Question

Докажите тождество: (sin2a + sin4a)^2 + (cos2a + cos4a)^2 = 4cos^2a

Answer 1

Sin²2a+2sin2asin4a+sin²4a+cos²2a+2cos2acos4a+cos²4a=
=2+2(cos2acos4a+sin2asin4a)=2+2cos(4a-2a)=2+2cos2a=
=2(1+cos2a)=2(1+2cos²a-1)=4cos²a

Answer 2

(sin2α+sin4α)² +(cos2α+cos4α)² =(2sin3α*cosα)² +(2cos3αcosα)² =
4cos²α(sin²3α+cos²3α) =4cos²α*1 = 4cos²α.