Question

спростіть вирaз sin 2 B cosB+sinBcos2B

Answer 1

Ответ:

cos2bcosb+sin2bsinb=cos(2b−b)=cosb

sin3ϕcosϕ−cos3ϕsinϕ=sin(3ϕ−ϕ)=sin2ϕ

sin(a+b)sin(a−b)=0,5(cos(a+b−a+b)−cos(a+b+a−b))=

=0,5(cos2b−cos2a)=0,5(cos2b−cos2a)=

=0,5(cos

2

b−sin

2

b−cos

2

a+sin

2

a)=

=0,5((1−sin

2

b)−sin

2

b−(1−sin

2

a)+sin

2

a)=

=0,5(1−sin

2

b−sin

2

b−1+sin

2

a+sin

2

a)=

=0,5(−2sin

2

b+2sin

2

a)=sin

2

a−sin

2

b