Question Number 157575 by quvonch3737 last updated on 24/Oct/21
$$ \\ $$ $$\sqrt{\frac{\mathrm{1}}{\mathrm{2}}โ\frac{\mathrm{1}}{\mathrm{2}}\sqrt{\frac{\mathrm{1}}{\mathrm{2}}+\frac{\mathrm{1}}{\mathrm{2}}\mathrm{cos}\:\boldsymbol{\alpha}}} \\ $$ $$\left(\Pi<\boldsymbol{\alpha}<\mathrm{2}\Pi\right) \\ $$
Commented bymr W last updated on 24/Oct/21
$${this}\:{is}\:{not}\:{a}\:{question}! \\ $$ $${besides}\:{if}\:{you}\:{ask}\:{for}\:{help}\:{and}\:{other} \\ $$ $${people}\:{have}\:{helped}\:{you},\:{you}\:{should} \\ $$ $${at}\:{least}\:{give}\:{a}\:{feedback}. \\ $$
Commented byquvonch3737 last updated on 24/Oct/21
$${no}??? \\ $$
Commented bymr W last updated on 25/Oct/21
$${what}\:{do}\:{you}\:{mean}\:{with}\:\natural{no}???\varepsilon? \\ $$ $${if}\:{you}\:{mean}\:{i}\:{didn}'{t}\:{say}\:{the}\:{true},\: \\ $$ $${did}\:{you}\:{response}\:{to}\:{answer}\:{in}\:\: \\ $$ $${Q}\mathrm{157379}?\:{did}\:{you}\:{response}\:{to}\: \\ $$ $${answer}\:{to}\:{this}\:{post}? \\ $$
Answered by ajfour last updated on 25/Oct/21
$${i}\:{think}\:{simplify}: \\ $$ $$\sqrt{\frac{\mathrm{1}}{\mathrm{2}}โ\frac{\mathrm{1}}{\mathrm{2}}\mid\mathrm{sin}\:\left(\frac{\alpha}{\mathrm{2}}\right)\mid} \\ $$ $$=\sqrt{\frac{\mathrm{1}โ\mathrm{cos}\:\left(\frac{\alpha}{\mathrm{2}}โ\frac{\pi}{\mathrm{2}}\right)}{\mathrm{2}}} \\ $$ $$=\mid\mathrm{sin}\:\left(\frac{\alpha}{\mathrm{4}}โ\frac{\pi}{\mathrm{4}}\right)\mid \\ $$
Commented byajfour last updated on 25/Oct/21
$${and}\:{to}\:{know}\:{if}\:{its}\:{really}\:{okay} \\ $$ $${some}\:{comprehendable} \\ $$ $${feedback}\:{usually}\:{is}\:{helpful}... \\ $$ $${Yes}\:\:\:{or}\:\:\:{No}\:??? \\ $$