tex]...

Нюта1000 @Нюта1000

October 2021 1 17 Report

[tex]\sqrt{32}cos^{2}\frac{13\pi}{8}-\sqrt{32}sin^{2}\frac{13\pi}{8}[/tex]

Answers & Comments

Yena

$\sqrt{32}cos^{2}\frac{13\pi}{8}-\sqrt{32}sin^{2}\frac{13\pi}{8}=\sqrt{32}(cos^{2}\frac{13\pi}{8}-sin^{2}\frac{13\pi}{8})=\\\ \\\ =\sqrt{32}*cos(2*\frac{13\pi}{8})=\sqrt{32}*cos\frac{13\pi}{4}=\sqrt{32}*cos(\frac{8\pi}{4}+\frac{5\pi}{4})=\\\ \\\ =\sqrt{32}*cos(2\pi+\frac{5\pi}{4})=\sqrt{32}*cos\frac{5\pi}{4}=\sqrt{32}*cos(\frac{4\pi}{4}+\frac{\pi}{4})=\\\ \\\ =\sqrt{32}*cos(\pi+\frac{\pi}{4})=-\sqrt{32}*cos\frac{\pi}{4}=-\sqrt{32}*\frac{\sqrt{2}}{2}=\\\ \\\ =-\frac{\sqrt{64}}{2}=-\frac{8}{2}=-4$

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