Question -
Answer -
Let us consider LHS:
sin2 42o – cos2 78o =sin2 (90o – 48o) – cos2 (90o –12o)
= cos2 48o – sin2 12o [since,sin (90 – A) = cos A and cos (90 – A) = sin A]
We know, cos (A + B) cos (A – B) = cos2A –sin2B
Then the above equation becomes,
= cos2 (48o + 12o)cos (48o – 12o)
= cos 60o cos 36o [since,cos 36o = (√5 + 1)/4]
= 1/2 × (√5 + 1)/4
= (√5 + 1)/8
= RHS
Hence proved.