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

Example 29 - Chapter 3 Class 11 Trigonometric Functions - Part 2
Example 29 - Chapter 3 Class 11 Trigonometric Functions - Part 3
Example 29 - Chapter 3 Class 11 Trigonometric Functions - Part 4
Example 29 - Chapter 3 Class 11 Trigonometric Functions - Part 5

  1. Chapter 3 Class 11 Trigonometric Functions (Term 2)
  2. Serial order wise

Transcript

Example 29 Prove that cos2 π‘₯+cos2 (π‘₯+πœ‹/3) + cos2 (π‘₯βˆ’πœ‹/3) = 3/2 Lets first calculate all 3 terms separately We know that cos 2x = 2 cos2 x βˆ’ 1 cos 2x + 1 = 2cos2 x π‘π‘œπ‘ β‘γ€–2π‘₯ + 1γ€—/2 = cos2 x So, cos2 x = πœπ¨π¬β‘γ€–πŸπ’™ + πŸγ€—/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 βˆ’" πœ‹/3) in cos2 x = cos⁑〖2π‘₯ + 1γ€—/2 cos2 ("x" βˆ’πœ‹/3) = cos⁑〖2(π‘₯ βˆ’ πœ‹/3)+ 1γ€—/2 = cos⁑〖(2π‘₯ βˆ’ 2πœ‹/3)+ 1γ€—/2 Solving L.H.S cos2 x + cos2 (π‘₯+ πœ‹/3) + cos2 (π‘₯βˆ’πœ‹/3) = (1 + cos⁑2π‘₯)/2 + (1 + cos⁑(2π‘₯ + 2πœ‹/3))/2 + (1 + cos⁑(2π‘₯ βˆ’ 2πœ‹/3))/2 = 1/2 [1+cos⁑〖2π‘₯+1+π‘π‘œπ‘ (2π‘₯+2πœ‹/3)+1+π‘π‘œπ‘ (2π‘₯βˆ’2πœ‹/3)γ€— ] = 1/2 [3+cos⁑〖2π‘₯+π‘π‘œπ‘ (2π‘₯+2πœ‹/3)+π‘π‘œπ‘ (2π‘₯βˆ’2πœ‹/3)γ€— ] = 1/2 [3+cos⁑〖2π‘₯+2π‘π‘œπ‘ ((2π‘₯ + 2πœ‹/3 + 2π‘₯ βˆ’ 2πœ‹/3)/2).π‘π‘œπ‘ ((2π‘₯ + 2πœ‹/3 βˆ’(2π‘₯ βˆ’ 2πœ‹/3))/2)γ€— ] Using cos x + cos y = 2 cos ((π‘₯ + 𝑦)/2). cos ((π‘₯ βˆ’ 𝑦)/2) Replace x by ("2" π‘₯" + " 2πœ‹/3) & y by ("2x βˆ’ " 2πœ‹/3) = 1/2 [3+cos⁑〖2π‘₯+2π‘π‘œπ‘ ((4π‘₯ + 0)/2).π‘π‘œπ‘ ((0 + 4πœ‹/3)/2)γ€— ] = 1/2 [3+cos⁑〖2π‘₯+2π‘π‘œπ‘ (4π‘₯/2).π‘π‘œπ‘ ((4πœ‹/3)/2)γ€— ] = 1/2 [3+cos⁑〖2π‘₯+2 cos⁑2π‘₯ cos⁑〖2πœ‹/3γ€— γ€— ] = 1/2 [3+cos⁑〖2π‘₯+2 cos⁑2π‘₯ cos⁑(πœ‹βˆ’πœ‹/3) γ€— ] = 1/2 [3+cos⁑〖2π‘₯+2 cos⁑2π‘₯ γ€— (γ€–βˆ’cos〗⁑(πœ‹/3) ) ] = 1/2 [3+cos⁑〖2π‘₯+2 cos⁑2π‘₯ γ€— (βˆ’1/2) ] (As cos (Ο€ βˆ’ πœƒ) = βˆ’cos πœƒ) = 1/2 [3+cos⁑〖2π‘₯βˆ’2 Γ—1/2Γ—cos⁑2π‘₯ γ€— ] = 1/2 [3+cos⁑〖2π‘₯βˆ’cos⁑2π‘₯ γ€— ] = 1/2 [3+0] = 3/2 = R.H.S. Hence, L.H.S. = R.H.S. Hence Proved

About the Author

Davneet Singh's photo - Teacher, Engineer, Marketer
Davneet Singh
Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 10 years. He provides courses for Maths and Science at Teachoo.