Question

1) 6sin(pi/12)cos(pi/12) =
2) sin^2(5A)-cos^2(5A) =
3) (2tg2A) / (1-tg^2(2A)) =
4) sin(A/6)cos(A/6) =
5) 20sin^2(4A)cos^2(4A) =
6) sin3Acos3A =

Answer 1

1) (6sin(π/12))*(cos(π/12))=3sinπ/6=3*0.5=1.5

2) sin²(5α)-cos²(5α)= -cos10α

3)(2tg2α)/(1-tg²(2α))=tg4α

4)sin( A/6)cos(A/6)=0.5sin (A/3)

5) 20sin²(4α)cos²(4α)=(5*(2sin(4α))*cos(4α))²=5sin²8α

6)(sin3α)(cos3α)=0.5sin6α

Answer 2

можете пожалуйста помочь с математикой?