Finding the range of $f(x)=\cos x\left(\sin x+\sqrt{\smash[b]{\frac12+\sin^2x}}\right)$

Question

How do I find the number of integers in the range of $$f(x)= \cos x\left( \sin x + \sqrt{\dfrac 12 +\sin^2 x} \right)?$$

I set the derivative equal to $0$ but the method isn't efficient here because it gives a very complicated trigonometric equation. What's the proper way to do this then?

Answer 1

Done it!

Let $f(x)= y$

$\implies y\sec x - \sin x = \sqrt{\sin^2x +\dfrac 12} $

$\implies y^2 \sec^2 x - 2y\sec x \sin x = \dfrac 12 \implies 2y^2(\tan^2x)+ - 4y \tan x + 2y^2 -1 = 0$

For $\tan x $ to be real, $\Delta \ge0$

$\implies 6y^2- 4y^4 \ge 0 \implies -\sqrt{\dfrac 32}\le y \le\sqrt{\dfrac 32} $

Which is the correct range according to the graph as well.

Hence, the number of integers in the range of $f(x)$ is 3: $0$, $-1$ and $1$

Answer 2

hint: Put $u = \cos x, v = \sin x$, then the problem becomes: Find the min/max of the function $f(u,v) = u\left(v+\sqrt{1/2+v^2}\right)$, subject to $u^2+v^2=1$ by Lagrange Multiplier method. I think it is doable this way. Can you try ? It is two variable function but it is much nicer than the original function indeed !

Answer 3

Since you are asking about integers in the range, we do not need to fully know the end points of the range, and we don't need calculus.

Clearly the integer $0$ is in the range, considering $x=\pi/2$.

Since $f(\pi/4)=\frac{\sqrt{2}}{2}\left(\frac{\sqrt{2}}{2}+1\right)=\frac{1+\sqrt{2}}{2}>1$, the integer $1$ is in the range as well (by the Intermediate Value Theorem).

Is $2$ in the range? If so, then $$ \begin{align} 2&=\cos x\left( \sin x + \sqrt{\dfrac 12 +\sin^2 x} \right)\\ 2-\sin(x)\cos(x)&=\cos(x)\sqrt{\dfrac 12 +\sin^2 x}\\ 4-4\sin(x)\cos(x)+\sin^2(x)\cos^2(x)&=\cos^2(x)\left(\dfrac 12 +\sin^2 x\right)\\ 4-4\sin(x)\cos(x)&=\frac{1}{2}\cos^2(x)\\ 4-2\sin(2x)&=\frac{1}{2}\cos^2(x) \end{align} $$ This has no solutions since the smallest the left side can be is $2$, and the largest the right side can be is $\frac{1}{2}$.

So $2$ is not in $f$'s range. Again using the Intermediate Value Theorem, no integer larger than $2$ is in the range either, or else $2$ would be in the range.

Lastly, note that $x\mapsto\pi-x$ preserves $\left( \sin x + \sqrt{\dfrac 12 +\sin^2 x} \right)$ but negates $\cos(x)$. So $f(\pi-x)=-f(x)$. And this means that $f$ takes negative values just as it takes positive values. That is, $-1$ is in the range, but not $-2$ or anything smaller.

So the integers in the range of $f$ make up the set $\{-1,0,1\}$.