Find maximum value of $(1 + \sin x)(1 + \cos x)$. The question in my textbook was to find the minimum value of function, which was pretty simple, $0$. But that led me to thinking if we could find it's range.
I tried everything from AM-GM to differentiation to number theory and pure intuition.
The differentiation led me no where. By using AM GM and number theory, I was getting to $\left[0, \frac{3+2\sqrt{2}}{2}\right]$. I think this might be wrong, but I am writing my number theory solution here:
We know if $a+b=c$, then max value of $ab$ is when $a=b=c/2$. So max value of $ab= c^2/4$. If we say $(1 + \sin x) = a$ and $b=(1 + \cos x)$ and $(1 + \sin x)+(1 + \cos x)=c$, then max value of $ab$ would be $\frac{c^2}{4}$, we know max value of $c$ is $2 + \sqrt{2}$, so max value of $\frac{c^2}{4}$ would be on simplification $\frac{3 + 2\sqrt{2}}2$.
I had searched a lot for this on the internet but I could not find it, my solution seems right to me but I am not sure. Any help would be appreciated.
 A: $$(1+\sin x)(1+\cos x)=\frac12(\sin x+\cos x+1)^2=\left(\cos\left(x-\frac\pi4\right)+\frac1{\sqrt2}\right)^2$$
and the extrema must be
$$0\text{ and }\left(1+\frac1{\sqrt 2}\right)^2.$$
A: Let $y=(1+\sin x)(1+\cos x)=1+\sin x+\cos x+\sin x\cos x$
If $\sin x+\cos x=u, u^2=1+2\sin x\cos x\le2\implies u\le\sqrt2$
$y=1+u+\dfrac{u^2-1}2=\dfrac{u^2+2u+1}2=\dfrac{(u+1)^2}2$
Now $u+1\le\sqrt2+1$
A: By AM-GM and C-S we obtain:
$$(1+\sin{x})(1+\cos{x})\leq\left(\frac{1+\sin{x}+1+\cos{x}}{2}\right)^2\leq$$
$$\leq\left(\frac{2+\sqrt{(1+1)(\sin^2x+\cos^2x)}}{2}\right)^2=\left(1+\frac{1}{\sqrt2}\right)^2.$$
The equality occurs for $x=45^{\circ},$ which says that we got a maximal value.
As you said, the minimal value is $0$ and since our expression is continuous,
we got a range, which you wrote.
The inequality $\sin{x}+\cos{x}\leq\sqrt2$ we can prove by using AM-GM:
$$\sin{x}+\cos{x}\leq|\sin{x}|+|\cos{x}|=\sqrt{1+2|\sin{x}|\cdot|\cos{x}|}\leq$$
$$\leq\sqrt{1+|\sin{x}|^2+|\cos{x}|^2}=\sqrt2.$$
A: $$\frac d{dx}(1+\sin x)(1+\cos x)$$
$$=\frac d{dx}[1+\cos x](1+\sin x)+\frac d{dx}[1+\sin x](1+\cos x)$$
$$=-\sin x(1+\sin x)+\cos x(1+\cos x)$$
$$=\cos^2x-\sin^2x+\cos x-\sin x$$
$$=(\cos x-\sin x)(\cos x+\sin x+1)$$
Now when either $\cos x-\sin x=0$ or $\cos x+\sin x+1=0$, the expression equals 0. You can find the solutions to $x$ and check if each of them results in a maximum or minimum.
