Sin^2 4α, если cosα=1/√5

Решите задачу:

$cos\alpha =\frac{1}{\sqrt5}\\\\sin\alpha=\pm \sqrt{1-cos^2\alpha}=\pm \sqrt{1-\frac{1}{5}}=\pm \frac{2}{\sqrt5}\; ,\; \; sin^2\alpha =\frac{4}{5}\\\\sin^24\alpha =(2sin2\alpha \cdot cos2\alpha )^2=4\cdot (sin2\alpha )^2\cdot (cos^2\alpha -sin^2\alpha)^2=\\\\=4\cdot (2\, sin\alpha \, cos\alpha )^2\cdot (cos^2\alpha -(1-cos^2\alpha ))^2=\\\\=16\cdot sin^2\alpha \cdot cos^2\alpha \cdot (2cos^2\alpha -1)^2\; ;\\\\\\sin^24\alpha =16\cdot \frac{4}{5}\cdot \frac{1}{5}\cdot (\frac{2}{5}-1)^2=\frac{64}{25}\cdot (-\frac{3}{5})^2=\\\\=\frac{64}{25}\cdot \frac{9}{25}=\frac{576}{625}=0,9216$