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

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

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Anzante2m
Hint: $$\displaystyle{\left({1}+{\sin{{x}}}\right)}{\left({1}+{\cos{{x}}}\right)}={1}+{\sin{{x}}}+{\cos{{x}}}+{\frac{{12}}{{\sin{{2}}}}}{x}={1}+\sqrt{{2}}{\sin{{\left({x}+{\frac{{\pi}}{{{4}}}}\right)}}}+{\frac{{12}}{{\sin{{2}}}}}{x}$$
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Jenny Sheppard

$$\displaystyle{\left({1}+{\sin{{x}}}\right)}{\left({1}+{\cos{{x}}}\right)}={1}+{\sin{{x}}}+{\cos{{x}}}+{\sin{{x}}}{\cos{{x}}}$$
Note that $$\displaystyle{\sin{{x}}}{\cos{{x}}}={\frac{{{1}}}{{{2}}}}{\sin{{2}}}{x}$$
Note also that
$$\sin x +\cos x=\sqrt2(\frac{\sqrt2}{2}\sin x+ \frac{\sqrt2}{2} \cos x) =\sqrt{2}(\cos \frac{\pi}{4} \sin x+\sin \frac{\pi}{4}\cos x)=\sqrt2 \sin(x+\frac{\pi}{4})$$
Both of these factors reach their maximum when $$\displaystyle{x}={\frac{{\pi}}{{{4}}}}$$, so the maximum is $$\displaystyle{\frac{{32}}{+}}\sqrt{{2}}$$

Vasquez

$$\begin{array}{}Note \\A=(1+\sin x)(1+\cos x) \\=1+\sin x+\cos x+\frac12 \sin 2x \\=1+\sqrt2 \cos(\frac{\pi}{4}-x)+\frac12 \cos (\frac{\pi}{2}-2x) \\=\frac12+\sqrt2 \cos (\frac{\pi}{4}-x)+ \cos^2 (\frac{\pi}{4}-x) \\=(\cos (\frac{\pi}{4}-x)+\frac{1}{\sqrt2})^2 \\Thus \\0 \leq A \leq (1+\frac{1}{\sqrt2})^2 \end{array}$$