Cos3B-ctg6Bsin3B=1/2cos3B

Question

Cos3B-ctg6Bsin3B=1/2cos3B

Answer 1

Cos3β-ctg6βsin3β=1/2cos3β

Решение

Cos3β-ctg6βsin3β= Cos3β-cos6β/Sin6β *sin3β=

= (Cos3β*sin6β - cos6β*sin3β)/sin6β=

=sin3β/sin6β = sin3β/2sin3β Cos3β = 1/Cos3β