Maths · Trigonometrical identities and trigonometrical functions
\( \mathbf{A}=\cos 20^{0} \cos 40^{0} \cos 60^{0} \cos 80^{0} \) \( \mathbf{B}=\cos 6^{0} \cos 42^{0} \cos 66^{0} \cos 78^{0} \) \( \mathbf{C}=\cos \mathbf{3} \mathbf{6}^{\mathbf{0}} \cos \mathbf{7} \mathbf{2}^{\mathbf{0}} \cos \mathbf{1 0} \mathbf{8}^{\mathbf{0}} \cos \mathbf{1} \mathbf{4} \mathbf{4}^{\mathbf{0}} \)
- A. \( A>B>C \)
- B. \( B>C>A \)
- C. \( C>A>B \)
- D. \( A=B=C \)
Step-by-step solution
We evaluate each product using trigonometric identities. A = cos20° cos40° cos60° cos80°. Using cos20° cos40° cos80° = 1/8 (since cos20° cos40° cos80° = (1/4) cos60° = 1/8) and cos60° = 1/2, we get A = 1/16. B = cos6° cos42° cos66° cos78°. Pairing as (cos6° cos66°) and (cos42° cos78°) and using product-to-sum gives B = 1/16. C = cos36° cos72° cos108° cos144°. Using cos108° = -cos72°, cos144° = -cos36°, so C = (cos36° cos72°)^2. Since cos36° cos72° = 1/4, we get C = 1/16. Thus A = B = C.
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