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Question Number 172817 by Mikenice last updated on 01/Jul/22
Answered by som(math1967) last updated on 02/Jul/22
$$\sqrt{\mathrm{2}+\sqrt{\mathrm{2}+\sqrt{\mathrm{2}\left(\mathrm{1}+{cos}\mathrm{8}\theta\right)}}} \\ $$$$\sqrt{\mathrm{2}+\sqrt{\mathrm{2}+\mathrm{2}{cos}\mathrm{4}\theta}} \\ $$$$\sqrt{\mathrm{2}+\sqrt{\mathrm{2}\left(\mathrm{1}+{cos}\mathrm{4}\theta\right)}} \\ $$$$\sqrt{\mathrm{2}+\mathrm{2}{cos}\mathrm{2}\theta}=\sqrt{\mathrm{2}\left(\mathrm{1}+{cos}\mathrm{2}\theta\right)}=\mathrm{2}{cos}\theta \\ $$
Answered by CElcedricjunior last updated on 02/Jul/22
/8);(𝛑/8)[ k=(√(2+2cos2𝚯 )) si𝚯∈]((−3𝛑)/4);(𝛑/4)[ k=2cos𝚯 si 𝚯∈]((−3𝛑)/2);(𝛑/2)[ ........Le ce^ le^ bre cedric junior......](Q172868.png)
$$\sqrt{\mathrm{2}+\sqrt{\mathrm{2}+\sqrt{\mathrm{2}+\mathrm{2}\boldsymbol{\mathrm{cos}}\mathrm{8}\boldsymbol{\Theta}}}}=\boldsymbol{\mathrm{k}} \\ $$$$\boldsymbol{\mathrm{cos}}\mathrm{2}\boldsymbol{\mathrm{x}}=\mathrm{2}\boldsymbol{\mathrm{cos}}^{\mathrm{2}} \boldsymbol{\mathrm{x}}−\mathrm{1}=>\mathrm{4}\boldsymbol{\mathrm{cos}}^{\mathrm{2}} \boldsymbol{\mathrm{x}}=\mathrm{2}+\mathrm{2}\boldsymbol{\mathrm{cos}}\mathrm{2}\boldsymbol{\mathrm{x}} \\ $$$$\boldsymbol{\mathrm{k}}=\sqrt{\mathrm{2}+\sqrt{\mathrm{2}+\sqrt{\mathrm{4}\boldsymbol{\mathrm{cos}}^{\mathrm{2}} \mathrm{4}\boldsymbol{\Theta}}}} \\ $$$$\left.\boldsymbol{\mathrm{k}}=\sqrt{\mathrm{2}+\sqrt{\mathrm{2}+\mathrm{2}\boldsymbol{\mathrm{cos}}\mathrm{4}\boldsymbol{\Theta}}}\:\boldsymbol{\mathrm{si}}\:\boldsymbol{\Theta}\in\right]\frac{−\mathrm{3}\boldsymbol{\pi}}{\mathrm{8}};\frac{\boldsymbol{\pi}}{\mathrm{8}}\left[\right. \\ $$$$\left.\boldsymbol{\mathrm{k}}=\sqrt{\mathrm{2}+\mathrm{2}\boldsymbol{\mathrm{cos}}\mathrm{2}\boldsymbol{\Theta}\:\:}\:\:\:\:\:\boldsymbol{{si}\Theta}\in\right]\frac{−\mathrm{3}\boldsymbol{\pi}}{\mathrm{4}};\frac{\boldsymbol{\pi}}{\mathrm{4}}\left[\right. \\ $$$$\left.\boldsymbol{\mathrm{k}}=\mathrm{2}\boldsymbol{\mathrm{cos}\Theta}\:\:\:\boldsymbol{{si}}\:\boldsymbol{\Theta}\in\right]\frac{−\mathrm{3}\boldsymbol{\pi}}{\mathrm{2}};\frac{\boldsymbol{\pi}}{\mathrm{2}}\left[\right. \\ $$$$ \\ $$$$........\mathscr{L}{e}\:{c}\acute {{e}l}\grave {{e}bre}\:{cedric}\:{junior}...... \\ $$$$ \\ $$