Question

plz answer as soon as possible

Answer 1

First: solving (sin2θ+cos2θ)3=(1)3
(sin2θ+cos2θ)(sin4θ+2sin2θ⋅cos2θ+cos4θ)=1

(sin6θ+2sin4θ⋅cos2θ+sin2θ⋅cos4θ)+(cos2θ⋅sin4θ+2sin2θ⋅cos4θ+cos6θ)=1

(sin6θ+cos6θ)+[(2sin4θ⋅cos2θ+2sin2θ⋅cos4θ)+(sin2θ⋅cos4θ+cos2θ⋅sin4θ)]=1

(sin6θ+cos6θ)+[sin2θ⋅cos2θ⋅((2sin2θ+2cos2θ)+(cos2θ+sin2θ))]=1

(sin6θ+cos6θ)+[sin2θ⋅cos2θ⋅((2)+(1))]=1
(sin6θ+cos6θ)+3[sin2θ⋅cos2θ]=1

∴sin6θ+cos6θ=1−3[sin2θ⋅cos2θ]

Using a the same method starting with (sin2θ+cos2θ)2=(1)2
you get:
sin4θ+cos4θ=1−2[sin2θ⋅cos2θ]

The left side of the equation =
2(sin6θ+cos6θ)−3(sin4θ+cos4θ)+1
=2(1−3sin2θ⋅cos2θ)−3(1−2sin2θ⋅cos2θ)+1
=2−6sin2θ⋅cos2θ−3+6sin2θ⋅cos2θ+1
=2−3+1=0