Question

Given two functions $f\left( x \right)$ and $g\left( x \right)$. For what $x$ does the equation ${f}'\left( x \right)=g\left( x \right)$ hold true?
$f\left( x \right)={{\sin }^{3}}2x$ and $g\left( x \right)=4\cos 2x-5\sin 4x$.

Answer 1

We will find the derivative of $f\left( x \right)$. Then we will simplify the obtained derivative. We will use the identities involving double angles for simplification like, $\sin 2x=2\sin x\cos x$. After simplifying the derivative, we will look for the value of $x$ such that the equation ${f}'\left( x \right)=g\left( x \right)$ holds true.

Complete step by step answer:
The given functions are $f\left( x \right)={{\sin }^{3}}2x$ and $g\left( x \right)=4\cos 2x-5\sin 4x$. Now, let us calculate the derivative of the function $f\left( x \right)$ as follows,
${f}'\left( x \right)=3{{\sin }^{2}}2x\times 2\cos 2x$
Now, we will simplify the above equation in the following manner,
${f}'\left( x \right)=3\sin 2x\times 2\sin 2x\cos 2x$
We know that $\sin 2x=2\sin x\cos x$.
Therefore, we have ${f}'\left( x \right)=3\sin 2x\times \sin 4x$.
Now, we want to find the value of $x$ such that the equation ${f}'\left( x \right)=g\left( x \right)$. Substituting the values of ${f}'\left( x \right)$ and $g\left( x \right)$, we get the following equation,
$3\sin 2x\sin 4x=4\cos 2x-5\sin 4x$
The angles of the trigonometric functions in the above equation are $2x$ and $4x$. If we consider $x=\dfrac{\pi }{4}$, we have $2x=2\times \dfrac{\pi }{4}=\dfrac{\pi }{2}$ and $4x=4\times \dfrac{\pi }{4}=\pi$. We will check the LHS of the above equation for $x=\dfrac{\pi }{4}$ as follows,
$\begin{aligned} & \text{LHS =}3\sin \left( 2\times \dfrac{\pi }{4} \right)\sin \left( 4\times \dfrac{\pi }{4} \right) \\\ & =3\sin \left( \dfrac{\pi }{2} \right)\sin \left( \pi \right) \end{aligned}$
Now, we know that $\sin \pi =0$. Therefore, $\text{LHS = 0}$. Next, let us check the RHS of the above equation for $x=\dfrac{\pi }{4}$,
$\begin{aligned} & \text{RHS =}4\cos \left( 2\times \dfrac{\pi }{4} \right)-5\sin \left( 4\times \dfrac{\pi }{4} \right) \\\ & =4\cos \left( \dfrac{\pi }{2} \right)-5\sin \left( \pi \right) \end{aligned}$
We know that $\cos \dfrac{\pi }{2}=0$ and $\sin \pi =0$. Therefore, $\text{RHS = 0}$. Since $\text{LHS = RHS = 0}$, we have ${f}'\left( x \right)=g\left( x \right)$ for $x=\dfrac{\pi }{4}$.

Note: Simplifying the given equations using trigonometric identities is useful in such type of questions. After simplification, we can guess the value of the angle by looking at the simplified equation. It is useful to know the values of the trigonometric functions for standard angles. The calculations need to be done carefully and to avoid confusion, it is essential to write the calculations explicitly.