Помогите доказать тождество

cos2a*2cos2a-1=cos4a

1) cos2a*2cos2a-1 = (2cos^2 (a) - 1) * 2 (2cos^2 (a) - 1) - 1=2 (2cos^2 (a) - 1) ^2 - 1=

2 (4cos^4 (a) - 4cos^2 (a) + 1) - 1 = 8cos^4 (a) - 8cos^2 (a) + 1

2) cos4a=cos (2a+2a) = cos2a*cos2a-sin2a*sin2a=

(2cos^2 (a) - 1) (2cos^2 (a) - 1) - 2sina*cosa*2sina*cosa=

(4cos^4 (a) - 4cos^2 (a) + 1) - 4sin^2 (a) * cos^2 (a) =

(4cos^4 (a) - 4cos^2 (a) + 1) - 4 (1-cos^2 (a)) * cos^2 (a) =

(4cos^4 (a) - 4cos^2 (a) + 1) - 4 (cos^2 (a) - cos^4 (a)) =

4cos^4 (a) - 4cos^2 (a) + 1-4cos^2 (a) + 4cos^4 (a) = 8cos^4 (a) - 8cos^2 (a) + 1

Получаем: 1) = 2), т. е. cos2a*2cos2a-1=cos4a