Question

1) 4sin⁶a + 4cos⁶a - 3cos²2α;
помогите пожалуйста даю 40 баллов

Answer 1

$\displaystyle\bf\\4Sin^{6} \alpha +4Cos^{6} \alpha -3Cos^{2} 2\alpha =4(Sin^{6} \alpha +Cos^{6} \alpha) -3Cos^{2} 2\alpha =\\\\\\=4\cdot\Big[(Sin^{2} \alpha )^{3} +(Cos^{2} \alpha )^{3} \Big]-3Cos^{2} 2\alpha =\\\\\\=4\cdot\Big[(\underbrace{Sin^{2} \alpha +Cos^{2} \alpha}_{1} )\cdot(Sin^{4} \alpha -Sin^{2} \alpha Cos^{2}\alpha +Cos^{4} \alpha\Big]-3Cos^{2} 2\alpha =\\\\\\=4\cdot\Big[(Sin^{2} \alpha +Cos^{2} \alpha )^{2}-3Sin^{2} \alpha Cos^{2}\alpha \Big] -3Cos^{2} 2\alpha =$

$\displaystyle\bf\\=4\cdot(1-3Sin^{2} \alpha Cos^{2} \alpha )-3Cos^{2} 2\alpha =\\\\\\=4-12Sin^{2} \alpha Cos^{2} \alpha -3Cos^{2} 2\alpha =4-3\cdot 4Sin^{2} \alpha Cos^{2} \alpha -3Cos^{2} 2\alpha =\\\\\\=4-3Sin^{2} 2\alpha -3Cos^{2} 2\alpha=4-3\cdot(Sin^{2} 2\alpha +Cos^{2} 2\alpha)=\\\\\\=4-3\cdot 1=4-3=1$