Question

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Answer 1

Ответ:

${}\qquad \qquad \boxed{\ cos^2t=\dfrac{1+cos2t}{2}\ \ \Rightarrow \ \ \ 1+cos2t=2cos^2t\ }\\\\\\\\\sqrt{2+\sqrt{2+2cos2t}}=\sqrt{2+\sqrt{2\cdot (1+cos2t)}}=\sqrt{2+\sqrt{2\cdot 2cos^2t}}=\sqrt{2+2\cdot |\, cost\, |}=\\\\\star \ t\in \Big (\dfrac{3\pi}{2}\ ;\ 2\pi \Big)\ \ \Rightarrow \ \ cost>0\ \ \ \Rightarrow \ \ |\, cost\, |=cost\ \star \\\\\\=\sqrt{2+2cost}=\sqrt{2\, (1+cost)}=\sqrt{2\cdot 2cos^2\frac{t}{2}}=2\cdot |\, cos\frac{t}{2}\, |=$

$\star \ t\in \Big (\dfrac{3\pi}{2}\ ;\ 2\pi \Big)\ \ \Rightarrow \ \ \dfrac{t}{2}\in \Big(\dfrac{3\pi }{4}\ ;\ \pi \ \Big)\ \ \Rightarrow \ \ cos\dfrac{t}{2}<0\ \ \Rightarrow \ \ \Big|\, cos\dfrac{t}{2}\, \Big|=-cos\dfrac{t}{2}\ \star \\\\\\=-2cos\dfrac{t}{2}$