Q.211. Trigonometry

Question 211.
Prove that cos 6°cos 42°cos 66°cos 78° = 1/16.

Answer 211.

Method 1:
cos 6° cos 42° cos 66° cos 78°

= (1/4) * [2cos6° cos66°] * [2cos42° cos78°]
= (1/4) * (cos72° + cos60°) * (cos120° + cos36°)
= (1/4) (cos72° + 1/2) * (- 1/2 + cos36°)
= (1/4) [- 1/4 + (1/2) (cos36° - cos72°) + cos36° cos72°]
= (1/4) [- 1/4 + sin54° sin18° + sin54° sin18°]
= (1/4) [- 1/4 + (2sin54° sin18° cos18°) / cos18°]
= (1/4) [- 1/4 + (2sin54° sin36°) / cos18°]
= (1/4) [- 1/4 + (cos18° - cos90°) / 2cos18°]
= (1/4) (- 1/4 + 1/2)
= 1/16

Method 2:
cosx cos (60° - x) cos (60° + x)
= cosx [cos^2 60° - sin^2x]
= cosx [(1/4) - 1 + cos^2 x]
= cosx [ - 3/4 + cos^2 x]
= (1/4) (4cos^3 x - 3cosx)
= (1/4) cos3x.

Using the above identity,
cos6° cos54° cos66°
= cos6° cos (60° - 6°) cos (60° + 6°)
= (1/4) cos18° ... (1)

cos18° cos42° cos78°
= cos18 cos (60° - 18°) cos (60° + 18°)
= (1/4) cos54° ... (2)

Multiplying (1) and (2),
(cos6° cos54° cos66°) * (cos18° cos42° cos78°) = (1/16) cos18° cos54°
=> cos6° cos42° cos66° cos78° = 1/16.

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