Question

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Answer 1

$\displaystyle\bf\\Sin2\alpha =-\frac{1}{5} \\\\\\20Sin^{2}\alpha \cdot Cos^{2} \alpha +2=5\cdot\underbrace{(4Sin^{2} \alpha Cos^{2} \alpha ) }_{Sin^{2} 2\alpha }+2=5Sin^{2} 2\alpha +2=\\\\=5\cdot\Big(-\frac{1}{5} \Big)^{2} +2=5\cdot\frac{1}{25}+2=\frac{1}{5} +2=0,2+2=2,2$