prawa trygonometria

[logs4]

udowodnij:\(\displaystyle{ 2sin2 -3cos2 -3sin +3=sin (4cos +6sin -3)}\)

[mmoonniiaa]

\(\displaystyle{ L=2 2 sin\alpha cos\alpha-3(1-2sin^2\alpha)-3sin\alpha+3= \\ =4sin\alpha cos\alpha - 3 +6sin^2\alpha-3sin\alpha+3=sin\alpha(4cos\alpha+6sin\alpha-3)=P}\)