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Prove that: sin2 42° – cos2 78° = (√5 + 1)/8 |
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Answer» Let us consider the LHS sin2 42° – cos2 78° = sin2 (90° – 48°) – cos2 (90° – 12°) = cos2 48° – sin2 12° [since, sin (90 – A) = cos A and cos (90 – A) = sin A] As we know, cos (A + B) cos (A – B) = cos2A – sin2B Now the above equation becomes, = cos2 (48° + 12°) cos (48° – 12°) = cos 60° cos 36° [since, cos 36° = (√5 + 1)/4] = 1/2 × (√5 + 1)/4 = (√5 + 1)/8 = RHS Thus proved. |
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