Question

100 раз прошу уже


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Answer 1

$\star \ \ cos2a=cos^2a-sin^2a=(1-sin^2a)-sin^2a=1-2sin^2a\ \ \Rightarrow \\\\2sin^2a=1-cos2a\ \ \ \Rightarrow \ \ \ sin^2a=\dfrac{1-cos2a}{2}\ \ \ \star \\\\\\1)\ \ \sqrt{\dfrac{1}{2}-\dfrac{1}{2}\, cosa}=\sqrt{\dfrac{1}{2}\cdot (1-cosa)}=\sqrt{\dfrac{1}{2}\cdot 2sin^2\dfrac{a}{2}}=\Big|\, sin\dfrac{a}{2}\, \Big|$

$2)\ \ \sqrt{\dfrac{1}{2}-\dfrac{1}{2}\, sina}=\sqrt{\dfrac{1}{2}\cdot (1-sina)}=\sqrt{\dfrac{1}{2}\cdot \Big(1-cos(\dfrac{\pi}{2}-a)\Big)}=\\\\\\=\sqrt{\dfrac{1}{2}\cdot 2sin^2\Big(\dfrac{\pi}{4}-\dfrac{a}{2}\Big)}=\Big|\, sin\Big(\dfrac{\pi}{4}-\dfrac{a}{2}\Big)\, \Big|$

$\star \ \ cos2a=cos^2a-sin^2a=cos^2a-(1-cos^2a)=2cos^2a-1\ \ \Rightarrow \\\\2cos^2a=1+cos2a\ \ \ \Rightarrow \ \ \ cos^2a=\dfrac{1+cos2a}{2}\ \ \ \star \\\\\\3)\ \ \sqrt{\dfrac{1+cos4a}{2}}=\sqrt{cos^22a}=|\, cos2a\, |$

$4)\ \ \sqrt{\dfrac{1-cos6a}{8}}=\sqrt{\dfrac{1}{4}\cdot sin^23a}=\dfrac{1}{2}\cdot |\, sin3a\, |$

$5)\ \ 0\leq a\leq \dfrac{\pi}{2}\ \ \ \Rightarrow \ \ 0\leq 2a\leq \pi \\\\\sqrt{2+\sqrt{2+2cos4a}}=\sqrt{2+\sqrt{2\, (1+cos4a)}}=\sqrt{2+\sqrt{2\cdot 2cos^22a}}=\\\\=\sqrt{2+2\cdot |cos2a|}=\\\\=\left\{\begin{array}{l}\sqrt{2+2\, cos2a}=\sqrt{2(1+cos2a)}=\sqrt{2\cdot 2cos^2a}=2\cdot cosa\ ,\ 0\leq 2a\leq \dfrac{\pi}{2}\ ,\\\sqrt{2-2\, cos2a}=\sqrt{2\, (1-cos2a)}=\sqrt{2\cdot 2sin^2a}=2\cdot sina\ ,\ \dfrac{\pi}{2}<2a\leq \pi \ ,\end{array}\right=$

$=\left\{\begin{array}{l}2\, cosa\ ,\ esli\ \ \ 0\leq a\leq \dfrac{\pi}{4}\ ,\\2\, sina\ ,\ \ esli\ \ \dfrac{\pi}{4}<a\leq \dfrac{\pi}{2}\end{array}\right$