Question

( sin x+sin ^{2} x=1 )
( sin x=1-sin ^{2} x )
( sin x=cos ^{2} x )
( sin ^{2} x=cos ^{4} x )
SO, ( cos ^{wedge} 8 x+2 cos ^{wedge} 6 x+cos ^{4} x )
( =cos ^{4} xleft(cos ^{4} x+2 cos ^{2} x+1right) )
( =sin ^{2} xleft(sin ^{2} x+2 sin x+1right) )
( =sin ^{4} x+2 sin ^{3} x+sin ^{2} x )
( =sin ^{4} x+sin ^{3} x+sin ^{3} x+sin ^{2} x )
( =sin ^{2} xleft(sin ^{2} x+sin xright)+sin xleft(sin ^{2} x+sin xright) )
( =sin ^{2} x times(1)+sin x(1) )
( =sin ^{2} x+sin x )
( =1 )
İs your final answer

# TULICOL Ullege bas If sin x + sin2 x = 1, then cos8 x + 2 cos6 x + cos4 x = .... (A) 0 (B) - 1 (C) 2 (D) 1

Solution