Extrema of $(1+\sin x)(1+\cos x)$

Question

Find the extrema of $(1+\sin x)(1+\cos x)$ without using calculus.

I was able to figure out the minima by observing that each of the brackets range from $0$ to $2$. Therefore the minima has to be $0$ when either one of the brackets is zero.

However I couldnt figure out the maxima. I tried expanding it to complete the square but it didnt quite work out well.

Any hint is appreciated!

Answer 1

Hint: $(1+\sin x)(1+\cos x)=1+\sin x+\cos x+\frac 12\sin 2x=1+\sqrt 2\sin(x+\frac {\pi}{4})+\frac 12\sin 2x$

Answer 2

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

Note that $\sin x \cos x = \frac 12 \sin 2x.$

Note also that $\sin x + \cos x = \sqrt 2 \left ( \frac {\sqrt 2}{2} \sin x + \frac {\sqrt 2}{2} \cos x \right) = \sqrt 2 \left ( \cos \frac {\pi}{4} \sin x + \sin \frac{\pi}{4} \cos x \right )= \sqrt 2 \sin (x+ \frac{\pi}{4}).$

Both of these factors reach their maximum when $x=\frac{\pi}{4}$, so the maximum is $\frac 32 + \sqrt 2$.

Answer 3

We can also use the inequality \begin{gather*} AM\geq GM\\ So,\ \\ \sqrt{( 1+\sin x)( 1+\cos x)} \leq \frac{1+\sin x+1+\cos x}{2}\\ Now,\ the\ function\ on\ the\ RHS\ is\\ 1+\frac{\sin x+\cos x}{2} =1+\frac{1}{\sqrt{2}}\sin\left( x+\frac{\pi }{4}\right)\\ The\ maxima\ of\ this\ function\ is\ 1+\frac{1}{\sqrt{2}} ,\\ because\ maximum\ value\ of\ \sin\left( x+\frac{\pi }{4}\right) =1\\ \sqrt{( 1+\sin x)( 1+\cos x)} \leq 1+\frac{1}{\sqrt{2}}\\ ( 1+\sin x)( 1+\cos x) \leq \left( 1+\frac{1}{\sqrt{2}}\right)^{2} =\frac{3+2\sqrt{2}}{2} =\frac{3}{2} +\sqrt{2} \end{gather*}

Answer 4

Note

\begin{align} A&=(1+\sin x)(1+\cos x)\\ &= 1+\sin x+ \cos x+ \frac12\sin2x\\ &= 1+\sqrt2 \cos(\frac\pi4-x)+ \frac12\cos(\frac\pi2-2x)\\ &= \frac12+\sqrt2 \cos(\frac\pi4-x)+ \cos^2(\frac\pi4-x)\\ &= \left( \cos(\frac\pi4 -x) + \frac1{\sqrt2}\right)^2 \end{align} Thus $$0\le A \le (1+\frac1{\sqrt2})^2$$