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I want to find the minima of $f(x) = \sin (1+ \sin x)$ for $0<x<6$

So a minimum value of $\sin \theta$ occurs at $\frac{3 \pi }{2}$, so I would have thought to set $1+\sin x = \frac{3 \pi}{2}$, but this is not valid.

The answers are two minima at $\frac{\pi}{2}$ and $\frac{3 \pi}{2}$ but I can’t seem to justify them. I can work out the maxima.

I don’t want to use calculus.

Any help would be appreciated.

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    $\begingroup$ Hint: notice that $0\leq 1+\sin x\leq 2 (<\pi)$. $\endgroup$
    – Gary
    Feb 9, 2021 at 12:33
  • $\begingroup$ Ah, sorry. I was not attentive. $\endgroup$
    – kolobokish
    Feb 9, 2021 at 12:40
  • $\begingroup$ The minimum value of $\sin\theta$ occurs at $\color{Red}{\theta=}\frac{3\pi}{2}$... $\endgroup$
    – runway44
    Feb 9, 2021 at 12:41
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    $\begingroup$ $\pi/2$ is only a local minimum; the global minima are at $3\pi/2+2n\pi$. How do you expect to locate local minima without calculus? (I'm not saying it's impossible, I just wonder whether you have thought this through.) $\endgroup$
    – TonyK
    Feb 9, 2021 at 12:42

3 Answers 3

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$$-1\le\sin x\le 1,\forall x\in\mathbb{R}$$ Thus we know that $0\le\sin x +1\le 2$ therefore $$\sin (\sin x+1)\ge 0$$ More precisely $\sin (\sin x+1)=0$ at $x=\frac32\pi$ where $\sin x=-1$ and we have the global minimum.

To get the local minimum at $x=\frac{\pi}{2}$ consider that

  • $\sin x$ increases for $0<x<\frac{\pi}{2}$ and decreases for $\pi>x>\frac{\pi}{2}$
  • $\sin (\sin x+1)$ is decreasing when $\frac{\pi}{2}<x<2$, reaches a minimum $\sin 2$ when $\sin x = 1$ at $x=\frac{\pi}{2}$ then increases again up to $1$ when $x=\pi$

Hope it is clear. It's quite complicated to explain without derivatives :)

a graph can help


$$...$$ enter image description here

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  • $\begingroup$ Did you want to write $0 \leq 1 + \sin(x) \leq 2$ by chance? Because otherwise $0\leq \sin(x)\leq 2$ sounds a bit weird :D $\endgroup$
    – Laplacian
    Feb 9, 2021 at 12:47
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    $\begingroup$ @Turing Thank you! $\endgroup$
    – Raffaele
    Feb 9, 2021 at 12:49
  • $\begingroup$ What about the local minimum $\pi/2$? $\endgroup$ Feb 9, 2021 at 13:00
  • $\begingroup$ @MartinVesely I misread the question. Thought they wanted the global minimum. I'll edit my answer $\endgroup$
    – Raffaele
    Feb 9, 2021 at 13:05
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Notice that $0\leq 1 +\sin x\leq 2$ ($<\pi$). If $0<w<2$, then $\sin w >0$ and if $w=0$, then $\sin w = 0$. Thus the (global) minimum will occur when $1+\sin x =0$, i.e., $\sin x =-1$. Since $0<x<6$, this means $x=\frac{3\pi}{2}$.

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I understand that you do not want to use calculus but it seems to be the only way how to identify all minima:

A first derivative of $f(x)$ is $f'(x) = \cos(x)\cos(1+\sin(x))$. From $f'(x)=0$ we have two equations:

$$ \cos(x) = 0 $$

$$ \cos(1+\sin(x))=0 $$

On given interval the first equation has solution $x \in \{\pi/2; 3\pi/2\}$.

The second one can be turned to $1+\sin(x) = \frac{\pi}{2}$ since cosinus is zero in $\pi/2$, hence we have either $$ x = \arcsin(\pi/2-1) $$ or $$ x = \pi - \arcsin(\pi/2-1) $$

You can show that $\pi/2$ adn $3\pi/2\$ are minima and others are maxima.

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