Example 29 - Prove cos2 x + cos2 (x + pi/3) + cos2 (x - pi/3) - Examples

  1. Chapter 3 Class 11 Trigonometric Functions
  2. Serial order wise
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Eπ‘₯ample 29 Prove that cos2 π‘₯+cos2 (π‘₯+"Ο€" /3) + cos2 (π‘₯ βˆ’ "Ο€" /3) = 3/2 Lets first calculate all 3 terms separately We know that cos 2x = 2cos2 x – 1 cos 2x + 1 = 2cos2 x π‘π‘œπ‘ β‘γ€–2π‘₯ + 1γ€—/2 = cos2 x So, cos2x = cos⁑〖2x + 1γ€—/2 Replacing x with x + πœ‹/3 is about cos2 (x + πœ‹/3) = cos⁑〖2( π‘₯ + πœ‹/3) + 1γ€—/2 = cos⁑〖 (2π‘₯ + 2πœ‹/3) + 1γ€—/2 Similarly , replacing x with x + πœ‹/3 in cos2x = π‘π‘œπ‘ β‘γ€–2π‘₯ + 1γ€—/2 cos2 (x – πœ‹/3) = cos⁑〖2( π‘₯ βˆ’ πœ‹/3) + 1γ€—/2 = cos⁑〖( 2π‘₯ βˆ’ 2πœ‹/3) + 1γ€—/2 Now we solve L.H.S cos2x + cos2 ( π‘₯+ 2πœ‹/3) + cos2 ( π‘₯ βˆ’ 2πœ‹/3) = (1 + cos⁑2π‘₯)/2 + (1 + cos⁑( 2π‘₯ + 2πœ‹/3))/2 + γ€–1 + cos〗⁑( 2π‘₯ βˆ’ 2πœ‹/3)/2 = 1/2 [1+cos⁑2π‘₯ + 1+ cos⁑〖( 2π‘₯+2πœ‹/3) γ€— + γ€–1+ cos〗⁑( 2π‘₯βˆ’2πœ‹/3) ] = 1/2 [ 1 + 1 + 1 + cos 2x + cos ( 2π‘₯+2πœ‹/3) + γ€– cos〗⁑( 2π‘₯βˆ’2πœ‹/3)] = 1/2 [ 3 + cos 2x + cos ( 2π‘₯ + 2πœ‹/3) + γ€– cos〗⁑( 2π‘₯ βˆ’ 2πœ‹/3)] = 1/2 ["3 + cos 2x + 2cos " ( (2π‘₯ + 2πœ‹/3 + 2π‘₯ βˆ’ 2πœ‹/3)/2)" ." γ€– cos〗⁑( (2π‘₯ + 2πœ‹/3 βˆ’ (2π‘₯ +2πœ‹/3))/2) ] = 1/2 ["3 + cos 2x + 2cos " ( (2π‘₯ + 2π‘₯ + 2πœ‹/3 βˆ’ 2πœ‹/3)/2)" ." γ€– cos〗⁑( (2π‘₯ βˆ’ 2π‘₯ + 2πœ‹/3 + 2πœ‹/3)/2) ] = 1/2 ["3 + cos 2x + 2cos " ( (4π‘₯+0)/2)" ." γ€– cos〗⁑( (0 + 4πœ‹/3)/2) ] = 1/2 [ 3 + cos 2x + 2cos ( (4π‘₯+0)/2) .γ€– cos〗⁑( (0 + 4πœ‹/3)/2) = 1/2 ["3 + cos 2x + 2cos " ( 4π‘₯/2)" ." γ€– cos〗⁑( (4πœ‹/3)/2) ] = 1/2 ["3 + cos 2x + 2cos 2x cos " 2πœ‹/3] = 1/2 ["3 + cos 2x + 2cos 2x cos " (πœ‹βˆ’ πœ‹/3)] = 1/2 ["3 + cos 2x + 2cos 2x " (βˆ’ "cos " (πœ‹/3))] = 1/2 [ 3 + cos 2x + 2cos 2x (βˆ’ 1/2)] = 1/2 [ 3 + cos 2x – 2 Γ— 1/2 Γ— cos 2x ] = 1/2 [ 3 + cos 2x – cos 2x ] = 1/2 [ 3 + 0] = 3/2 = R.H.S. Hence, L.H.S. = R.H.S. Hence proved

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