Prove that $f(x)=\cos(2 \pi x)-\sin(2-\pi x) <0$ in $\left[\frac{1}{4},\frac{1}{2}\right]$

Question

As part of one of my exercises I have to argue that the function $f(x) = \cos(2 \pi x)+ 4\sin(2-\pi x)$ has a unique reversing function on the interval $\left[\frac{1}{4},\frac{1}{2}\right]$.

I want to use the argument, that f is strictly monotonously falling and continuous.

The continuity is clear, but I'm struggling with a formal proof of the monotony. I tried to show that $f'(x) = 2 \pi(4\cos(2 \pi x)-\sin(2-\pi x)) <0$ on the given interval and it's "intuitively" clear to me that this has to be the case, but I don't know how to substantiate this with formal arguments.

I'd be thankful for any help.

Answer 1

So first off, we have $$f(x) = \cos(2\pi x) + 4\sin(2 - \pi x),\ x \in \left[\frac{1}{4},\frac{1}{2} \right] $$ This is equivalent to considering the function $f(y)$ where $y = x/\pi$ and $$f(y) = \cos(2y) + 4\sin(2 - y),\ y \in \left[\frac{\pi}{4},\frac{\pi}{2} \right]$$ That just makes it a little easier on the eyes. Next, $$f'(y) = -2\sin(2y) - 4\cos(2 - y) $$ You want to show that $$\begin{eqnarray} f'(y) &<& 0\\ \iff-2\sin(2y) - 4\cos(2 - y) &<& 0\\ \iff \sin(2y) + 2\cos(2 - y) &>& 0\\ \iff 2\cos(y)\sin(y) + 2\cos(2)\cos(y) + 2\sin(2)\sin(y) &>& 0\\ \iff \cos(y)\sin(y) + \cos(2)\cos(y) + \sin(2)\sin(y) &>& 0\\ \iff \cos(y)(\sin(y) + \cos(2)) + \sin(2)\sin(y) &>& 0 \end{eqnarray}$$ Where in the 4th line down, I've used some basic trigonometric identities (sine double angle and the cosine difference of angles formula). If we can show that the very bottom inequality is true, then the top (desired) result will follow.

Notice that for $\frac{\pi}{4} \le y \le \frac{\pi}{2}$, $\frac{1}{2} \le \sin(y) \le 1$ and $0 \le \cos(y) \le \frac{1}{2}$. Moreover, you can use a calculator or Wolfram Alpha to verify that $\sin(2) \approx 0.9092\dots\ $, and $\cos(2) \approx -0.4161\dots$.

Since $\sin(2) > 0$ and $\sin(y) > 0$, we have that $\sin(2)\sin(y) > 0$ and we're good for that term.

Now notice that since $\sin(y) \ge 0.5$ and $-0.50 < \cos(2)$, we have that $\sin(y) + \cos(2) > 0 \Rightarrow \cos(y)(\sin(y) + \cos(2)) > 0$ since $\cos(y) > 0$.

It now follows that $\cos(y)(\sin(y) + \cos(2)) + \sin(2)\sin(y) > 0$, and hence $f'(y) < 0 \Rightarrow f'(x) < 0$.

Answer 2

There's actually no need to consider the derivative, provided you are familiar with the basic properties of the sine and cosine function, in particular where they are increasing and where they are decreasing.

For ${1\over4}\le x\le{1\over2}$, we have ${\pi\over2}\le2\pi x\le\pi$. Since the cosine function is strictly decreasing from $0$ at $\pi/2$ to $-1$ at $\pi$, $\cos2\pi x$ is strictly decreasing for ${1\over4}\le x\le{1\over2}$.

Similarly, for ${1\over4}\le x\le{1\over2}$, we have $0\lt2-{\pi\over2}\le2-\pi x\le2-{\pi\over4}\lt{\pi\over2}$, where the first and last inequalities reflect the simple inequalities $\pi\lt4$ and $8\lt3\pi$, respectively. Since the sine function is strictly increasing from $0$ at $0$ to $1$ at $\pi/2$, $4\sin(2-\pi x)$ is strictly decreasing for ${1\over4}\lt x\lt{1\over2}$. (Because of the minus sign in $2-\pi x$, you're starting at the largest value when $x={1\over4}$ and ending at the smallest value when $x={1\over2}$.)

Finally, the sum of two strictly decreasing functions is strictly decreasing. That (and continuity) is all it takes for $f(x)=\cos(2\pi x)+4\sin(2-\pi x)$ to have a unique inverse on the interval $[{1\over4},{1\over2}]$.