Shirley Thompson

## Answered question

2022-01-02

Find the extrema of $\left(1+\mathrm{sin}x\right)\left(1+\mathrm{cos}x\right)$ without using calculus.

### Answer & Explanation

Anzante2m

Beginner2022-01-03Added 34 answers

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

Jenny Sheppard

Beginner2022-01-04Added 35 answers

$\left(1+\mathrm{sin}x\right)\left(1+\mathrm{cos}x\right)=1+\mathrm{sin}x+\mathrm{cos}x+\mathrm{sin}x\mathrm{cos}x$
Note that $\mathrm{sin}x\mathrm{cos}x=\frac{1}{2}\mathrm{sin}2x$
Note also that
$\mathrm{sin}x+\mathrm{cos}x=\sqrt{2}\left(\frac{\sqrt{2}}{2}\mathrm{sin}x+\frac{\sqrt{2}}{2}\mathrm{cos}x\right)=\sqrt{2}\left(\mathrm{cos}\frac{\pi }{4}\mathrm{sin}x+\mathrm{sin}\frac{\pi }{4}\mathrm{cos}x\right)=\sqrt{2}\mathrm{sin}\left(x+\frac{\pi }{4}\right)$
Both of these factors reach their maximum when $x=\frac{\pi }{4}$, so the maximum is $\frac{32}{+}\sqrt{2}$

Vasquez

Expert2022-01-08Added 669 answers

$\begin{array}{}Note\\ A=\left(1+\mathrm{sin}x\right)\left(1+\mathrm{cos}x\right)\\ =1+\mathrm{sin}x+\mathrm{cos}x+\frac{1}{2}\mathrm{sin}2x\\ =1+\sqrt{2}\mathrm{cos}\left(\frac{\pi }{4}-x\right)+\frac{1}{2}\mathrm{cos}\left(\frac{\pi }{2}-2x\right)\\ =\frac{1}{2}+\sqrt{2}\mathrm{cos}\left(\frac{\pi }{4}-x\right)+{\mathrm{cos}}^{2}\left(\frac{\pi }{4}-x\right)\\ =\left(\mathrm{cos}\left(\frac{\pi }{4}-x\right)+\frac{1}{\sqrt{2}}{\right)}^{2}\\ Thus\\ 0\le A\le \left(1+\frac{1}{\sqrt{2}}{\right)}^{2}\end{array}$

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