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Question:

Prove that: 2cos(π/13)cos(9π/13) + cos(3π/13) + cos(5π/13) = 0

Solution:

LHS = 2cos(π/13)cos(9π/13) + cos(3π/13) + cos(5π/13)
= cos(10π/13) + cos(8π/13) + cos(3π/13) + cos(5π/13) [∵ 2cosAcosB = cos(A+B) + cos(A-B)]
= cos(π - 3π/13) + cos(π - 5π/13) + cos(3π/13) + cos(5π/13) [∵ cos(π - θ) = -cosθ]
= -cos(3π/13) - cos(5π/13) + cos(3π/13) + cos(5π/13)
= 0
= RHS
Hence proved