Minima of $\sin (1+\sin x)$

Question

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.

Answer 1

$$-1\le\sin x\le 1,\forall x\in\mathbb{R}$$ 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

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Answer 2

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 $x=\frac{3\pi}{2}$.

Answer 3

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.