Example 29 - Class 11
Last updated at May 29, 2018 by Teachoo
Transcript
Eπ₯ample 29 Prove that cos2 π₯+cos2 (π₯+"Ο" /3) + cos2 (π₯ β "Ο" /3) = 3/2 Lets first calculate all 3 terms separately We know that cos 2x = 2cos2 x β 1 cos 2x + 1 = 2cos2 x πππ β‘γ2π₯ + 1γ/2 = cos2 x So, cos2x = cosβ‘γ2x + 1γ/2 Replacing x with x + π/3 is about cos2 (x + π/3) = cosβ‘γ2( π₯ + π/3) + 1γ/2 = cosβ‘γ (2π₯ + 2π/3) + 1γ/2 Similarly , replacing x with x + π/3 in cos2x = πππ β‘γ2π₯ + 1γ/2 cos2 (x β π/3) = cosβ‘γ2( π₯ β π/3) + 1γ/2 = cosβ‘γ( 2π₯ β 2π/3) + 1γ/2 Now we solve L.H.S cos2x + cos2 ( π₯+ 2π/3) + cos2 ( π₯ β 2π/3) = (1 + cosβ‘2π₯)/2 + (1 + cosβ‘( 2π₯ + 2π/3))/2 + γ1 + cosγβ‘( 2π₯ β 2π/3)/2 = 1/2 [1+cosβ‘2π₯ + 1+ cosβ‘γ( 2π₯+2π/3) γ + γ1+ cosγβ‘( 2π₯β2π/3) ] = 1/2 [ 1 + 1 + 1 + cos 2x + cos ( 2π₯+2π/3) + γ cosγβ‘( 2π₯β2π/3)] = 1/2 [ 3 + cos 2x + cos ( 2π₯ + 2π/3) + γ cosγβ‘( 2π₯ β 2π/3)] = 1/2 ["3 + cos 2x + 2cos " ( (2π₯ + 2π/3 + 2π₯ β 2π/3)/2)" ." γ cosγβ‘( (2π₯ + 2π/3 β (2π₯ +2π/3))/2) ] = 1/2 ["3 + cos 2x + 2cos " ( (2π₯ + 2π₯ + 2π/3 β 2π/3)/2)" ." γ cosγβ‘( (2π₯ β 2π₯ + 2π/3 + 2π/3)/2) ] = 1/2 ["3 + cos 2x + 2cos " ( (4π₯+0)/2)" ." γ cosγβ‘( (0 + 4π/3)/2) ] = 1/2 [ 3 + cos 2x + 2cos ( (4π₯+0)/2) .γ cosγβ‘( (0 + 4π/3)/2) = 1/2 ["3 + cos 2x + 2cos " ( 4π₯/2)" ." γ cosγβ‘( (4π/3)/2) ] = 1/2 ["3 + cos 2x + 2cos 2x cos " 2π/3] = 1/2 ["3 + cos 2x + 2cos 2x cos " (πβ π/3)] = 1/2 ["3 + cos 2x + 2cos 2x " (β "cos " (π/3))] = 1/2 [ 3 + cos 2x + 2cos 2x (β 1/2)] = 1/2 [ 3 + cos 2x β 2 Γ 1/2 Γ cos 2x ] = 1/2 [ 3 + cos 2x β cos 2x ] = 1/2 [ 3 + 0] = 3/2 = R.H.S. Hence, L.H.S. = R.H.S. Hence proved
Examples
Example 1
Example 2
Example 3
Example 4
Example 5
Example 6
Example 7
Example 8
Example 9
Example 10
Example 11
Example 12
Example 13
Example 14
Example 15
Example 16
Example 17
Example 18
Example 19
Example 20
Example 21
Example 22
Example 23
Example 24 Important
Example 25
Example 26
Example 27 Important
Example 28 Important
Example 29 You are here
Chapter 3 Class 11 Trigonometric Functions
Serial order wise
About the Author
Davneet Singh
Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 8 years. He provides courses for Maths and Science at Teachoo. You can check his NCERT Solutions from Class 6 to 12.
27