Question

найти sinx*cosx если cosx=7sinx

Answer 1

$cosx=7sinx\\ =>1=7tgx=>tgx=\frac{1}{7}$

$sinx*cosx=\frac{1}{2}*2*sinx*cosx=\frac{1}{2}sin2x$

$po formule\\ sin2x=\frac{2tgx}{1+tg^2x}$

$=> \frac{1}{2}sin2x=\frac{1}{2}*\frac{2tgx}{1+tg^2x}=\\ \frac{1}{2}*\frac{2*\frac{1}{7}}{1+\frac{1}{49}}=\frac{7}{50}=\frac{14}{100}=0.14$