Prove that (cos2 48∘−sin2 12∘)=(√5+1)8
We have
LHS = (cos2 48∘−sin2 12∘)
=12(2cos2 48∘−2sin2 12∘)
=12(1+cos96∘)−(1−cos24∘)
[∵2cos2θ=(1+cos 2θ) and 2sin2 θ=(1−cos 2θ)]
=12(cos96∘ +cos24∘)
=12 [2 cos(96∘+24∘2)cos(96∘−24∘2)]
=12 ×2cos 60∘ 36∘ =12× (√5+1)4 [∵ cos 36∘=(√5+1)4]
= (√5+1)8=RHS