Review question

# Can we sketch $y=3 \cos x - \cos 3x$? Add to your resource collection Remove from your resource collection Add notes to this resource View your notes for this resource

Ref: R5251

## Solution

Prove that $\begin{equation*} \frac{d}{dx} (3 \cos x - \cos 3x)= 6 \sin x \cos 2x. \end{equation*}$
Using the chain rule or by thinking about horizontal scaling, we have that $\begin{equation*} \frac{d}{dx} \cos 3x = - 3 \sin 3x. \end{equation*}$ By the compound angle formula, \begin{align*} \sin 3x = \sin(x + 2x) &= \sin x \cos 2x + \sin 2x \cos x \\ &= \sin x \cos 2x + 2 \sin x \cos^2 x \\ &= \sin x \cos 2x + \sin x (1 + \cos 2x) \\ &= 2 \sin x \cos 2x + \sin x \end{align*} so that \begin{align*} \frac{d}{dx} \left( 3 \cos x - \cos 3x \right) &= - 3 \sin x + 3 \sin 3x \\ &= - 3 \sin x + 6 \sin x \cos 2x + 3 \sin x \\ &= 6 \sin x \cos 2x. \end{align*}

Alternatively, we can use the standard formula, $\sin P-\sin Q = 2\cos\left(\dfrac{P+Q}{2}\right)\sin\left(\dfrac{P-Q}{2}\right)$.

For what two ranges of values of $x$ between $0$ and $\pi$ is $\begin{equation*} 3 \cos x - \cos 3x \end{equation*}$

an increasing function of $x$?

A function is increasing exactly when its derivative is positive, so we are looking for those ranges of values of $x$ such that $\begin{equation*} \frac{d}{dx} \left( 3 \cos x - \cos 3x \right) > 0 \end{equation*}$ which, from the above, is equivalent to the inequality $\begin{equation*} \sin x \cos 2x > 0. \end{equation*}$

This holds exactly when $\sin x$ and $\cos 2x$ have the same sign. We know that $\sin x$ is positive when $0 < x < \pi$, so the required inequality can only hold when $\cos 2x > 0$. We know that:

• $\cos 2x > 0$ when $0 < x < \pi/4$,
• $\cos 2x < 0$ when $\pi/4 < x < 3\pi/4$, and
• $\cos 2x > 0$ when $3\pi/4 < x < \pi$.

The regions are therefore $0< x < \pi/4$ and $3\pi/4 < x < \pi$.

Draw a rough sketch-graph of $y = 3 \cos x - \cos 3x$ for values of $x$ from $0$ to $\pi$, indicating clearly any maximum or minimum points.

From the above, we know that $y$ is

• equal to $2$ when $x = 0$;
• increasing in the region $0 < x < \pi/4$;
• stationary when $x = \pi/4$;
• decreasing in the region $\pi/4 < x < 3\pi/4$;
• stationary when $x = 3\pi/4$;
• increasing when $3\pi/4 < x < \pi$; and
• equal to $-2$ when $x = \pi$.
The only aspects left to consider are the location of the $x$-intercept in the decreasing phase and the value of the function at the stationary points. Symmetry suggests that the $x$-intercept might be at $x=\frac{\pi}{2}$ so we start by trying that. \begin{align*} \text{At }x=\frac{\pi}{2},\quad y &= 3 \cos \frac{\pi}{2} - \cos \frac{3\pi}{2} = 0,\\ \text{At }x=\frac{\pi}{4},\quad y &= 3 \cos \frac{\pi}{4} - \cos \frac{3\pi}{4} = \frac{3\sqrt{2}}{2} + \frac{\sqrt{2}}{2} = 2\sqrt{2},\\ \text{At }x=\frac{3\pi}{4},\quad y &= 3 \cos \frac{3\pi}{4} - \cos \frac{9\pi}{4} = -\frac{3\sqrt{2}}{2} - \frac{\sqrt{2}}{2} = -2\sqrt{2}. \end{align*}

We also have rotational symmetry about $\left(\dfrac{\pi}{2},0\right)$, as $y(x) = -y\left(\pi - x \right)$ for any $0 < x < \dfrac{\pi}{2}$. We thus have the following sketch graph.