Find the range of $f(x)=\cos(\sin x)+\sin(\cos x)$

Question

Find the range of $$f(x)=\cos(\sin x)+\sin(\cos x)$$ My try: Evident that $f$ is $2\pi$ periodic.

Let us assume $\cos x=t$ $\implies$ $t \in [-1,1]$ So the equivalent function of $f(x)$ is now: $$g(t)=\sin t+\cos\left(\sqrt{1-t^2}\right)$$ Case $1.$ Let $t \in [0,1]$ Now $$g'(t)=\cos t+t \times\frac{\sin\left(\sqrt{1-t^2}\right)}{\sqrt{1-t^2}}$$ $\implies$ $g'(t) >0$ So $g$ is increasing from $\cos 1$ to $1+\sin 1$ in $t \in [0,1]$

Case $2.$ When $t \in [-1,0]$ I came to know from the graph of $g(t)$ that it is decreasing. But how to prove it is decreasing formally i could't do it.

Answer 1

$$f(x)=\cos(\sin x)+ \sin (\cos x) <2. $$, is bounded periodic with period $2\pi$, $f(x)$ is an even function. So it would suffice if we study it in $[0,\pi]$. $$f'(x)=-\sin (\sin x) \cos x-\cos(\cos x)\sin x \implies f'(0)=0, f'(\pi)=0. $$ $f'(x)=0$ also implies $$\frac{\sin(\sin x)}{\cos(\cos x)}=-\tan x,$$ that is satisfied by $x=2.4488=x_0$ $$f(0)=1+\sin 1, f(x_0)=0.1071, f(\pi)=1-\sin[1]=1.8414$$ indicating local max, min and max at $x=0,3\pi/4, \pi$. Therefor the range of $f(x)$ is$$ [~0.1071,~1+\sin 1 ~]$$ See the fig for $f(x):

Answer 2

You really only need to worry about $x \in [0, \pi/2]$. As you said, the period is $2\pi$, so let's only worry about $[-\pi, \pi]$. Since the function is even, it follows that we can just consider $[0, \pi]$. That said, for each $x_1 \in (\pi/2, \pi]$, consider $x_2 = \pi - x_1 \in [0, \pi]$. We have $\cos(x_1) \in [-1, 0)$, hence $\sin(\cos(x_1)) \in [-\sin(1), 0)$ and so \begin{align*} f(x_1) &= \cos(\sin(x_1)) + \sin(\cos(x_1)) \\ &= \cos(\sin(x_2)) - \sin(\cos(x_2)) \\ &< \cos(\sin(x_2)) + \sin(\cos(x_2)) \\ &= f(x_2) \end{align*} Therefore, every point in $(\pi/2, \pi]$ is lesser than some point in $[0, \pi/2]$, hence the maximum must occur in $[0, \pi/2]$, where $t \in [0, 1]$.

Answer 3

Show that $f(-x)=f(x)$. So you have an even function, and that handles your case 2.

Answer 4

It's not decreasing on $t \in [-1, 0]$ actaully. There are at least 2 intervals $[-1, t_0]$ and $[t_0, 0]$ (maybe more). On the first one it's decreasing and on the second one it's increasing. If you take a $\lim\limits_{t \rightarrow -1} g^{'}(t)$ it would be equal to $cos(-1) - 1$ (because $\lim\limits_{y \rightarrow 0} \frac{sin y}{y} = 1$). And that's a negative value. Then $g^{'}(0) = 1$. And finding a point where $g^{'}(t) = 0$ doesn't look like an easy task. I'm not even sure it can be done analytically. There are always some problems with in analytical derivation with transcendental function.

Answer 5

What I am gonna present isn't a very rigorous mathematical solution, but some intuition-based and observational approach.

First of all, the function is continuous and differentiable in its domain.

Second, it is an even function, i.e. it is symmetric about y-axis. Every even function, which is differentiable, has zero slope at $x=0$. Otherwise, it would have a sharp corner at $x=0$ which would render it non differentiable. So $x=0$ is a critical point.

Now, to find the other critical points, we need to take the derivative and equate it to zero, i.e. $$\frac{d}{dx}\left(\cos\left(\sin x\right)+\sin\left(\cos x\right)\right)=0$$ Which yields this beautiful equation $$\frac{\sin\left(\sin x\right)}{\sin x}=-\frac{\cos\left(\cos x\right)}{\cos x}$$

The points where $\sin x$ is zero satisfies the equation(here is is an indetereminate form but in the original derivative, it would yield zero)

Analytically, on plotting the function, the points that correspond to sinx=0 represent local maxima, out of which $x∈2n\pi$ represents the global maxima.

The points of minima are almost equal to $x∈(2n+1)\pi±\frac{\pi}{4}$ but not exactly.

$$f(x=2n\pi)=1+\sin 1$$ $$f(x=(2n+1)\pi±\frac{\pi}{4})=\cos\left(\frac{1}{\sqrt{2}}\right)-\sin\left(\frac{1}{\sqrt{2}}\right)$$

On plotting $y = \cos\left(\frac{1}{\sqrt{2}}\right)-\sin\left(\frac{1}{\sqrt{2}}\right)$ along with the function and with $y=1+\sin 1$, we get the the following graphs

However on zooming in,

SO for approximation purposes you can assume the minima to be $y = \cos\left(\frac{1}{\sqrt{2}}\right)-\sin\left(\frac{1}{\sqrt{2}}\right)$, depending upon your error tolerance.