Question Number 157575 by quvonch3737 last updated on 24/Oct/21
$$ \\ $$$$\sqrt{\frac{\mathrm{1}}{\mathrm{2}}β\frac{\mathrm{1}}{\mathrm{2}}\sqrt{\frac{\mathrm{1}}{\mathrm{2}}+\frac{\mathrm{1}}{\mathrm{2}}\mathrm{cos}\:\boldsymbol{\alpha}}} \\ $$$$\left(\Pi<\boldsymbol{\alpha}<\mathrm{2}\Pi\right) \\ $$
Commented by mr 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 by quvonch3737 last updated on 24/Oct/21
$${no}??? \\ $$
Commented by mr W last updated on 25/Oct/21
$${what}\:{do}\:{you}\:{mean}\:{with}\:``{no}???''? \\ $$$${if}\:{you}\:{mean}\:{i}\:{didn}'{t}\:{say}\:{the}\:{true},\: \\ $$$${did}\:{you}\:{response}\:{to}\:{answer}\:{in}\:\: \\ $$$${Q}\mathrm{157379}?\:{did}\:{you}\:{response}\:{to}\: \\ $$$${answer}\:{to}\:{this}\:{post}? \\ $$
Answered by ajfour last updated on 25/Oct/21
$${i}\:{think}\:{simplify}: \\ $$$$\sqrt{\frac{\mathrm{1}}{\mathrm{2}}β\frac{\mathrm{1}}{\mathrm{2}}\mid\mathrm{sin}\:\left(\frac{\alpha}{\mathrm{2}}\right)\mid} \\ $$$$=\sqrt{\frac{\mathrm{1}β\mathrm{cos}\:\left(\frac{\alpha}{\mathrm{2}}β\frac{\pi}{\mathrm{2}}\right)}{\mathrm{2}}} \\ $$$$=\mid\mathrm{sin}\:\left(\frac{\alpha}{\mathrm{4}}β\frac{\pi}{\mathrm{4}}\right)\mid \\ $$
Commented by ajfour last updated on 25/Oct/21
$${and}\:{to}\:{know}\:{if}\:{its}\:{really}\:{okay} \\ $$$${some}\:{comprehendable} \\ $$$${feedback}\:{usually}\:{is}\:{helpful}... \\ $$$${Yes}\:\:\:{or}\:\:\:{No}\:??? \\ $$