Question

Срочно номер 3 под цифрой 2

Answer 1

$frac{sin5alpha-sin3alpha}{2cos4alpha}=sinalpha\sinalpha-sinbeta=2sinfrac{alpha-beta}2cdotcosfrac{alpha+beta}2\frac{sin5alpha-sin3alpha}{2cos4alpha}=frac{2sinfrac{5alpha-3alpha}2cdotcosfrac{5alpha+3alpha}2}{2cos4alpdha}=frac{2sinalphacdotcos4alpha}{2cos4alpha}=sinalpha$