How do you use the sum and double angle identities to find \[\sin 3x\]?

Hint: The trigonometric functions here given here that we have to solve here by using the sum and double angle identities. The following are the identities used for solving the above trigonometric function. The identities of the angle sum of \[\sin \]:-
\[\sin \left( \alpha +\beta \right)\,=\,\sin \,x\,\,\cos \beta +\,\cos \alpha \,\,sin\beta \] and the double angle identity is:-
\[{{\cos }^{2\alpha }}\,=\,{{\cos }^{2}}\alpha \,-\,{{\sin }^{2}}\alpha \,=\,2\,\,{{\cos }^{2}}\alpha -1=1-2{{\sin }^{2}}\alpha \]
By using the above condition we have to solve the given problem.

Complete step by step solution:
Here, we have to use the sum and double angle identities to find \[\sin \,3x\].
So, the angle sum formula is \[\sin \left( \alpha +\beta \right)=\sin \alpha cos\beta +cos\,\alpha \,sin\beta \].
By using the formula we get, \[\sin \,3x\,\,=\,\sin \,\left( 2x\,+\,x \right)\]
Now, we have to use the double angle formula.
\[\cos 2\alpha \,\,=\,{{\cos }^{2}}\alpha -{{\sin }^{2}}\alpha =2{{\cos }^{2}}\alpha \,\,-1\,\,=\,1-2{{\sin }^{2}}\alpha \]
Applying this in the problem we get,
\[\sin \,3x\,=\,\sin \,\left( 2x\,+\,x \right)\,=\,\sin \,2x\,\,{{\cos }{x}}+{{\cos }{2x}}\,\,\sin x\]
\[=2\sin x\,\,{{\cos }^{2}}x\,\,+\,\sin x\,-2{{\sin }^{3}}x\]
\[=\,2\sin \,x\,\left( 1-{{\sin }^{2}}x \right)+\sin x\,-\,2{{\sin }^{3}}x\]
Now simplifying it we get,
\[=2\,\sin x-\,2{{\sin }^{3}}x\,+\,\sin x\,\,-\,2{{\sin }^{3}}x\]
\[=3\sin x\,\,-\,4{{\sin }^{3}}x\]

By using the sum and double angle identities for finding the \[{{\sin }^{3}}x\] we get \[3\sin x-4{{\sin }^{3}}x\].

Note: Always start to solve the example from the complex side. Which is more difficult to solve. Also we have to express everything in the \[\sin \] and cosine. Check all the terms where we have to apply the double angle formula where it is needed. If there is \[{{\sin }^{2}}x\] and \[{{\cos }^{2}}x\] we use the Pythagoras identities to transform it.