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Misc 1 Prove that: 2cos 𝜋/13 cos 9𝜋/13 + cos 3𝜋/13 + cos 5𝜋/13 = 0 Solving L.H.S 2cos 𝜋/13 cos 9𝜋/13 + cos 3𝜋/13 + cos 5𝜋/13 = ("cos " 𝟏𝟎𝝅/𝟏𝟑 " + cos " 𝟖𝝅/𝟏𝟑) + cos 3𝜋/13 + cos 5𝜋/13 We know that 2 cos x cos y = cos (x + y) + cos (x – y) Putting x = 9𝜋/13 and y = 𝜋/13 2cos 𝟗𝝅/𝟏𝟑 cos 𝝅/𝟏𝟑 = cos (9𝜋/13 " + " 𝜋/13) + cos(9𝜋/13 " + " 𝜋/13) = cos (𝟏𝟎𝝅/𝟏𝟑) + cos ((𝟖 𝝅)/𝟏𝟑) = ("cos " 10𝜋/13 " + cos " 3𝜋/13) + ("cos " 8𝜋/13 " + cos " 5𝜋/13) = ("2 cos " ((10𝜋/13 + 3𝜋/13)/2)" . cos " ((10𝜋/13 − 3𝜋/13)/2)) + ("2cos " ((8𝜋/13 + 5𝜋/13)/2)" . cos " ((8𝜋/13 − 5𝜋/13)/2)) = ("2 cos " ((𝟏𝟑𝝅/𝟏𝟑)/𝟐)" . cos " ((𝟕𝝅/𝟏𝟑)/𝟐)) + ("2 cos " (𝟏𝟑𝝅/𝟏𝟑)/𝟐 " . cos " (𝟑𝝅/𝟏𝟑)/𝟐) = ("2 cos " 𝜋/2 " . cos " 7𝜋/26) + ("2 cos " 𝜋/2 " . cos " 3𝜋/26) = 2 cos 𝝅/𝟐 ("cos " 7𝜋/26 " + cos " 3𝜋/26) = 2 × 0 ("cos " 7𝜋/26 " + cos " 3𝜋/26) = 0 = R.H.S. Hence L.H.S. = R.H.S. Hence proved

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Davneet Singh

Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 14 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.