tex]...

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[tex]\displaystyle
\sqrt{\frac{1}{2}+\frac{1}{2}\sqrt{\frac{1}{2}+\frac{1}{2}\sqrt{\frac{1}{2}+\frac{1}{2}\cos2\alpha}}}[/tex] при [tex]0 \leq \alpha \leq \pi/2[/tex]

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NNNLLL54

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$\frac{1}{2} + \frac{1}{2}cos2 \alpha = \frac{1}{2} \cdot (1+cos2 \alpha )=\frac{1}{2}\cdot 2cos^2 \alpha =cos^2 \alpha \\\\Tak\; kak\; \; 0 \leq \alpha \leq \frac{\pi}{2}\; ,\; to\; cos \alpha \ \textgreater \ 0\; \; i\; \; \sqrt{cos^2 \alpha }=|cos \alpha |=cos \alpha .\\\\ \sqrt{\frac{1}{2}+ \frac{1}{2}\sqrt{\frac{1}{2}+\frac{1}{2}\sqrt{\frac{1}{2}+\frac{1}{2}cos2 \alpha }} } = \sqrt{\frac{1}{2}+\frac{1}{2} \sqrt{\frac{1}{2}+\frac{1}{2}\sqrt{cos^2 \alpha }} } =$

$= \sqrt{ \frac{1}{2} + \frac{1}{2} \sqrt{ \frac{1}{2} +\frac{1}{2} \cdot cos \alpha } } = \sqrt{ \frac{1}{2} + \frac{1}{2}\sqrt{\frac{1}{2}\cdot (1+cos \alpha )} } =\\\\= \sqrt{ \frac{1}{2} + \frac{1}{2} \sqrt{\frac{1}{2}\cdot 2cos^2\frac{ \alpha }{2}} } = \sqrt{ \frac{1}{2} + \frac{1}{2}cos \frac{ \alpha }{2} } = \sqrt{ \frac{1}{2}\cdot (1+cos\frac{ \alpha }{2}) } =$

$= \sqrt{ \frac{1}{2}\cdot 2cos^2\frac{ \alpha }{4} } = \sqrt{cos^2 \frac{ \alpha }{4}}=|cos \frac{ \alpha }{4} } |=cos\frac{ \alpha }{4}$

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