Question

sin^2 4α, если cosα=1/√5

Answer 1

cosa=1/√5

sin²4a=(2sin2a*cos2a)²=

4sin²2a•cos²2a==16sin²a•cos²a•cos²2a

1)cos²2a=(cos²a-sin²a)²=(2cos²a-1)²=
(2*1/5-1)²=((2-5)/5)²=9/25

2)16sin²a•cos²a•cos²2a=
16(1-1/5)*1/5*9/25=16*4/5*1/5*9/25=
16*36/625=576/625

Answer 2

$cosalpha =frac{1}{sqrt5}\\sinalpha=pm sqrt{1-cos^2alpha}=pm sqrt{1-frac{1}{5}}=pm frac{2}{sqrt5}; ,; ; sin^2alpha =frac{4}{5}\\sin^24alpha =(2sin2alpha cdot cos2alpha )^2=4cdot (sin2alpha )^2cdot (cos^2alpha -sin^2alpha)^2=\\=4cdot (2, sinalpha , cosalpha )^2cdot (cos^2alpha -(1-cos^2alpha ))^2=\\=16cdot sin^2alpha cdot cos^2alpha cdot (2cos^2alpha -1)^2; ;\\\sin^24alpha =16cdot frac{4}{5}cdot frac{1}{5}cdot (frac{2}{5}-1)^2=frac{64}{25}cdot (-frac{3}{5})^2=\\=frac{64}{25}cdot frac{9}{25}=frac{576}{625}=0,9216$