Question

What is \[\sin \left( 4x \right)\] in terms of \[\sin \left( 2x \right)\text{ and cos}\left( 2x \right)\]?

Answer 1

Hint: In the given question we need to split a \[\sin \] function into \[\sin \text{ and cos functions}\] of just the half angles. We will proceed by very basic formula that is \[\sin \left( \alpha +\beta \right)=\sin \alpha \cos \beta +\cos \alpha \sin \beta \] and then we will generalise it to form a new formula so that we can directly use it further for splitting \[\sin \text{ and cos functions}\] into half angles.

Complete step by step solution:
So, let’s just look at the formula once again,
\[\sin \left( \alpha +\beta \right)=\sin \alpha \cos \beta +\cos \alpha \sin \beta \]
Since we need to find an expression for \[\sin \left( 2\alpha \right)\], we will split \[2\alpha \text{ into }\alpha +\alpha \]
So, proceeding further we got the left-hand side term as \[\sin \left( \alpha +\alpha \right)\], putting this in the formula we get
\[\begin{align}
  & \sin \left( \alpha +\alpha \right)=\sin \left( \alpha \right)\cos \left( \alpha \right)+\cos \left( \alpha \right)\sin \left( \alpha \right) \\
 & \sin \left( \alpha +\alpha \right)=2\sin \left( \alpha \right)\cos \left( \alpha \right) \\
 & \sin \left( 2\alpha \right)=2\sin \left( \alpha \right)\cos \left( \alpha \right)...............(i) \\
\end{align}\]
We have obtained a generalized formula in which we just split the given term of \[\sin \]into another term of \[\sin \text{ and cos}\] in just the half angles.
So, we can call the obtained equation \[(i)\] as Half angle formula for \[\sin \] functions
Now coming back to our question, we have \[\sin \left( 4x \right)\] on the left-hand side Let us do a simple transformation for our ease
Let \[2x=y\]
So, our left-hand side expression becomes \[\sin \left( 2y \right)\] for which we have a direct formula
\[\sin \left( 2y \right)=2\sin \left( y \right)\cos \left( y \right).............\left( ii \right)\]
Now putting the value of \[y\]in the equation \[\left( ii \right)\] we get
\[\sin \left( 4x \right)=2\sin \left( 2x \right)\cos \left( 2x \right)\]
Which is the required result.

Note: Note that here we did a transformation of x in y just for better understanding and our ease, we can directly apply the half-angle formula in x also, also do remember the half-angle formula always because it saves your time in the exam, and helps in solving the questions very easily. You can try finding out a similar half-angle formula for \[\cos \] function. The final result will look something like this \[\cos \left( 2x \right)={{\cos }^{2}}\left( x \right)-{{\sin }^{2}}\left( x \right)\]