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Misc 1 Prove that: 2cos πœ‹/13 cos 9πœ‹/13 + cos 3πœ‹/13 + cos 5πœ‹/13 = 0 Solving L.H.S 2cos πœ‹/13 cos 9πœ‹/13 + cos 3πœ‹/13 + cos 5πœ‹/13 We know that 2 cos x cos y = cos (x + y) + cos (x – y) Putting x = 9πœ‹/13 and y = πœ‹/13 2cos πŸ—π…/πŸπŸ‘ cos 𝝅/πŸπŸ‘ = cos (9πœ‹/13 " + " πœ‹/13) + cos(9πœ‹/13 " + " πœ‹/13) = cos (πŸπŸŽπ…/πŸπŸ‘) + cos ((πŸ– 𝝅)/πŸπŸ‘) = ("cos " πŸπŸŽπ…/πŸπŸ‘ " + cos " πŸ–π…/πŸπŸ‘) + cos 3πœ‹/13 + cos 5πœ‹/13 = ("cos " 10πœ‹/13 " + cos " 3πœ‹/13) + ("cos " 8πœ‹/13 " + cos " 5πœ‹/13) Using cos x + cos y = 2 cos (π‘₯ + 𝑦)/2 cos (π‘₯ βˆ’ 𝑦)/2 = ("2 cos " ((10πœ‹/13 + 3πœ‹/13)/2)" . cos " ((10πœ‹/13 βˆ’ 3πœ‹/13)/2)) + ("2cos " ((8πœ‹/13 + 5πœ‹/13)/2)" . cos " ((8πœ‹/13 βˆ’ 5πœ‹/13)/2)) = ("2 cos " ((13πœ‹/13)/2)" . cos " ((7πœ‹/13)/2)) + ("2 cos " (13πœ‹/13)/2 " . cos " (3πœ‹/13)/2) = ("2 cos " πœ‹/2 " . cos " 7πœ‹/26) + ("2 cos " πœ‹/2 " . cos " 3πœ‹/26) = 2 cos 𝝅/𝟐 ("cos " 7πœ‹/26 " + cos " 3πœ‹/26) = 2 Γ— 0 ("cos " 7πœ‹/26 " + cos " 3πœ‹/26) = 0 = R.H.S. Hence L.H.S. = R.H.S. Hence proved

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Davneet Singh

Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 12 years. He provides courses for Maths and Science at Teachoo.