1.

Prove that: cos 78° cos 42° cos 36° = 1/8

Answer»

Let us consider the LHS

cos 78° cos 42° cos 36°

Now let us multiply and divide by 2 we get,

cos 78° cos 42° cos 36° = 1/2 (2 cos 78° cos 42° cos 36°)

As we know, 2 cos A cos B = cos (A + B) + cos (A – B)

Now the above equation becomes,

= 1/2 (cos (78° + 42°) + cos (78° – 42°)) × cos 36°

= 1/2 (cos 120° + cos 36°) × cos 36°

= 1/2 (cos (180° – 60°) + cos 36°) × cos 36°

= 1/2 (-cos (60°) + cos 36°) × cos 36° [since, cos(180° – A) = – A]

= 1/2 (-1/2 + (√5 + 1)/4) ((√5 + 1)/4) [since, cos 36° = (√5 + 1)/4]

= 1/2 (√5 + 1 – 2)/4 ((√5 + 1)/4)

= 1/2 (√5 – 1)/4) ((√5 + 1)/4)

= 1/2 ((√5)2 – 12)/16

= 1/2 (5 - 1)/16

= 1/2 (4/16)

= 1/8

= RHS

Thus proved.



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