How can I simplify \[\sin (2x)\cos (x) + \cos (2x)\sin (x)\]?

Hint: To solve this we need to know the formula of sum and difference formula of sine function. We know that sum formula of sine is \[\sin (a + b) = \sin (a)\cos (b) + \cos (a)\sin (b)\] and the difference formula of sine is \[\sin (a - b) = \sin (a)\cos (b) - \cos (a)\sin (b)\]. By comparing we can see that we can use the sum formula of sine function.

Complete step-by-step answer:
Given, \[\sin (2x)\cos (x) + \cos (2x)\sin (x)\].
We know the sum formula of sine is \[\sin (a + b) = \sin (a)\cos (b) + \cos (a)\sin (b)\], by comparing with the given problem we can say that \[a = 2x\] and \[b = x\].
Then we have,
\[ \Rightarrow \sin (2x)\cos (x) + \cos (2x)\sin (x) = \sin (2x + x)\]
\[ = \sin (3x)\]
Thus we have
\[ \Rightarrow \sin (2x)\cos (x) + \cos (2x)\sin (x) = \sin (3x)\]
So, the correct answer is “sin (3x)”.

Note: Whenever we are asked to simplify a given trigonometric expression, it is important to note that the trigonometric identities and equations that are applicable to the given expression must be used. For the same given problem if they are given options and the obtained answer is not matching with the options we can further simplify this. That is we have the identity: \[\sin (3x) = 3\sin (x) - 4{\sin ^3}(x)\]. Using this we have