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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

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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

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Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 12 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.