Question

Sin^4b-cos^4b+cos^2b

Answer 1

(1-cos2b)²/4-(1+cos2b)²/4+(1+cos2b)/2=
=1/4-1/2*cos2b+1/4*cos²2b-1/4-1/2*cos2b-1/4*cos²2b+1/2+1/2*cos2b=
=1/2-1/2*cos2b

Answer 2

Упростить можно и дальше

Answer 3

$sin^4b-cos^4b+cos^2b=\\=(sin^2b-cos^2b)underbrace {(sin^2b+cos^2b)}_1+cos^2b=-cos2b+cos^2b=\\=[, sin^2 alpha = frac{1-cos2 alpha }{2} ; to ; ; 2sin^2 alpha -1=-cos2 alpha , ]=\\=2sin^2b-1+cos^2b=sin^2b-1+(sin^2b+cos^2b)=\\=sin^2b-1+1=sin^2b$