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

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

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.

• Hint: notice that $0\leq 1+\sin x\leq 2 (<\pi)$.
– Gary
Feb 9, 2021 at 12:33
• Ah, sorry. I was not attentive. Feb 9, 2021 at 12:40
• The minimum value of $\sin\theta$ occurs at $\color{Red}{\theta=}\frac{3\pi}{2}$... Feb 9, 2021 at 12:41
• $\pi/2$ 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.) Feb 9, 2021 at 12:42

$$-1\le\sin x\le 1,\forall x\in\mathbb{R}$$ 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 $$0 and decreases for $$\pi>x>\frac{\pi}{2}$$
• $$\sin (\sin x+1)$$ is decreasing when $$\frac{\pi}{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

$$...$$

• 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 Feb 9, 2021 at 12:47
• @Turing Thank you! Feb 9, 2021 at 12:49
• What about the local minimum $\pi/2$? Feb 9, 2021 at 13:00
• @MartinVesely I misread the question. Thought they wanted the global minimum. I'll edit my answer Feb 9, 2021 at 13:05

Notice that $$0\leq 1 +\sin x\leq 2$$ ($$<\pi$$). If $$0, then $$\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, this means $$x=\frac{3\pi}{2}$$.

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.