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)

Question

Алгебра: 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 =

raisaaaaaaaaaaaa · Answer

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

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 =

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