Miscellaneous

Chapter 3 Class 11 Trigonometric Functions
Serial order wise

This video is only available for Teachoo black users

This video is only available for Teachoo black users

Get live Maths 1-on-1 Classs - Class 6 to 12

### Transcript

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 We know that 2 cos x cos y = cos (x + y) + cos (x β y) Putting x = 9π/13 and y = π/13 2cos ππ/ππ cos π/ππ = cos (9π/13 " + " π/13) + cos(9π/13 " + " π/13) = cos (πππ/ππ) + cos ((π π)/ππ) = ("cos " πππ/ππ " + cos " ππ/ππ) + cos 3π/13 + cos 5π/13 = ("cos " 10π/13 " + cos " 3π/13) + ("cos " 8π/13 " + cos " 5π/13) Using cos x + cos y = 2 cos (π₯ + π¦)/2 cos (π₯ β π¦)/2 = ("2 cos " ((10π/13 + 3π/13)/2)" . cos " ((10π/13 β 3π/13)/2)) + ("2cos " ((8π/13 + 5π/13)/2)" . cos " ((8π/13 β 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/π ("cos " 7π/26 " + cos " 3π/26) = 2 Γ 0 ("cos " 7π/26 " + cos " 3π/26) = 0 = R.H.S. Hence L.H.S. = R.H.S. Hence proved