Question

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

Answer 1

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

Answer 2

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

Answer 3

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