Question

$2( \sin ( ^ { 6 } \theta ) + \cos ( ^ { 6 } \theta ) )-3( \sin ( ^ { 4 } \theta ) + \cos ( ^ { 4 } \theta ) )+1 = 0$
Prove the identity​

Answer 1

Answer:

LHS=2(sin 6θ+cos 6 θ)−3(sin 4θ+cos 4 θ)+1

=2{(sin 2 θ+cos 2θ) 3 −3sin 2θcos 2 θ(sin 2 θ+cos 2θ)}

−3(sin 2θ+cos 2 θ) 2−2(sin 2θcos 2θ)}+1

We know, [sin²x+cos²x=1]

=2{1−3sin 2θcos 2θ}−3{1−2sin 2 θcos 2 θ}+1

=2−6sin 2θcos 2 θ−3+6sin 2 θcos 2 θ+1

=0

=RHS