Question

2cos(-3П- В )+sin(-П/2+В) ________________________ 3cos(B+П)

Answer 1

$\frac{2cos(-3 \pi - \beta )+sin(- \pi /2+ \beta )}{3cos( \beta + \pi) }= \frac{2cos(-(3 \pi + \beta ))+sin(-( \pi /2- \beta ))}{3cos( \pi + \beta )}=\\\\= \frac{2cos(3 \pi + \beta )-sin( \pi /2- \beta )}{3cos( \pi + \beta )}= \frac{-2cos \beta-cos \beta }{-3cos \beta }= \frac{-3cos \beta }{-3cos \beta }=1$