Question

найдите максимум функции f(x)=cos(2018x)+sin(2018x)

Answer 1

$f(x)=cos(2018x)+sin(2018x)=cos(y)+sin(y)=\\\\ =[cos(y)*\frac{\sqrt{2}}{2}+sin(y)*\frac{\sqrt{2}}{2}]*\frac{2}{\sqrt{2}}=\\\\ =[cos(y)*cos(\frac{\pi}{4})+sin(y)*sin(\frac{\pi}{4})]*\sqrt{2}=\\\\ =cos(y-\frac{\pi}{4})*\sqrt{2}=\sqrt(2)*cos(2018x-\frac{\pi}{4})\\\\ -1 \leq cos(2018x-\frac{\pi}{4}) \leq 1\\\\ -\sqrt{2} \leq \sqrt{2}*cos(2018x-\frac{\pi}{4}) \leq \sqrt{2}\\\\ -\sqrt{2} \leq f(x) \leq \sqrt{2}$

Максимального значения $\sqrt{2}$ функция $f(x)$ достигает при условии $cos(2018x-\frac{\pi}{4})=1$

$2018x-\frac{\pi}{4}=2\pi n,\ n\in Z\\\\ 2018x=\frac{\pi}{4}+2\pi n,\ n\in Z\\\\ x=\frac{\pi}{4*2018}+\frac{2\pi n}{2018},\ n\in Z\\\\ x=\frac{\pi}{8072}+\frac{\pi n}{1009},\ n\in Z$
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множество точек  $(\frac{\pi}{8072}+\frac{\pi n}{1009};\ \sqrt{2}),\ n\in Z$ есть максимумами функции $f(x)$