Помогите решить, пожалуйста Log_2 tg П/8 + log_2 cos^2 П/8. Created by Krid2013. algebra-ru

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Krid2013 @Krid2013

July 2022 1 1 Report

Помогите решить, пожалуйста
Log_2 tg П/8 + log_2 cos^2 П/8

Answers & Comments

nKrynka Log_2 [(tgπ/8)*(cos^2 (π/8)] = log_2 [(sin{π/8) /cos(π/8)] * [cos^2(π/8)] =
log_2 [(sin{π/8)*cos(π/8)] log_2 (1/2)*2* [(sin{π/8)*cos(π/8)] =
log_2 [(1/2)*sin(2π/8)] = log_2(1/2)*sin(π/4) = log_2 (1/2)*(√2/2) =
log_2[2^(-3/2)] = - 3/2

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Krid2013 а там же вроде log_2 sin π/8 * cos π/8 получается

nKrynka og_2 [(tgπ/8)*(cos^2 (π/8)] = log_2 [(sin{π/8) /cos(π/8)] * [cos^2(π/8)] =
log_2 [(sin{π/8)*cos(π/8)] log_2 (1/2)*2* [(sin{π/8)*cos(π/8)] =
log_2 [(1/2)*sin(2π/8)] = log_2(1/2)*sin(π/4) = log_2 (1/2)*(√2/2) =
log_2[2^(-3/2)] = - 3/2

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