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

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.

$$f(x)=\cos(\sin x)+ \sin (\cos x) <2.$$, 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):

• Thanks, I have corrected it now. Sep 13, 2020 at 10:07

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 $$t \in [0, 1]$$.

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

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.

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.