formula.
We will use the formula $\sin a - \sin b = 2\cos \left( {\dfrac{{a + b}}{2}} \right)\sin \left( {\dfrac{{a - b}}{2}} \right)$ to simplify our term \[sin5x - sinx\]
We have equation
$ \Rightarrow sin5x - sinx$
We know that $\sin a - \sin b = 2\cos \left( {\dfrac{{a + b}}{2}} \right)\sin \left( {\dfrac{{a - b}}{2}} \right)$
We will substitute a equal to 5x and b equals to x
$ \Rightarrow sin5x - sinx = 2\cos \left( {\dfrac{{5x + x}}{2}} \right)\sin \left( {\dfrac{{5x - x}}{2}} \right)$
$ \Rightarrow sin5x - sinx = 2\cos \left( {3x} \right)\sin \left( {2x} \right)$
We will use the formula and expand $\cos 3x = (4cos3x - 3cosx)$ and $\sin 2x = (2sinxcosx)$
$ = 2(4co{s^3}x - 3cosx)(2sinxcosx)$
$ = 4\sin x{\cos ^2}x(4co{s^2}x - 3)$
We know that ${\cos ^2}\theta = 1 - {\sin ^2}\theta $, we will substitute
$ = 4\sin x(1 - {\sin ^2}x)(4(1 - {\sin ^2}x) - 3)$
We will multiply them and open the brackets
$ = 4\sin x(1 - {\sin ^2}x)(1 - 4{\sin ^2}x)$
$ = 4\sin x(1 - 5{\sin ^2}x + 4{\sin ^4}x)$
Hence, the equation \[sin5x - sinx\] is $4\sin x(1 - 5{\sin ^2}x + 4{\sin ^4}x)$

Note: An algebraic sum of two or more angles is known as a compound angle. Through trigonometric functions, we may denote compound angles using trigonometric identities. Similarly, using the formula, we may convert the form from product to sum \[2\sin A\sin B = \cos \left( {A - B} \right) - \cos \left( {A + B} \right)\]