Question

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

Answer 1

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

& \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{aligned}$$ We have obtained a generalized formula in which we just split the given term of $$\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)$$