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